NCERT Solutions for Class 11 Physics Chapter 6 Systems of Particles and Rotational Motion

Centre of Mass 

In the first part of mechanics, we have discussed all about point objects. In this chapter, we shall deal with the cases of large objects or systems of point objects. To deal with such scenarios, we need to know about the centre of mass of an object or a system.

• Definition

Centre of mass is a hypothetical point where the whole mass of the object is assumed to be concentrated mathematically. It is the weighted mean of the positions of all the point objects with masses M1, M2, M3, ......,Mn respectively

Example :

• Motion of Centre of Mass

Kinematic of the System of Particles

System of particles can move in different ways as observed by us in daily life. To understand this, we need to understand few new parameters. 

Rigid body: A body in which distance between any two particles remain same regardless of any external changes. 

• Various Types of Motion

(i) Translational Motion 

A system is said to be in translational motion, if all the particles within the system have same linear velocity

(ii) Rotational Motion 

An object is said to be in pure rotational motion, when all the points lying on the system are in circular motion about one common fixed axis.

In pure rotational motion, angular velocity of all the points is same about the fixed axis.

(iii) Rotational + Translational motion 

An object is said to be in rotational + translational motion, when the particle is rotating with some angular velocity about a movable axis.

For Example

• v = velocity of axis. 

• ω = Angular velocity of system about O.

Rotational Dynamics 

• Torque

Similar to force, the cause of rotational motion is a physical quantity called a torque/moment of force/angular force.

Torque about point O,

$\tau \overrightarrow{}=r\overrightarrow{}\times F\overrightarrow{}$

$\tau =r.Fsin\mathrm{\Theta }$

Where,

r = distance from the point O to point of application of force. 

F = force 

θ = angle between $r\overrightarrow{}andF\overrightarrow{}$

Direction of Torque: 

Direction of torque is given by right hand thumb rule. If we curl the fingers of right hand from first vector $(r\overrightarrow{})$ to the second vector $(f\overrightarrow{})$  then right-hand thumb gives us direction of their cross product, i.e., the torque.

Newton's Law in Rotation 

∑τ = Iα 

Where,  I = moment of Inertia

α = Angular Acceleration

Moment of Inertia 

Moment of inertia gives the measure of mass distribution about an axis.

$I=\sum m_{1}r{{}_1}^2$

Where $r{}_1=$ Perpendicular distance of the $i_{th}$ mass from the axis of rotation.

SI unit → kg-m2 

Gives the measure of rotational inertia and is analogous to mass in linear motion. 

• Moment of Inertia of Continuous Bodies

When the distribution of mass of a system of particle is continuous, the discrete sum $I=\sum m_{1}r{{}_1}^2$ is replaced by an integral. The moment of inertia of the whole body takes the form 

$I=\int r^{2}dm$

• Theorems on Moment of Inertia

(i) Parallel Axis Theorem: Let $I_{CM}$ be the moment of inertia of a body about an axis through its centre of mass and let $I_P$ be the moment of inertia of the same body about another axis which is parallel to the first one. If d is the distance between these two parallel axes and M is the mass of the body then according to the parallel axis theorem :

$I_P=I_{CM}+M{d}^2$

(ii) Perpendicular Axis Theorem :  Consider a planar body (i.e., a body of zero thickness) of mass M. Let X and Y axes be two mutually perpendicular lines in the plane of the body. The axes intersect at origin O.

Let $I_x$=moment of inertia of the body about X–axis.

Let $I_y$=moment of inertia of the body about Y–axis. Then the moment of inertia of the body about Z– axis (Passing through O and perpendicular to the plane of the body) is given by :  $$I_z=I_x+I_y$$

The above result is known as the perpendicular axis theorem.

