Topics

Unit vectors along Radial direction ( \(\widehat{\mathbf{r}}\) ) and tangential direction ( \(\widehat{\mathbf{v}}\) )

Suppose a particle moving in a cirular path of radius r in \(xy\) plane with origin as centre and makes angle \(\theta\) with \(x\)-axis as shown, at any instant. Its position vector at this instant can be given by
\[\Rightarrow \ \widehat{\mathbf{r}} = \frac{\overrightarrow{\mathbf{r}}}{\mathbf{r}} = cos\theta\widehat{\mathbf{i}} + sin\theta\widehat{\mathbf{j}}\]

Also \(\ \overrightarrow{v} = - vsin\theta\widehat{i} + vcos\theta\widehat{j}\)
\[\therefore\ \widehat{\mathbf{v}} = \frac{\overrightarrow{\mathbf{v}}}{\mathbf{v}} = - sin\theta\widehat{\mathbf{i}} + cos\theta\widehat{\mathbf{j}} \]

Uniform circular motion

When a particle is moving with constant speed in a circular path, its motion is called unifrom circular motion. Although magnitude of velocity is constant but its direction changes continusly. It means it is continuonsly having some acceleration. This acceleration is always directing towards the centre. This is called radial acceleration ( \(a_{t}\) ) or centripetal acceleration ( \(a_{c}\) ).

As we have discussed in the previous topic

\[\overrightarrow{\mathbf{r}} = \mathbf{r}(cos\theta\widehat{\mathbf{i}} + sin\theta\widehat{\mathbf{j}})\]

and

\[\overrightarrow{v} = - vsin\theta\widehat{i} + vcos\widehat{j} = v( - sin\theta\widehat{i} + cos\theta\widehat{j})\]

\[\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = v( - cos\theta\widehat{i} - sin\theta j)\left( \frac{d\theta}{dt} \right) + v(sin\theta\widehat{i} + cos\theta j)\frac{dv}{dt}\]

\(|\overline{v}|\) is constant \(\Rightarrow \ \frac{d|\overline{v}|}{dt} = 0\ \) i.e. \(\ \frac{dv}{dt} = 0\)

Also \(\frac{d\mathbf{\theta}}{dt} = \mathbf{\omega}\)

\[\begin{matrix} \therefore\ \overrightarrow{a} & \ = - v\omega(cos\theta\widehat{i} + sin\theta\widehat{j}) \\ \overrightarrow{a} & \ = v\omega_{\overrightarrow{a}} \end{matrix}\]

here \(\ \widehat{a} = - (cos\theta\widehat{i} + sin\theta\widehat{j}) = - \widehat{j}\)
Thus direction of this acceleration is opposite to \(\widehat{\mathbf{r}}\) i.e. radially inwards.
Also \(\ |\overrightarrow{a}| = v\omega\)
\(\Rightarrow \ a_{r} = v\omega\) or \(\frac{v^{2}}{r}\) or \(\omega^{2}r\ \lbrack\because v = r\omega\rbrack\)
Magnitude of this acceleration is constant but direction changes continuounsly, always being normal to velcoity as shown in the figure above.

Non-uniform circular motion

For a particle moving in non-uniform cicular motion both direction as well as magnitude of velocity change.
Now, \(\ \overrightarrow{v} = v( - sin\theta\widehat{i} + cos\theta\widehat{j})\)

\[\begin{matrix} \therefore\ \overrightarrow{a} & \ = \frac{d\overrightarrow{v}}{dt} = - v(sin\theta\widehat{i} + cos\widehat{j}) \cdot \frac{d\theta}{dt} + ( - sin\theta\widehat{i} + cos\widehat{j}) \cdot \frac{dv}{dt} \\ & \ = - v\omega\widehat{r} + \frac{dv}{dt}\overset{˙}{v} \end{matrix}\]

we can write \(\overline{\mathbf{a}} = {\overline{\mathbf{a}}}_{t} + {\overline{\mathbf{a}}}_{t}\)
where \({\overrightarrow{a}}_{\mathbf{z}} = v\omega( - \overset{˙}{r})\) i.e. radial acceleration.
and \({\overrightarrow{a}}_{1} = \left( \frac{dv}{dt} \right)\widehat{v}\) having unit vector equal to \(\widehat{v}\) i.e. it is also in tangential direction and is called tangential acceleration.

Thus in this case acceleration has two componants one along velocity i.e. tangential acceleration and another normal to velocity i.e. radial acceleration as shown.

If net acceleration ( \(\overrightarrow{\mathbf{a}}\) ) makes angle \(\phi\) with tangential direction, we may write

\[tan\phi = \frac{a_{f}}{a_{1}}\]

also

\[|\overrightarrow{a}| = \sqrt{a_{1}^{2} + a_{1}^{2}}\]

Some points to remember

\[\begin{matrix} & a_{t} = \frac{dv}{dt} = r\frac{d\omega}{dt} = r\alpha \\ \therefore & |\overrightarrow{a}| = \sqrt{\left( \omega^{2}r \right)^{2} + (r\alpha)^{2}} \end{matrix}\]

• Students need not to be confused with \(\left| \frac{d\overrightarrow{v}}{dt} \right|\) and \(\frac{d}{dt}|\overrightarrow{v}|.\overrightarrow{v}\) contains both magnitude and driection. Thus \(\frac{d\overrightarrow{v}}{dt}\) means \({\overrightarrow{a}}_{mt}\) i.e. \(\left| \frac{d\overrightarrow{v}}{dt} \right| = \left| {\overrightarrow{a}}_{nt} \right|\).

Also \(|\overrightarrow{v}|\) means magnitude of velocity only which changes because of tangential acceleration.
\[\therefore\ \frac{d|\overrightarrow{v}|}{dt} = a_{1}\]

Radius of curvature

If a body is moving in any curvilinear path, then at different locations, the curvature would be different, thus the radius would be different.