• Radius of Gyration 

If M is the mass and I is the moment of inertia of a rigid body about a given axis then the radius of gyration (K) of the body about that axis is given by :

$K=\sqrt{\frac{I}{M}}$

Angular Momentum and Angular Impulse

• Angular Momentum

(i) For a particle

Angular momentum about a point (O) is given as

$L\overrightarrow{}=r\overrightarrow{}\times p\overrightarrow{}=r\overrightarrow{}\times \left(mv\overrightarrow{}\right)=m\left(r\overrightarrow{\times v\overrightarrow{}}\right)$

where $r\overrightarrow{}$ is position vector of the particle w.r.t. O and v\vec{} is velocity of particle

(ii) For a particle moving in a circle

For a particle moving in a circle of radius r with a speed v, its linear momentum is mv, magnitude of angular momentum (L) is given as :

$L=mv{r}_1=mvr$

As $\mathrm{\Theta }$ being ${90}^{\circ }$ , sin ${90}^{\circ }$ = 1

Direction of  $L\overrightarrow{}$is out of the plane of circle.

• Conservation of Angular Momentum

If $\tau {\overrightarrow{}}_{net}=0$

$\Rightarrow \frac{dL\overrightarrow{}}{dt}=0$

$\Rightarrow L\overrightarrow{}=Constant$

$\Rightarrow L{\overrightarrow{}}_f=L\overrightarrow{{}_i}$

• Angular Impulse

$J\overrightarrow{}=\int \tau \overrightarrow{}dt=\mathrm{\Delta }L\overrightarrow{}$

• Rotational Kinetic Energy

Rotational kinetic energy of the system rotating about a fixed axis

Hence rotational kinetic energy of the system $=\frac{1}{2}I{\omega }^2$

Where I = Moment of inertia about the axis.

Rolling

Rolling motion is a combination of rotation and translation. In case of rolling all point of a rigid body have same angular speed but different linear speed. 

• Pure Rolling (without Slipping)

For a rolling motion to be pure rolling the velocity of point of contact of body with platform should be equal for both rolling body and platform.

(i) General case (when surface is moving) 

$V_A=V_B$

$\Rightarrow V_{cm}-\omega R=V_b$

In terms of acceleration: $a_{cm}-\alpha R=a_b$

(ii) Special case (when VB = 0)  

$v_{cm}-\omega R=0$

$\Rightarrow V_{cm}=\omega R$

• Total KE of Rolling Body

• Rolling Motion On an Inclined Plane

(i) Pure rolling on an incline plane 

$a_R=\frac{gsin\mathrm{\Theta }}{1+\frac{I}{m{R}^2}}$

NCERT Solutions for Class 11 Physics Chapter 6 – Free PDF Download

1. Give the location of the centre of mass of a 

1. sphere, 

2. cylinder, 

3. ring, and 

4. cube, each of uniform mass density.

Does the centre of mass of a body necessarily lie inside the body?

Ans: The centre of mass (C.M.) can be defined as a point where the mass of a body is supposed to be concentrated.

For the above listed geometric shapes having a uniform mass density, the centre of mass lies at their respective geometric centres.

The centre of mass of a specific body need not necessarily lie inside of the body. For example, the centre of mass of bodies such as a ring, a hollow sphere, etc., lie outside the respective body.

2. In the $$HCl$$molecule, the separation between the nuclei of the two atoms is about $$1.27\stackrel{0}{A}$$ $$20kg$$. Find the approximate location of the CM of the molecule, given that a chlorine atom is about $$35.5$$times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Ans: The provided situation can be expressed as:

Distance between $$H$$and $$Cl$$atoms$$1.27\stackrel{0}{A}$$

Mass of $$H$$atom $$=m$$

Mass of $$Cl$$atom $$=35.5m$$

Let the centre of mass of the given system be at a distance $$x$$ from the $$Cl$$atom.

Distance between the centre of mass and the $$H$$atom $$=(1.27-x)$$

Let us suppose that the centre of mass of the given molecule lies at the origin. Therefore, it can be written as:

$$\frac{m(1.27-x)+35.5mx}{m+35.5m}=0$$

$$\Rightarrow m(1.27-x)+35.5mx=0$$

$$\Rightarrow 1.27-x=-35.5x$$

$$\Rightarrow x=\frac{-1.27}{(35.5-1)}=\text{[}-0.37\stackrel{0}{A}$$

Here, the negative sign gives an indication that the centre of mass lies at the left side of the molecule.

Therefore, the centre of mass of the $$HCl$$ molecule lies $$0.37\stackrel{0}{A}$$ from the $$Cl$$ atom.