For general curvilinear motion, when the particle crosses a point A , it is satisfying condition of moving on an imaginary circle. At this instant, if \(a_{\bot} = \frac{V^{2}}{R_{c}}\) (where \(R_{c}\) is radius of curvature at this instant.)
\[{R_{c} = \frac{v^{2}}{a_{\bot}} }{R_{c} = \frac{\text{~}\text{(speed)}\text{~}\ ^{2}}{\text{~}\text{comp. of acceleration }\text{perpendicudar}\text{ to velocity}\text{~}} }\]

Conical Pendulum

It consists of a small sphare (bob) connected to a light string. The bob is given some speed such that it revolve in a horizontal circular path while the string makes constant angle (say \(\theta\) ) with the vertical as shown.
Resolving in horizontal (radial) direction and vertical direction,

\[{\left( F_{ma} \right), = 0 \Rightarrow \text{ }Tcos\theta = mg }{\Rightarrow T = \frac{mg}{cos\theta} }\]Also \(\left( F_{ma} \right)_{a} = ma \Rightarrow Tsin\theta = \frac{mV^{2}}{r}\)

\[\Rightarrow \ \left( \frac{mg}{cos\theta} \right)sin\theta = \frac{{mV}^{2}}{\mathcal{l}sin\theta} \]

\[\Rightarrow \ V = sin\theta\sqrt{\frac{g\mathcal{l}}{cos\theta}} \]Time period of revolution \((t) = \frac{2\pi r}{V} = \frac{2\pi\mathcal{l}sin\theta}{sin\theta\sqrt{\frac{g\mathcal{l}}{cos\theta}}} = 2\pi\sqrt{\frac{\mathcal{l}cos\theta}{g}}\)

Results obtained

Tension \((T) = \frac{mg}{cos\theta}\)
Speed \((V) = sin\theta\sqrt{\frac{g\mathcal{l}}{cos\theta}}\)
Time period \((t) = 2\pi\sqrt{\frac{\mathcal{l}cos\theta}{\text{ }g}}\)

Banking of road

When a vehicle is taking on horizontal circular tum, friction on tyres due to the road provides centripetal acceleration. Many times the friction is not sufficient enough, for example when speed is too large or the turn is too narrow. In such cases, instant of keeping the road horizontal, it is made tilted with the outerside as higher side. This is called banking of road and the angle with which the road is kept inclined is called angle of banking. This enables the normal force to provide a horizontal component to help friction to provide centripetal acceleration. This can be explained below mathematically.

\[\left( F_{y} \right) = 0 \Rightarrow \text{ }Ncos\theta + fsin\theta = mg \]

Under limithing condition, i.e. when the car is about to skid, \(f = \mu N\)
\[\Rightarrow \ Ncos\theta + \mu Nsin\theta = mg\]

\[\begin{array}{r} N = \frac{mg}{cos\theta + \mu sin\theta}\#(i) \end{array}\]

Also \(\ \left( F_{s} \right)_{mat} = ma_{t}\)
\[{\Rightarrow \ Nsin\theta + fcos\theta = m\left( \frac{V^{2}}{R} \right) }{\Rightarrow \ N(sin\theta + \mu cos\theta) = \frac{mV^{2}}{R} }\]substituting value of N found in equation (i), we get

\[\frac{mg(sin\theta + \mu cos\theta)}{cos\theta + \mu sin\theta} = \frac{{mV}^{2}}{R}\]

\[\Rightarrow \ V = \sqrt{\frac{gR(sin\theta + \mu cos\theta)}{(cos\theta + \mu sin\theta)}} \]This is the speed at which friction achieves its maximum value i.e. it is the maximum speed with which the vehicle can take turn safely on banked road.

Special cases :

I If friction is neglected i.e. \(\mu\) is taken zero.

\[V = \sqrt{gRtan\theta}\]

II If banking is not provided i.e. \(\theta = 0^{\circ}\)

\[V = \sqrt{\mu gR}\]

Centrifugal force

This the pseudo force which has to be taken under account when the frame of reference is rotating
object itself, since the revolving object is accelerated i.e. non-inertial frame of refemce. The magnitude of this forceis \(\frac{mV^{2}}{r}\) or \(m\omega^{2}r\) (where \(r\) - radius of circular path of the object) and its direction is always radially outwards.

Solving problems with the concept of centrifugal force

Tension in rotating ring

Diagram shown top view of a uniforming of mass \(m\) roating in horizontal plane.

Here tension inside the ring is an internal force for the ring. So to find it, we take a very small part of the ring which subtends angle \(d\theta\) as shown.

we take \(y\)-axis towards the centre and \(x\)-axis in tangential direction and origin as the mid-point of the element
\[\therefore\ \left( F_{nef} \right)_{y} = (dm)\omega^{2}R\]

\[\begin{array}{r} \therefore\ 2\text{ }Tsin\left( \frac{\text{ }d\theta}{2} \right) = (dm)\omega^{2}R\#(i) \end{array}\]

but \(sin\left( \frac{d\theta}{2} \right) \simeq \frac{d\theta}{2}\ (\because\text{ }d\theta\) is very small)
Also \(\mathbf{dm}\mathbf{= (}\) Mass per unit length \(\mathbf{\circ}\) f ring \(\mathbf{\times}\) length of element
\[\therefore\ \mathbf{dm} = \left( \frac{\mathbf{M}}{2\pi\mathbf{R}} \right)(\mathbf{Rd}\mathbf{\theta}) = \left( \frac{\mathbf{M}}{2\pi} \right)\mathbf{d}\mathbf{\theta} \]Putting these values in equation (i), we get

\[\begin{matrix} & 2T\left( \frac{d\theta}{2} \right) = \left( \frac{M}{2\pi} \right)d\theta \times \omega^{2}R \\ \therefore & T = \frac{M\omega^{2}R}{2\pi} \end{matrix}\]

Solved Example

Q. 1 The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed \(\omega\) varies with time ' \(t\) ' as.
(A) 2 a sinct
(B) \(2asin\frac{\omega t}{2}\)
(C) 2a \(cos\omega t\)
(D) \(2acos\frac{\omega t}{2}\)

Sol. In time t, let the particle moves from A to B and rotates by angle \(\theta = \omega t\)

Magnitude of displacement

\[\begin{matrix} & S = |\overrightarrow{S}| = AB \\ \Rightarrow \ & S/2 = asin(\theta/2) \\ \therefore\ & S = 2asin(\theta/2) = 2asin\left( \frac{\omega t}{2} \right) \end{matrix}\]

Q. 2 The speed of an object undergoing uniform circular motion is \(4\text{ }m/s\). Find the minimum possible centripetal acceleration (in \(m/s^{2}\) ) of the object. [ Take \(\pi = 25/8\) ]
Sol. Let the particle rotates by angle \(\theta( = \omega t)\) in the given time interval
\[\therefore\ |\Delta\overline{V}| = \left| {\overline{V}}_{r} - {\overline{V}}_{\text{in}\text{~}} \right| = \sqrt{\left| {\overline{V}}_{r} \right|^{2} + \left| {\overline{V}}_{\text{in}\text{~}} \right|^{2} - 2\left| {\overline{\text{ }V}}_{r} \right| \times \left| {\overline{V}}_{\text{in}\text{~}} \right| \times cos\theta} \]but \(\ \left| {\overrightarrow{V}}_{i} \right| = \left| {\overrightarrow{V}}_{in} \right| = V\) and also \(|\Delta\overrightarrow{V}| = V\)
\[{\therefore\ V = \sqrt{V^{2} + V^{2} - 2{\text{ }V}^{2}cos\theta} }{\Rightarrow \ \mathbf{\cos}\mathbf{\theta} = \frac{1}{2} }\]

\(\therefore\ \theta\) i.e. \(\omega t = \frac{\pi}{3},\frac{5\pi}{3}\) etc.
\[{\Rightarrow \ \omega_{\text{min}\text{~}}t = \frac{\pi}{3} }{\therefore\ \omega_{\max} = \frac{\pi}{3t} = \frac{\pi}{3(0.5)} = \frac{2\pi}{3} }\]Also \(\ a_{r} = \omega V\)
\[{\Rightarrow \ a_{r} = \frac{2\pi}{3} \times 4 = \frac{8\pi}{3}\text{ }m/s^{2} }{\therefore\ a_{r} = \frac{25}{3}\text{ }m/s^{2} = 8.33\text{ }m/s^{2} }\]Q. 3 Two strings of length \(/ = 0.5\text{ }m\) each are connected to a block of mass \(\mathbf{m} = 2\mathbf{kg}\) at one end and their ends are attached to the point \(A\) and \(B\mathbf{0}\mathbf{.}\mathbf{5}\text{ }\mathbf{m}\) apart on a vertical pole which rotates with a constant angular velocity \(\omega = 7rad/sec\). Find the ratio \(\frac{T_{1}}{{\text{ }T}_{2}}\) of tension in the upper string \(\left( T_{1} \right)\) and the lower string ( \(T_{2}\) ). [ Use \(g = 9.8\text{ }m/s^{2}\) ]

Sol. \(\ Sin\theta = \frac{0.25}{0.50} = \frac{1}{2}\ \Rightarrow \ \theta = \frac{\pi}{6}\)
\[{r = 0.5Cos\theta = 0.5\left( \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{4} }{\left( F_{\text{max}\text{~}} \right),0 \Rightarrow {\text{ }T}_{1}sin\theta - T_{2}sin\theta - mg = 0 }\]

\[\begin{array}{r} \begin{matrix} \therefore & \left( T_{1} - T_{2} \right)sin\theta = mg \\ & \left( F_{ne} \right)_{x} = m\omega^{2}r \\ \therefore & \left( T_{1} + T_{2} \right)cos\theta = m\omega^{2}r \end{matrix}\#(ii) \end{array}\]

Dividing equation (ii) by (i)

\[\begin{matrix} & \left( \frac{T_{1} + T_{1}}{T_{1} - T_{2}} \right)cot\theta = \frac{\omega^{2}r}{g} \\ \therefore & \frac{T_{1} + T_{1}}{T_{1} - T_{2}} = \frac{\omega^{2}rtan\theta}{g} = \frac{(7)^{2} \times (\sqrt{3}/4) \times 1/\sqrt{3}}{9.8} \\ \therefore & \frac{T_{1} + T_{1}}{T_{1} - T_{2}} = \frac{5}{4} \\ \therefore & \frac{T_{1}}{T_{2}} = \frac{5 + 4}{5 - 4} = 9 \end{matrix}\]

Q. 4 A long horizontal rod has a bead which can slide along its length, and initially placed at a distance \(L\) from one end of \(A\) of the rod. The rod is set in angular motion about \(A\) with constant angular acceleration \(\alpha\). If the coefficient of friction between the rod and the bead is \(\mu\) and gravity is neglected, then the time after which the bead starts slipping is
(A) \(\sqrt{\frac{\mu}{\alpha}}\)
(B) \(\frac{\mu}{\sqrt{\alpha}}\)
(C) \(\frac{1}{\sqrt{\mu\alpha}}\)
(D) infinitesimal

Sol

	\[N = ma_{1} = m\alpha L\]
and	\[f = ma_{1} = m\omega^{2}L\]
but	\[\omega = 0 + \alpha t = \alpha t\]
\[\therefore\]	\[f = \mu\alpha^{2}t^{2}L\]

Also at the instant of slipping, \(f = \mu N\)

\[\begin{matrix} \therefore & m\alpha^{2}t^{2}L = \mu(m\alpha L) \\ \therefore & t = \sqrt{\frac{\mu}{\alpha}} \end{matrix}\]

Q. 5 A bead of mass \(m = 300gm\) moves in gravity free region along a smooth fixed ring of radius \(R = 2\text{ }m\). The bead is attached to a spring having natural length \(R\) and spring constant \(k = 10\text{ }N/m\). The other end of spring is connected to a fixed point O on the ring.
\(AB = \frac{6R}{5}\). Line OB is diameter of ring.

Find (a) Speed of bead at A if normal reaction on bead due to ring at A is zero.
(b) The rate of change in speed at this instant.

Sol. The length of the spring, when bead is at \(A\), is
\[OA = \sqrt{(OB)^{2} - (AB)^{2}} = \sqrt{(2R)^{2} - \left( \frac{6R}{5} \right)^{2}} = \frac{8R}{5} \]Elongation in the spring \(= OA -\) natural length

\[= \frac{8R}{5} - R = \frac{3R}{5}\]

Also \(cos\theta = \frac{(8R/5)}{2R} = \frac{4}{5}\) and \(sin\theta = \left( \frac{6R/5}{2R} \right) = \frac{3}{5}\)
Radial component of spring force \(= kxcos\theta = \frac{mv\ ^{2}}{R}\)
\[{\Rightarrow \ k\left( \frac{3R}{5} \right)\left( \frac{4}{5} \right) = \frac{mv^{2}}{R} }{\Rightarrow \ v = \sqrt{\left( \frac{12{kR}^{2}}{25m} \right)} = 8\text{ }m/s}\]

Tangential component of force \(= kxsin\theta = k\left( \frac{3R}{5} \right)\left( \frac{3}{5} \right) = \frac{9kR}{25}\)
\[\Rightarrow \ m\frac{dv}{dt} = \frac{9kR}{25} \]Rate of change in speed \(\ \frac{dv}{dt} = \frac{9kR}{25\text{ }m} = 24\text{ }m/s^{2}\)
Q. 6 A bead of mass \(m\) is attached to one end of a spring of natural length \(\sqrt{3}R\) and spring constant \(k = \frac{(\sqrt{3} + 1)mg}{R}\). The other end of the spring is fixed at point \(A\) on a smooth fixed vertical ring of radius \(R\) as shown in the figure. What is the normal reaction at B just after the bead is released?

Sol.