3. A child sits stationary at one end of a long trolley moving uniformly with a speed $$v$$on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Ans: There will not be any change in the speed of the centre of mass of the given system.

The child is running arbitrarily on a trolley that is moving forward with velocity $$v$$. However, the running of the child will have no effect on the velocity of the centre of mass of the trolley. This happens because the force due to the motion of the child is purely internal. Internal forces in a body produce no effect on the motion of the bodies on which they are acting. Because there is no external force involved in the (child + trolley) system, the child’s motion will not produce any change in the speed of the centre of mass of the trolley.

4. Show that the area of the triangle contained between the vectors $$a$$ and $$b$$ is one half of the magnitude of $$a\times b$$.

Ans: Let us consider two vectors $$\overrightarrow{OK}=\left|\overrightarrow{a}\right|$$ and $$\overrightarrow{OM}=\left|\overrightarrow{b}\right|$$ , which are inclined at an angle $$\theta$$, as shown in the following figure.

In $$\mathrm{\Delta }OMN$$, we can express the relation:

$$\sin \theta =\frac{MN}{OM}=\frac{MN}{\left|\overrightarrow{b}\right|}$$

$$\Rightarrow MN=\left|\overrightarrow{b}\right|\sin \theta$$

Now, 

$$|\overrightarrow{a}\times \overrightarrow{b}|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta$$

$$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=OK\cdot MN\times \frac{2}{2}$$

$$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=2\times Areaof\mathrm{\Delta }OMK$$

$$\Rightarrow Areaof\mathrm{\Delta }OMK=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|$$

5. Show that $$a\cdot (b\times c)$$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors, $$a$$, $$b$$ and $$c$$.

Ans: A parallelepiped with origin $$O$$ and sides $$a$$, $$b$$, and $$c$$is depicted in the following figure.

Volume of the given parallelepiped $$=abc$$

And 

$$\overrightarrow{OA}=\overrightarrow{a}$$

$$\overrightarrow{OB}=\overrightarrow{b}$$

$$\overrightarrow{OC}=\overrightarrow{c}$$

Let us suppose that $$\hat{n}$$ be a unit vector perpendicular to both $\overrightarrow{b}$ and $\overrightarrow{c}$. Therefore,

$$\hat{n}$$ and $$\overrightarrow{a}$$ have the same direction.

$$\overrightarrow{b}\times \overrightarrow{c}=bc\sin \theta \hat{n}$$

$$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=bc\sin {90}^{\circ }\hat{n}$$

$$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=bc\hat{n}$$

Now,

$$\overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=a\cdot \left(bc\hat{n}\right)$$

$$\Rightarrow \overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=abc\cos \theta \hat{n}$$

$$\Rightarrow \overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=abc\cos 0^{\circ }\hat{n}$$

$$\Rightarrow \overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=abc\cos 0^{\circ }$$

$$\Rightarrow \overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=abc$$

$$\Rightarrow \overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=abc=Volumeoftheparallelepiped$$

6. Find the components along the $$x$$, $$y$$, $$z$$axes of the angular momentum $$l$$ of a particle, whose position vector is $$r$$with components $$x$$, $$y$$, $$z$$and momentum is $$p$$with components $$p_x$$, $$p_y$$ and $$p_z$$. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Ans: Linear momentum of the particle, $$\overrightarrow{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}$$

Position vector of the particle, $$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$

Angular momentum,

$$\overrightarrow{l}=\overrightarrow{r}\times \overrightarrow{p}$$

$$\Rightarrow \overrightarrow{l}=(x\hat{i}+y\hat{j}+z\hat{k})\times (p_x\hat{i}+p_y\hat{j}+p_z\hat{k})$$

$\Rightarrow \overrightarrow{I}=\left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ p_x& p_y& p_z\end{array}\right)$

Now, 

$$l_x\hat{i}+l_y\hat{j}+l_z\hat{k}=\hat{i}(y{p}_z-z{p}_y)-\hat{j}(x{p}_z-z{p}_x)+\hat{k}(x{p}_y-z{p}_x)$$

On comparing the coefficients of $$\hat{i}$$, $$\hat{j}$$and $$\hat{k}$$, we can write:

$$l_x=y{p}_z-z{p}_y$$,

$$l_y=z{p}_x-x{p}_z$$,

$$l_z=x{p}_y-y{p}_x$$ …… (1)

The particle is moving in the x-y plane. Therefore, the z-component of the position vector and linear momentum vector is becoming zero, i.e., $$z=p_z=0$$

Thus, equation (1) reduces to:

$$l_x=0$$

$$l_y=0$$

$$l_z=x{p}_y-y{p}_x$$

Hence, when the particle is subject to move in the x-y plane, the direction of angular momentum will be along the z-direction. 