At the instant of release, compression in the spring is

\[x = \sqrt{3}R - (\sqrt{3} - 1)R\]

\(\therefore\ \) spring force, \(kx = \frac{(\sqrt{3} + 1)mg}{R} \times (\sqrt{3} - 1)R = 2mg\)
Also at the instant of release, speed \(= 0\)
\(\therefore\ \) radial acceleration, \(a_{r} = \frac{v^{2}}{R} = 0\)
\(\therefore\ \) Along radial direction, \(F_{\text{net}\text{~}} = 0\)
\[{\therefore\ N - mgcos30^{\circ} - kxcos60^{\circ} = 0 }{N = mg\frac{\sqrt{3}}{2} + (2mg) \times \frac{1}{2} = \frac{mg(\sqrt{3} + 2)}{2}}\]

Introduction

Figure (a) shows a skier starting from rest at the top of a uniform slope. What's the skier's speed at the bottom? You can solve this problem by applying Newton's second law to find the skier's constant acceleration and then the speed. But what about the skier in figure (b)? Here the slope is continuously changing and so is the acceleration. Constant-acceleration equations are not applicable here, so solving for the details of the skier's motion is difficult.

Figure (a)

Figure (b)

There are many cases where motion involves changing forces and accelerations. In this chapter, we introduce the important physical concepts of work and energy. These powerful concepts enable us to "shortcut" the detailed application of Newton's law to analyze these more complex situations. We begin with the concept of work.

Work

In day-today life, me oftencly use the word "Work". In physics, we define the work in quite different manner than we usually use the word "Work" in daily life. Work done by a constant force \(\overline{F}\) on an object when displaced by \(\overrightarrow{S}\) is given by

\[W = F^{'}S\]

Where \(F^{'}\) is the component of force which is in the direction of displacement

In above figure, \(F^{'} = Fcos\theta\)

\[\begin{matrix} \therefore & W = (Fcos\theta)S = FScos\theta \\ \therefore & W = \bar{F} \cdot \bar{S} \end{matrix}\]

So work may also be given as dot product of force and displacement.

• Its unit is Newton meter (N.m) or Joule (J)

• If \(\overrightarrow{F} = F_{x}\overrightarrow{i} + F_{y}\overrightarrow{j} + F_{z}\overset{˙}{k}\) and \(\overrightarrow{S} = \Delta x\widehat{i} + \Delta y\widehat{j} + \Delta z\widehat{k}\), then \(W = \overrightarrow{F} \cdot \overrightarrow{S} = F_{z}(\Delta x) + F_{y}(\Delta y) + F_{z}(\Delta z)\)

• Work done force is frame dependent as displacement is frame dependent

• Work can be positive or negative or zero. When a force speed up the praticle, it does positive work. A force acting at \(90^{\circ}\) to the motion does no work. And when a force slow down the motion, it does negative work.

e.g. According to equation the person pushing the car in figure (a) does work equal to the force he applies times the distance the car moves. But person pulling the suitcase in figure (b) does work equal to only the horizontal component of the force times the distance the suitcase moves. Furthermore, by our definition, the waiter of figure (c) does no work on the tray. Why not ? Because the force on the tray is vertical while the tray's displacement is horizontal; there's no component of force in the direction of the tray's motion.

(a)

(b)

(c)

Hilustration :

The airline passenger in above figure (b) exerts a \(80N\) force on his suitcase pulling at an angle \(60^{\circ}\) with the horizontal. What work he does on suitcase while pulling it 50 m on the floor?

Sol. \(\ W = (80\text{ }N)(50\text{ }m)cos60^{\circ}\)
\[= 2000\text{ }J = 2KJ\]

Work Done by friction

There is a misconception that the force of friction always does negative work. In reality, the work done by fricition may be zero, postive or negative depending upon the situation as shown in the figure.

(c)

(a) When a block is pulled by a force \(F\) and the block does not move, the work done by friction is zero.
(b) When a block is pulled by a force \(F\) on a stationary surface, the work done by the kinetic friction is negative.
(c) Block \(A\) is placed on the block \(B\). When the block \(A\) is pulled with force \(F\), the friction force does negative work on block \(A\) and positive work on block \(B\), which is being accelerated by a force \(F\). the displacement of A relative to the table is in the forward direction. The work done by kinetic friction on block \(B\) is positive.

Work Done by Gravity

Consider a block of mass \(m\) which slides down a smooth inclined plane of angle \(\mathbf{\theta}\) as shown in figure.

Let us assume the coordinate axes as shown in the figure to specify the components of the two vector although the value of work will not depend onthe orientation of the axes.
Now, the force of gravity, \({\overrightarrow{F}}_{i} = - mg\widehat{j}\)
and the displacement is given by

\[\overrightarrow{s} = \Delta x\widehat{i} + \Delta y\widehat{j} + \Delta z\widehat{k}\]

The work done by gravity is

\[W_{t} = \overrightarrow{F},\overrightarrow{s} = - mg\widehat{j} \cdot (\Delta x\widehat{i} + \Delta y\widehat{j} + \Delta z\widehat{k})\]

or

\[W_{z} = - mg(\Delta y)\]

Since

\[\Delta y = y_{f} - y_{i} = - h\]

\[\therefore\ W_{z} = + mgh \]If the block moves in the upward direction, then the work done by gravity is negative and is given by

\[W_{g} = - mgh\]

Important

1. The work done by the force of gravity depends only on the initial and final vertical coordinates, not on the path taken.

2. The work done by gravity is zero for path that returns to its initial point.

Work done by a variable force

Often the force applied to an object varies with position. Important examples include electric and gravitational force, which vary with the distance between interacting objects. The force of a spring that we encountered in previous chapter provides another example: as the spring stretches, the force increases.
In this case we have difficulty to apply \(W = \overrightarrow{F} \cdot \overrightarrow{S}\). since \(\bar{F}\) is not same for complete \(\overrightarrow{S}\).
Thus, we take a very small part d \(\overrightarrow{S}\) of its path. This displacement \(d\overrightarrow{S}\) is so small that in variation force may be neglected during it.

So we may write, for the work done during this displacement as

\[\begin{matrix} dW & \ = \overrightarrow{F} \cdot \text{ }d\overrightarrow{\text{ }S} \\ & \ = FdScos\theta \end{matrix}\]

The total work done in going from \(A\) to \(B\) as shown may be calculated by summing up i.e. integrating the work done during its small fractions.
i.e. \(\ W_{A \rightarrow B} = \int_{A}^{B}\mspace{2mu}\overrightarrow{F} \cdot d\overrightarrow{S} = \int_{A}^{B}\mspace{2mu}(Fcos\theta)dS\)
In terms of rectangular components,

\[= F_{x}\widehat{i} + F_{y}\widehat{j} + F_{z}\widehat{k}\]

and

\[d\overrightarrow{S} = dx\widehat{i} + dy\widehat{j} + dz\widehat{k}\]

therefore,

\[W_{A \rightarrow B} = \int_{x_{A}}^{x_{B}}\mspace{2mu} F_{x}dx + \int_{y_{A}}^{y_{A}}\mspace{2mu} F_{y}dy + \int_{z_{A}}^{z_{A}}\mspace{2mu} F_{z}dz\]

Work done as Area under the force displacement graph

Suppose of particle moving along a straight line and a force acting on it varies with its displacement x as shown.

\[\begin{matrix} W & \ = \int_{i =}^{i}\mspace{2mu}\mspace{2mu} F \cdot dx \\ & \ = \text{~}\text{Area under}\text{~}F\text{~}\text{vs}\text{~}x\text{~}\text{graph from}\text{~}x = x_{in}\text{~}\text{to}\text{~}x_{r} \end{matrix}\]

In general, the work done by a point \(x_{n}\) to final point \(x_{r}\) is given by the area under the force - displacement curve as shown in the figure.