7. Two particles, each of mass $$m$$and speed $$v$$, travel in opposite directions along parallel lines separated by a distance $$d$$. Show that the vector angular momentum of the two-particle system is the same whatever be the point about which the angular momentum is taken.

Ans: Let us suppose that at a certain instant two particles be at points $$P$$and $$Q$$, as shown in the given figure.

Angular momentum of the system about point $$P$$ can be given as:

$$\overrightarrow{L_P}=mv\times 0+mv\times d$$

$$\Rightarrow \overrightarrow{L_P}=mvd$$ …… (1)

Angular momentum of the system about point Q can be given as:

$$\overrightarrow{L_Q}=mv\times d+mv\times 0$$

$$\Rightarrow \overrightarrow{L_Q}=mvd$$ …… (2)

Let us consider a point $$R$$, which is at a distance $$y$$ from point $$Q$$, i.e.,

$QR=y$

$\Rightarrow PR=d-y$

Angular momentum of the system about point $$R$$ can be given as:

$$\overrightarrow{L_R}=mv\times (d-y)+mv\times y$$

$$\Rightarrow \overrightarrow{L_R}=mvd-mvy+mvy$$

$$\Rightarrow \overrightarrow{L_R}=mvd$$…… (3)

On comparing equations (1), (2), and (3), we get:

$${\overrightarrow{L}}_P={\overrightarrow{L}}_Q={\overrightarrow{L}}_R$$ …… (4)

We can hence infer from equation (4) that the angular momentum of a system is independent of the point about which it is taken. 

8. A non-uniform bar of weight $$W$$is suspended at rest by two strings of negligible weight as shown in Fig.7.39. The angles made by the strings with the vertical are $$36.9{}^{\circ }$$and $$53.1{}^{\circ }$$ respectively. The bar is $$2m$$long. Calculate the distance $$d$$of the centre of gravity of the bar from its left end.

Ans: The free body diagram of the bar can be drawn as shown in the given figure.

Length of the bar is given, $$l=2m$$

$$T_1$$and $$T_2$$ are the tensions generated in the left and right strings respectively.

At translational equilibrium, we can express:

$$T_1\sin 36.9{}^{\circ }=T_2\sin 53.1{}^{\circ }$$

$$\Rightarrow \frac{T_1}{T_2}=\frac{\sin 53.1{}^{\circ }}{\sin 36.9{}^{\circ }}$$

$$\Rightarrow \frac{T_1}{T_2}=\frac{0.800}{0.600}=\frac{4}{3}$$

$$\Rightarrow T_1=\frac{4}{3}{T}_2$$

On taking the torque about the centre of gravity, for rotational equilibrium, we can write:

$$T_1\cos 36.9{}^{\circ }\times d=T_2\cos 53.1{}^{\circ }(2-d)$$

$$\Rightarrow T_1\times 0.800\times d=T_{2}0.600(2-d)$$

$$\Rightarrow \frac{4}{3}\times T_2\times 0.800\times d=T_2(0.600\times 2-0.600d)$$

$$\Rightarrow 1.067d+0.6d=1.2$$

$$\Rightarrow d=\frac{1.2}{1.67}$$

$$\Rightarrow d=0.72m$$

Therefore, the centre of gravity of the given bar lies $$0.72m$$ from the left end of the bar.

9. A car weighs $$1800kg$$. The distance between its front and back axles is $$1.8m$$. Its centre of gravity is $$1.05m$$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

Ans: Given that,

Mass of the car is given as, $$m=1800kg$$

Distance between the front and back axles, $$d=1.8m$$

Distance between the centre of gravity and the back axle $$=1.05m$$

The different forces acting on the car are shown in the given figure:

The forces in the figure, $$R_f$$ and $$R_b$$are the forces exerted by the level ground on the front wheels and back wheels respectively.