Area (work) above the \(x\)-axis is taken as positive, and below \(x\)-axis as negative.

Work Energy Theorem

Closely related to work is energy - One of the most important concepts in all of physics. Here we introduce the energy associated with motion i.e. kinetic energy. Our goal is to relate kinetic energy and work.

The net work is done by all the forces acting on an object, so we use the net force in our expression for work. We'll consider one-dimensional motion with force and displacement along the same line. In that case, the Equation below gives the net work :

\[W_{nxt} = \int_{}^{}\ F_{nta}dx\]

But net force can also be written as

\[\begin{matrix} \Rightarrow & F_{net} = ma = mv\frac{dv}{dx} \\ & F_{net}dx = mvdv \\ & W_{net} = \int_{v_{tw}}^{v_{t}}\mspace{2mu}\mspace{2mu} mvdv \\ \therefore & W_{net} = \frac{1}{2}m\left\lbrack v_{f}^{2} - v_{in}^{2} \right\rbrack \\ \therefore & W_{net} = \Delta K.E. \end{matrix}\]

Thus change in an object's kinetic energy is equal to the net work done on the object. This called Work Energy Theorm.

Important

(i) The kinetic energy of an object is a measure of the amount of work needed to increase it speed from zero to a given value.
(ii) The kinetic energy of a particle is the work it can do on its surroundings in coming to rest.
(iii) Since the velocity and displacement of a particle depend on the frome of reference, the numerical values of the work and the kinetic energy also depend on the frame.
(iv) If work done by net force is positive, kinetic energy of the system increases. If net work done is negative K.E. decreases and if net work is zero, K.E. remains contact

Answers

Q. \(135\text{ }m/s\)
Q. 2 (i) \(6\text{ }m/s\)
(ii) \(3\text{ }m/s\)
Q. 3 (i) \(\frac{2\text{ }F}{k}\)
(ii) \(\frac{F}{\sqrt{mk}}\)
Q. 4
(a) \(2.4\text{ }m/s\) (b) \(4.2\text{ }m/s\)
Q. \(5\ 10\text{ }cm\)

Consevative and Non-conservertive force

If we throw a body up along smooth incline plane with some speed \(v_{i}\), then it moves along the incline till it becomes stationary for a moment and then moves down the incline. It is observed the, when it reaches the point of projection, its speed is \(v_{0}\) again, which proves that during the journey the net work done on the block is zero. Two forces act on the block during its motion. One is Normal force ( N ) which is continuously perpendicular to the block's motion, so its work for any part of the path is zero. Another one is weight ( mg ) which does negative work while upward motion and positive work of same magnitude during downward motion, does zero net work when the body reaches the initially position again.

Now we throw the same body on a rough incline plane. When it reaches the initial position, its speed is lesser than the speed of projection since friction does negative work for motion during up and as well as down the incline.

Thus, here we find two categories of force .

Conservative force

When the total work done by a force F acting as an object moves over any closed path is zero, the force is conservative. Mathematically.

\[\text{∮}\ \ \overrightarrow{F} \cdot \text{ }d\overrightarrow{r} = 0\ \text{~}\text{(conservative force)}\text{~}\]

Suppose we move an object along the straight path between point \(A\) and \(B\) shown in figure, along which a conservative force acts; let the work done by the conservative force be \(W_{AB}\). Since the work done over any closed path is zero, the work \(W_{BA}\) done in moving back from \(B\) to \(A\) must be \(- W_{AB}\), whether we return along the straight path or the curved path or any other path. So, going from A to B involves work \(W_{AB}\) ' regardless of the path taken.
In other words : The work done by a conservative force in moving between two points is independent of the path taken;

mathematically. \(\int_{A}^{B}\mspace{2mu}\overrightarrow{F} \cdot d\overrightarrow{r}\) depends only on the endpoints \(A\) and \(B\), not on the path between them.
These include force due to gravity (mg), spring force, electrostatic force etc.

Non-conservative force

Work done by non-conservative forces depends on path also. Least work is done for straight line path and any curved path it involves different work. These include frictional force, viscous force etc.

Potential Energy

When we move an object upto some height against gravity and release, it gains kinetic energy while moving down. It means work done against a conservative force like gravity is somehow stored, in the sense that we can get it back again in the form of kinetic energy. We can consider the "stored work" as potential energy \(U\).
The change \(\Delta U_{AB}\) in potential energy associated with a conservative force is the negative of the work done by that force as it acts over any path from point \(A\) to point \(B\) :

\[\Delta U_{AB} = - \int_{A}^{s}\mspace{2mu}\overrightarrow{F} \cdot d\overrightarrow{r}\]

If a conservative force does positive work (as does gravity on a falling object), then potential energy must decrease - and that means \(\Delta U\) must be negative. Conversely. if a conservative force does negative work (as does gravity on a weight being lifted), then energy is stored and \(\Delta U\) must be positive.
Change in potential energy are all that ever matter physically; the actual value of potential energy is meaningless. All though optenly it is convinient to establish a refernce point at which the potential energy is defined to be zero. When we say "the potential energy \(U\)," we really mean the potential-energy difference \(\Delta U\) between the point we're considering and the reference point.