At translational equilibrium we can write:

$$R_f+R_b=mg$$

$$\Rightarrow R_f+R_b=1800\times 9.8$$

$$\Rightarrow R_f+R_b=17640N$$ …... (1)

For rotational equilibrium, on taking the torque about the centre of gravity, we can write:

$$R_f\left(1.05\right)=R_b(1.8-1.05)$$

$$\Rightarrow R_f\times \left(1.05\right)=R_b\times \left(0.75\right)$$

$$\Rightarrow \frac{R_f}{R_b}=\frac{0.75}{1.05}=\frac{5}{7}$$

$$\Rightarrow \frac{R_b}{R_f}=\frac{7}{5}$$

$$\Rightarrow R_b=1.4R_f$$ …… (2)

Solving equations (1) and (2), we obtain:

$$1.4R_f+R_f=17640N$$

$$\Rightarrow R_f=\frac{17640}{2.4}N=7350N$$

$$\Rightarrow R_b=(17640-7350)N=10290N$$

Therefore, the force exerted on each front wheel can be given as $$\frac{7350}{2}N=3675N$$ and 

the force exerted on each back wheel can be given as $$\frac{10290}{2}N=5145N$$

10.

1. Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be $$\frac{2M{R}^2}{5}$$where $$M$$is the mass of the sphere and $$R$$is the radius of the sphere.

Ans: The moment of inertia (M.I.) of a sphere about its diameter can be given as: $$\frac{2M{R}^2}{5}$$

According to the theorem of parallel axes, the moment of inertia of a body about any axis is same as the sum of the moment of inertia of a certain body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

The moment of inertia about a tangent of the sphere can be expressed as:

$$\frac{2M{R}^2}{5}+M{R}^2=\frac{7}{5}M{R}^2$$

2. Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be $$\frac{1}{4}M{R}^2$$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Ans: The moment of inertia of a disc about its diameter can be given as:$$\frac{1}{4}M{R}^2$$.

We can infer that, according to the theorem of the perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is the same as the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The moment of inertia of the disc about its centre is $$\frac{1}{4}M{R}^2+\frac{1}{4}M{R}^2=\frac{1}{2}M{R}^2$$

The position of the perpendicular axis is shown in the following figure.

On application of the theorem of parallel axes:

The moment of inertia about an axis normal to the disc and passing through a point on its edge is $$\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$$.

11. Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Ans: Let us assume that $$m$$and $$r$$ be the respective mass and radius of the hollow cylinder and the solid sphere. The moment of inertia of the hollow cylinder about its standard axis can be given as,

$$I_1=m{r}^2$$

The moment of inertia of the solid sphere about an axis that passes through its centre can be given as,

$$I_2=\frac{2}{5}m{r}^2$$

The formula for torque in terms of angular acceleration and moment of inertia can be expressed as:

$$\tau =I\alpha$$

Where,

$$\tau =$$ Torque

$$\alpha =$$ Angular acceleration

$$I=$$Moment of inertia

For the hollow cylinder the expression can be given as,

$${\tau }_1=I_1{\alpha }_1$$

For the solid sphere the expression can be given as,

$${\tau }_2=I_2{\alpha }_2$$

As an equal amount of torque is applied to both the bodies it can be stated as, $$\frac{{\alpha }_2}{{\alpha }_1}=\frac{I_1}{I_2}=\frac{m{r}^2}{\frac{2}{5}m{r}^2}=\frac{2}{5}$$

$${\alpha }_2>{\alpha }_1$$ …… (1)

Using the relation $$\omega ={\omega }_0+\alpha t$$

Where,

$$\alpha =$$ Angular acceleration 

$$t=$$Time of rotation

$${\omega }_0=$$ Initial angular velocity

$$\omega =$$Final angular velocity

For equal $${\omega }_0$$and $$t$$, we have:

$$\omega =\alpha$$ …... (2)

From equations (1) and (2), we can conclude:

$${\omega }_2>{\omega }_1$$

Therefore, the angular velocity ($$\omega$$) of the solid sphere will be greater than that of the hollow cylinder.

12. A solid cylinder of mass $$20kg$$ rotates about its axis with angular speed $$100rad{s}^{-1}$$. The radius of the cylinder is $$0.25m$$. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Ans: Mass of the cylinder is given, $$m=20kg$$

Angular speed of the cylinder, $$\omega =100rad{s}^{-1}$$

Radius of the solid cylinder, $$r=0.25m$$

The moment of inertia of the solid cylinder can be expressed as:

$$I=\frac{1}{2}m{r}^2$$

$$\Rightarrow I=\frac{1}{2}\times 20kg\times {\left(0.25\right)}^2$$

$$\Rightarrow I=6.25kg{m}^2$$

Kinetic energy of the cylinder$$=\frac{1}{2}I{\omega }^2$$

$$\Rightarrow K.E.=\frac{1}{2}\times 6.25\times {\left(100\right)}^2$$

Angular Momentum of the cylinder,

$$L=I\omega$$

$$\Rightarrow L=6.25\times 100$$

$$\Rightarrow L=62.5Js$$

13.

1. A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $$40rev/min$$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $$\frac{2}{5}$$times the initial value? Assume that the turntable rotates without friction.

Ans: Given that,

Initial angular velocity of turntable, $${\omega }_1=40rev/min$$

Final angular velocity of the given turntable is $${\omega }_2$$

The moment of inertia of the child with stretched hands can be given as $$I_1$$

The moment of inertia of the child with folded hands can be given as $$I_2$$

The two moments of inertia are related to each other as follows:

$$I_2=\frac{2}{5}{I}_1$$

Since no external force acts on the child, the angular momentum $$L$$is not varying.

Therefore, for the two circumstances, we can write:

$$I_1{\omega }_1=I_2{\omega }_2$$

$$\Rightarrow {\omega }_2=\frac{I_1}{I_2}{\omega }_1$$

$$\Rightarrow {\omega }_2=\frac{I_1}{\frac{2}{5}{I}_1}\times 40=\frac{5}{2}\times 40$$

$$\Rightarrow {\omega }_2=100rev/min$$

2. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Ans: Given that,

Initial Kinetic energy of rotation of the turntable, $$E_i=\frac{1}{2}{I}_1{{\omega }^2}_1$$

Final kinetic energy of rotation of the turntable, $$E_f=\frac{1}{2}{I}_2{{\omega }^2}_2$$

$$\frac{E_f}{E_i}=\frac{\frac{1}{2}{I}_2{{\omega }^2}_2}{\frac{1}{2}{I}_1{{\omega }^2}_1}$$

$$\Rightarrow \frac{E_f}{E_i}=\frac{\frac{1}{2}\times \frac{2}{5}\times I_1{{\omega }^2}_2}{\frac{1}{2}{I}_1{{\omega }^2}_1}$$

$$\Rightarrow \frac{E_f}{E_i}=\frac{2}{5}\times \frac{{{\omega }^2}_2}{{{\omega }^2}_1}$$

$$\Rightarrow \frac{E_f}{E_i}=\frac{2}{5}\times \frac{{\left(100\right)}^2}{{\left(40\right)}^2}$$

$$\Rightarrow \frac{E_f}{E_i}=2.5$$

$$\Rightarrow E_f=2.5\times E_i$$

The increase in the rotational kinetic energy is related to the internal energy of the boy on the turntable.

14. A rope of negligible mass is wound round a hollow cylinder of mass $$3kg$$ and radius $$40cm$$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $$30N$$? What is the linear acceleration of the rope? Assume that there is no slipping.