Gravitational Potential Energy (Near the Earth's Surface)

The work done by gravity on a particle of mass \(m\) whose vertical coordinate changes from \(y_{A}\) to \(y_{B}\) is

\[W_{g} = - mg\left( y_{B} - y_{A} \right)\]

From equation, we have \(W_{g} = - \Delta U_{g} = - \left( U_{B} - U_{A} \right)\)
Thus gravitational potential energy at the point \(B\) near the surface of the Earth is given by

\[U_{B} = U_{A} + mgh\]

If we assume potential energy at the point \(A\) to be zero, then potential energy at the point \(B\) is given by

\[U_{B} = 0 + mgh = mgh\]

Elastic Potential Energy

When you stretch or compress a spring, you do work against the spring force, and that work gets stored as elastic potential energy. For an ideal spring, the force is \(F = - kx\), where x is the distance the spring is stretched from equilibrium, and the minus sign shows that the force opposes the stretching or compression.

\[\Delta U = - \int_{x_{1}}^{x_{2}}\mspace{2mu} F(x)dx = - \int_{x_{1}}^{x_{2}}\mspace{2mu}( - kx)dx = \frac{1}{2}kx_{2}^{2} - \frac{1}{2}kx_{1}^{2}\]

where \(x_{1}\) and \(x_{2}\) are the initial and final values of the stretch. If we take \(U = 0\) when \(x = 0\)-that is, when the spring is neither stretched nor compressed - then we can use this result to write the potential energy \(\mathbf{U}_{\mathbf{2}} = \mathbf{U}\) at an arbitrary stretch (or compression) \(x_{\mathbf{2}} = x,U = \frac{1}{\mathbf{2}}kx^{\mathbf{2}}\)

Conservation of Mechanical Energy

The work-energy theorem, shows that the change \(\bigtriangleup KE\) in a body's kinetic energy is equal to the net work done on it :

\[\Delta KE = W_{net}\]

Consider separately the work \(W_{c}\) done by conservative force and the work \(W_{nc}\) done by nonconservative forces. Then

\[\Delta KE = W_{c} + W_{nc}\]

We've defined the change in potential energy \(\Delta U\) as the negative of the work done by conservative forces. So we can write

\[\Delta KE = - \Delta U + W_{m}\]

or

\[\Delta KE + \Delta U = W_{nc}\]

We define the sum of the kinetic and potential energy as the mechanical energy. Then Equation shown that the change in mechanical energy is equal to the work downe by non-conservative forces.
i.e.

\[\Delta E = W_{nc}\]

\(\therefore\ \Delta E = 0\ \) if \(W_{nc} = 0\)
Thus if work done by non-conservative forces is zero the mechanical energy of the system is unchanged. This called law of conservation of mechanical energy. It may also be written as

\[\Delta U + \Delta KE = 0\]

or

\[\Delta U = - \Delta KE\]

or

\[U + KE = \text{~}\text{constant}\text{~}\]

or

\[U_{in} + KE_{in} = U_{f} + KE_{f}\]

The surfaces shown in the figure are frictionless and horizontal surface is taken as reference level.

By conservation of mechanical energy
\[\frac{1}{2}{kx}_{0}^{2} + 0 = 0 + \frac{1}{2}{mv}_{0}^{2} = mgH + 0\]

Types of equilibrium on the basis of stability

(a)

(b)

(c)

Suppose a small ball is placed on a smooth track under three different situations as shown in figure (a), (b) & (c). In all the three situation, the ball, is in equilibrium
(A) In figure (a), when the ball is slightly displaced from its equilibrium position, it tries to attain the shown position again, such type of equilibrium is called stable equilibrium.
Here potential energy in equilibrium position is minimum as compares to its neighbouring points i.e. under stable equilibrium, potential energy is minimum
i.e. \(\frac{dU}{dr} = 0\) and \(\frac{d^{2}U}{{dr}^{2}} > 0\)
(B) In figure(b), when the ball is slightly displaced from its equilibrium position it tends to move farther form the shown equilibrium position. Such type of equilibrium is called unstable equilibrium. Here potential energy in equilibrium is maximum as compared to its near by points

\[\text{~}\text{i.e.}\text{~}\frac{dU}{dr} = 0\text{~}\text{and}\text{~}\frac{d^{2}U}{{dr}^{2}} < 0\]

(C) In figure (c), when the ball is displaced, it accepts the new position as equilibrium position, such type of equilibrium is called neutral equillibrlum.
Here potential energy remains uniform for the equilibrium position

\[\text{~}\text{i.e.}\text{~}\frac{dU}{dr} = 0\text{~}\text{and}\text{~}\frac{d^{2}U}{dr^{2}} = 0\]

Although, the above discussion is under the effect of gravity but the results observed is applicable in other situations also where a particle can move under the effect of the conservative force only.
The above result can also be studied with the help of the following graphs.
For a particle whose position ( \(\mathbf{r}\) ) varies along a straight line, the graphs below
Show variation of \(U\) vs \(r\) and \(F\) vs \(r\).

At Point \(A\ F = 0;\frac{dU}{dr} = 0\), but \(F = 0\) at its nearly points also. So when slightly displaced from \(A\), the new position is also equilibrium. Thus point, A shown is the position of neutral equilibrium.

At Point B \(\ F = 0;\frac{dU}{dr} = 0\). Now when it is slightly displaced towards left of \(B\), force is positive i.e. towards right and when it is slightly displaced towards right of \(B\), force is negative i.e. towards right. Thus force tries to bring the particle towards \(B\) again. This type of force is called restoring force and the point \(B\) is the position of stable equilibrium.

At Point C \(\ F = 0,\frac{dU}{dr} = 0\) but when particle is displaced slightly form it towards any direction, force acts in that direction only i.e. to move the particle away from C . Thus point C is the position of unstable equilibrium.

Power

Power is the work done per unit time. If \(\Delta W\) a work done in time \(\Delta t\), the average power is \(P_{av} = \frac{\Delta W}{\Delta t}\) Average power becomes intantancous power as \(\Delta t\) approches to zero.

\[\begin{matrix} & \ \therefore\ \text{~}\text{Instantancous}\text{ Power}\text{~}\ P = {Lim}_{\Delta t \rightarrow 0}\left( \frac{\Delta W}{\Delta t} \right) \\ & \ \Rightarrow \ P = \frac{dW}{dt} \end{matrix}\]

Due to a particular force on a particle

\[\begin{matrix} & \Delta W = \overrightarrow{F} \cdot \Delta\overrightarrow{S} \\ \therefore & P = {Lim}_{\Delta t \rightarrow 0}\overrightarrow{F} \cdot \frac{\Delta\overrightarrow{S}}{\Delta t} = \overrightarrow{F} \cdot \frac{d\overrightarrow{S}}{dt} \\ \Rightarrow & P = \overrightarrow{F} \cdot \overrightarrow{V} \end{matrix}\ \text{~}\text{(where}\text{~}\overrightarrow{V}\text{~}\text{is velocity)}\text{~}\]

Centripetal force is always perpendicular to velocity, so power due to it is zero.

Practice Exercise

Q. 1 A chain of length \(L\) and mass \(M\) is arranged as shown in following four cases. Arrange the potential energy (assumed zero at horizontal surface) in decreasing order.

(i)

(ii)

(iii)

(iv)

44
Q. 2 A particle is constrained to move in the region \(x \geq 0\) and its potential energy is given by

\[U = 2x^{4} - 3x^{2}\text{ }J\ (\text{~}\text{where}\text{~}x\text{~}\text{is in}\text{~}m)\]

For what values of \(x\), the particle is under
(i) Stable equilibrium
(ii) Unstable equilibrium
Q. 3 A force \(\overrightarrow{F} = - k(y\widehat{i} + x\widehat{j})\) (where \(K\) is a positive constant) acts on a particle moving in the \(X - Y\) plane.Starting from the origin, the particle is taken along the positive \(x\)-axis to the point \((a,0)\) and then parallel to the \(y\)-axis to the point ( \(a,a\) ). Find the total work done by the force \(F\) on the particle.
Q. 4 A block starts from rest at point A, slides down a frictionless incline that terminates in a ramp pointing up at \(45^{\circ}\) angle, as shown in figure. Find an expression for the horizontanl range \(x\) shown in the figure as a function of the heights \(h_{1}\) and \(h_{2}\).

Q. 5 A horse pulls a cart upto 200 m in 5 minutes, exerting a force of 750 N along the displacement of the cart. Find its power output measured in watts and in horsepower.

Answers

Where work done by tension. \(W_{T} = 0\) (since tension is continuously perpendicular to motion) and work done by gravity \(W_{g} = - mgh\)

\[= - mgR(1 - cos\theta)\]

\[\therefore\ - mgR(1 - cos\theta) = \frac{1}{2}m\left( v^{2} - u^{2} \right)\]

\[\begin{array}{r} \therefore\ v^{2} = u^{2} - 2gR(1 - cos\theta)\#(ii) \end{array}\]

\(\therefore\ \) from equation (i) and (ii)

\[T = mgcos\theta + \frac{m}{R}\left\lbrack u^{2} - 2gR(1 - cos\theta) \right\rbrack\]

\[\begin{array}{r} \therefore\ T = \frac{{mu}^{2}}{R} + 3mgcos\theta - 2mg\#(iii) \end{array}\]

Thus T is maximum when \(\theta = 0^{\circ}\) i.e. at the lower most position A

\[T_{\max} = T_{A} = mg + \frac{mu^{2}}{R}\]

Also T is minimum when \(\theta = 180^{\circ}\) i.e. at the top most point B

\[T_{\min} = T_{B} = \frac{{mu}^{2}}{R} - 5mg\]

Thus : \(T_{\text{max}\text{~}} - T_{\text{min}\text{~}} = 6mg\) (independent of initial speed provided the porticle is able to complete the circular path)
From equations (i) and (iii), following results are obtained

Case I

If \(u = \sqrt{2gR},V\) becomes zero exactly at point \(C\)
Also when it is at point C tension, \(T = m\frac{v^{2}}{R} = 0\)
Now it oscillates between points C and D .

Case II

If \(u < \sqrt{2gR},V\) becomes zero
even before \(\theta = \frac{\pi}{2}\), let at point E
At this point

\[\begin{matrix} & T - mgcos\theta = m\left( \frac{v^{2}}{R} \right) = 0 \\ \Rightarrow \ & \text{ }T = mgcos\theta \neq 0 \end{matrix}\]

Case III

If \(u = \sqrt{5gR}\) then at the top most point \(B\) where \(\theta = 180^{\circ}\), the speed of the particle is
\[v = \sqrt{5gR - 2gR(1 - ( - 1))} = \sqrt{5gR - 4gR} = \sqrt{gR} \]At this point, \(\ mg + T = \frac{m(\sqrt{gR})^{2}}{R}\)
\[\Rightarrow \ T = 0 \]

i.e. string is just slacked, but it still continues to move in circuler path, since it has got speed.

Case IV

If \(u > \sqrt{5gR}\) and \(T > 0\) even at the topmost point ( B ) and particle completes circular path

Case V

If \(\sqrt{5gR} > u > \sqrt{2gR}\)
T becomes zero at some point between \(C\) and \(B\) i.e. for \(\frac{\pi}{2} < \theta < \pi\). And that instant \(v\) is non zero [see equation (ii) and thus particle move in parabolic path after that due to gravity only.

Motion of a particle at the inner surface of a vertical circular track

Same results as above are obtained when a body is given some speed \(u\) at the bottom most point inside a fixed circular loop as shown. Here instead of tension; normal force due to inner surface of the loop comes into action.

Practice Exercise

Q. 1 An \(900 - kg\) roller-coaster car is launched from a giant spring of constant \(k = 31kN/m\) into a frictionless loop-the-loop track of radius 6.2 m as shown in Figure. What is the minimum amount that the spring must be compressed if the car is to stay on the track?

Q. 2 A small toy car of mass \(m\) slides with negligible friction on 'loop' the loop track as shown in figure. The toy car starts from rest at point A which is height \(H = 2R\) above
level of the lowest point of the track:
(i) What normal force is exerted by the track on the toy car at point \(B\) ?
(ii) What are the speed and normal

force at point C ?
(iii) At what height will the ball leave the track and to what maximum height will it riseafterwards?
Q. 3 A particleattached to a vertical string of length 1 m is projected horizontal ly with avelocity of \(5\sqrt{2}\text{ }m/s\).
(a) What is maximum height reached by the particle from the lowermost point of its trajectory.
(b) If the string breaks when it makes an angle of \(60^{\circ}\) with downward vertical, find maximum height reached by the particle from the lowermost point of its trajectory.

Solved Example

Q. 1 Two blocks A and B , cach having a mass of 320 g connected by a light string passing over a smooth light pulley. The block A is attached to a spring of spring constant \(40\text{ }N/m\) whose other end is fixed to a support \(\mathbf{40}\text{ }\mathbf{cm}\) above the horizontal surface as shown. Initially, the spring is vertical and ustretched when the system is released to move. Find the velocity of the block A at the instant it breaks off
the surface below it. Take \(g = 10\text{ }m/s^{2}\).

Sol. At the instant of break-off, \(N = 0\)
\[\Rightarrow \ kxcos\theta = mg \]Where extension in the spring

\[\begin{matrix} & x = \mathcal{l} - 0.4 = \frac{0.4}{cos\theta} - 0.4 \\ \therefore\ & k\left( \frac{0.4}{cos\theta} - 0.4 \right)cos\theta = mg \\ & k(0.4)(1 - cos\theta) = mg \\ \Rightarrow \ & cos\theta = 1 - \frac{mg}{0.4k} \\ & = 1 - \frac{0.320 \times 10}{0.4 \times 40} \\ & cos\theta = \frac{4}{5}\ \Rightarrow \ x = \frac{0.4 \times 5}{4} - 0.4 = 0.1\text{ }m \\ \text{~}\text{Also}\text{~} & tan\theta = \frac{3}{4}\ \Rightarrow \ \text{ }d = 0.4tan\theta = 0.3\text{ }m \end{matrix}\]

\(\therefore\ \) By work energy theorem for motion of both the blocks.

\[\begin{matrix} & W_{g} + W_{s} = \Delta KE \\ \Rightarrow & mgd - \frac{1}{2}{kx}^{2} = \left( \frac{1}{2}mv^{2} \right) \times 2 \\ \Rightarrow & 0.32 \times 10 \times 0.3 - \frac{1}{2}(40)(0.1)^{2} = (0.32)v^{2} \\ \Rightarrow & v = 1.54\text{ }m/s \end{matrix}\]

Q. 2 A small body starts sliding from the height H with zero velocity down a smooth hill which has horizontal portion as shown. What must be the height of the horizontal portion ' \(h\) ' to ensure the maximum distance ' \(s\) ' covered by the body? What is the maximum value of ' \(s\) '?

Sol.

Applying work energy theorem for motion from A to B

\[\begin{matrix} & mg(H - h) = \frac{1}{2}m\left( V^{2} - 0 \right) \\ \therefore\ & V = \sqrt{2g(H - h)} \end{matrix}\]

If time taken by the disc to move from point B to the point C on ground is t , then

\[\begin{array}{r} \begin{matrix} & \Delta y = u_{y}t - \frac{1}{2}gt^{2} = h \\ \therefore & 0 - \frac{1}{2}gt^{2} = - h \Rightarrow t = \sqrt{\frac{2h}{g}} \\ & S = \Delta x = vt \\ \therefore & S = \sqrt{2g(H - h) \times \frac{2h}{g}} = 2\sqrt{Hh - h^{2}} \end{matrix}\#(i) \end{array}\]

for sto be maximum, \(\left( Hh - h^{2} \right)\) is maximum
i.e. \(\ \frac{d}{dh}\left\lbrack Hh - h^{2} \right\rbrack = 0\)
\[\therefore\ H - 2\text{ }h = 0\ \Rightarrow \ \text{ }h = \frac{H}{2} \]Putting this value of \(h\) in equation (i)
\[\therefore\ S_{\text{max}\text{~}} = 2\sqrt{H\left( \frac{H}{2} \right) - \left( \frac{H}{2} \right)^{2}} = H \]Q. 3 A constant horizontal force \(F\) of magnitude equal to \(\frac{mg}{2}\) begins to act on the bob of pendulum shown when the bob was at rest. What maximum angle the string makes with the vertical?

Sol. Let the string makes angle \(\theta\) with the vertical when it comes to rest (momentarily) as shown
\[\therefore\ x = \mathcal{l}sin\theta \]and \(\ h = \mathcal{l} - \mathcal{l}cos\theta = \mathcal{l}(1 - cos\theta)\)
Here work done by gravity \(W_{t} = - mgh = - mg\mathcal{l}(1 - cos\theta)\)
Work done by tension, \(W_{T} = 0\lbrack\overrightarrow{\text{ }T}\bot\overrightarrow{v}\rbrack\)
Work done by force \(F,W_{F} = F \times (\) displacement in the direction of \(F) = Fx\)

\[= \frac{mg}{2}\mathcal{l}sin\theta\]

\(\therefore\ \) Applying work-energy theorem

\[{\therefore\ \frac{1}{2}sin\theta = 1 - cos\theta }{\therefore\ \frac{1}{2}\left\lbrack 2 \times sin\left( \frac{\theta}{2} \right)cos\left( \frac{\theta}{2} \right) \right\rbrack = 2sin\left( \frac{\theta}{2} \right) }{\therefore\ tan\left( \frac{\theta}{2} \right) = \frac{1}{2} }{\therefore\ \theta = 2\tan^{- 1}\left( \frac{1}{2} \right) }\]Q. 4 A small body is placed at rest at the bottom \(B\) of a smooth hemispherical surface of wedge as shown. If the wedge is shifted horizontally towards right with acceleration \(a_{0} = 3\text{ }g\). find speed of the body w.r.t the wedge at the instant the body reaches points A .
Sol. Here we need calculation w.r.t the wedge which is accelerated i.e. non-inertial frame of reference, So we have to consider Pseudo froce ( \(F_{p}\) ) and work done by the Pseudo force ( \(W_{p}\) ) also.
Where \(F_{p} = {ma}_{0} = 3mg\) and is towards left i.e. apposite to \(a_{0}\) By work-energy theorem for motion from \(B\) to \(A\)

\[\begin{matrix} & F_{P}R - mgR = \frac{1}{2}m\left( v^{2} - 0 \right) \\ \Rightarrow & 3mgR - mgR = \frac{1}{2}mv^{2} \\ \therefore & v = 2\sqrt{gR} \end{matrix}\]

Q. 5 A particle is suspended vertically from a point O by an inextensible massless string of length \(L\). A vertical line \(AB\) is at a distance \(L/8\) from \(O\) as shown. The object given a horizontal velocity \(u\). At some point, its motion ceases to be eircular and eventually the object passes through the line AB. At the instant of crossing AB , its velocity is horizontal. Find u .

Sol Let the string slacks when the particle is at point P as shown
At point \(P,\ T + mgcos\alpha = \frac{mV}{L}\)
where \(T = 0\) (as string slacks)
\[\therefore\ mgcos\alpha = \frac{{mV}^{2}}{\text{ }L} \]

\[\begin{array}{r} \Rightarrow \ V^{V} = gLcos\alpha\#(i) \end{array}\]

After this it undergoes parabolic path. When it passes through line AB its velocity is horizontal which implies that ( \(Lsin\alpha - L/8\) ) is half of horizontal range.
i.e. \(\ Lsin\alpha - \frac{L}{8} = \frac{1}{2}\left\lbrack \frac{V^{2}sin(2\alpha)}{g} \right\rbrack\)
\(\therefore\) from equation (i)
\[{Lsin\alpha - \frac{L}{8} = \frac{1}{2}\left\lbrack \frac{\text{ }g\text{ }Lcos\alpha(2sin\alpha cos\alpha)}{g} \right\rbrack }{\therefore\ sin\alpha - \frac{1}{8} = sin\alpha\cos^{2}\alpha }{\Rightarrow \ sin\alpha\left( 1 - \cos^{2}\alpha \right) = \frac{1}{8} }{\Rightarrow \ \sin^{3}\alpha = \frac{1}{8}\ \Rightarrow \ sin\alpha = \frac{1}{2}\ \Rightarrow \ \alpha = 30^{\circ} }{\therefore\ V^{2} = \frac{g\mathbf{L}\sqrt{3}}{2} }\]Also by applying work energy theorem

\[\begin{matrix} & - mgL(1 + cos\alpha) = \frac{1}{2}m\left( V^{2} - u^{2} \right) \\ = & - gL\left( 1 + \frac{\sqrt{3}}{2} \right) = \frac{1}{2}\left( \frac{gL\sqrt{3}}{2} - u^{2} \right) \\ \therefore & \text{~}\text{On solving, we get}\text{~}\ u = \sqrt{\frac{gL}{2}(4 + 3\sqrt{3})} \end{matrix}\]