Ans: Given that,

Mass of the hollow cylinder is given as, $$m=3kg$$

Radius of the hollow cylinder is given as, $$r=40cm=0.4m$$

Applied force on the given rope is given as, $$F=30N$$

The moment of inertia of the hollow cylinder about its geometric axis can be given as:

$I=m{r}^2$ 

$\Rightarrow I=3\times {\left(0.4\right)}^2$ 

$\Rightarrow I=0.48kg{m}^2$

Torque acting on the rope,

$$\tau =F\times r$$

$\Rightarrow \tau =30\times 0.4$ 

$\Rightarrow \tau =12Nm$

For angular acceleration $$\alpha$$, torque can also be given by the expression:

$$\tau =I\alpha$$

$$\Rightarrow \alpha =\frac{\tau }{I}=\frac{12}{0.48}$$

$$\Rightarrow \alpha =25rad{s}^{-2}$$

Linear acceleration of the rope can be stated as $$=ra=0.4\times 25=10m{s}^{-2}$$.

15. To maintain a rotor at a uniform angular speed of $$200rad{s}^{-1}$$ an engine needs to transmit a torque of $$180Nm$$. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is $$100$$efficient.

Ans: Given that,

Angular speed of the rotor is given as, $$200rad/s$$

Torque required by the rotor of the engine is given as $$180Nm$$.

The power of the rotor $$\left(P\right)$$ can be expressed in the relation of torque and angular speed by the formula $$P=\tau \omega$$

$$\Rightarrow P=180\times 200=30\times {10}^3$$

$$\Rightarrow P=36kW$$

Therefore, the power required by the engine is $$36kW$$.

16. From a uniform disk of radius $$R$$, a circular hole of radius $$\frac{R}{2}$$ is cut out. The centre of the hole is at $$\frac{R}{2}$$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Ans: Given that,

Mass per unit area of the original disc can be given as $$\sigma$$.

Radius of the original disc $$=R$$

Mass of the original disc,

$$M=\pi R^2\sigma$$

The disc with the cut portion is shown in the given figure:

Radius of the smaller disc is given $$=\frac{R}{2}$$

Mass of the smaller disc is given as $$M^{\prime }=\pi {\left(\frac{R}{2}\right)}^2\sigma =\frac{1}{4}\pi R^2\sigma =\frac{M}{4}$$

Let us suppose that $$O$$ and $$O^{\prime }$$  be the respective centres of the original disc and the disc cut off from the original. As per definition of the centre of mass, the centre of mass of the original disc is assumed to be concentrated at $$O$$, while that of the smaller disc is assumed to be concentrated at $$O^{\prime }$$.

It is provided that:

$$O{O}^{\prime }=\frac{R}{2}$$

After the smaller disc has been cut from the original disc, the remaining portion left over after cutting is considered to be a system of two masses. The two masses can be expressed as:

$$M\left(concentratedatO\right)-M^{\prime }=\left(\frac{M}{4}\right)concentratedat{O}^{\prime }$$

(The negative sign in the above statement indicates that this portion has been removed from the original disc.)

Let us suppose that $$x$$be the distance through which the centre of mass of the remaining portion shifts from point $$O$$.

The relation between the centres of masses of two masses is given as:

$$x=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$$

For the given system, it can be written as:

$$\Rightarrow x=\frac{M\times 0-M^{\prime }\times \left(\frac{R}{2}\right)}{M+(-M^{\prime })}$$

$$\Rightarrow x=\frac{\frac{-M}{4}\times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-MR}{8}\times \frac{4}{3M}=\frac{-R}{6}$$

(The negative sign in the above statement indicates that the centre of mass gets shifted toward the left of point $$O$$.)

The centre of gravity of the resulting flat body can be located from the original centre of the body and opposite to the centre of the cut portion.

17. A meter stick is balanced on a knife edge at its centre. When two coins, each of mass $$5g$$ are put one on top of the other at the $$12cm$$ mark, the stick is found to be balanced at $$45cm$$. What is the mass of the meter stick?

Ans: Let us assume that $$W$$ and $$W^{\prime }$$ be the respective weights of the meter stick and the coin.

The mass of the meter stick is supposed to be concentrated at its mid-point, i.e., at the $$50cm$$ mark.

Mass of the meter stick is $$m^{\prime }$$

Mass of each coin is $$m=5g$$

When the coins are placed $$12cm$$ away from the end $$P$$, the centre of mass gets shifted by $$5cm$$ from point $$R$$ toward the end $$P$$. The centre of mass is located at a distance of $$45cm$$ from point $$P$$.

The net torque will be thus, conserved for rotational equilibrium about point $$R$$.

This can be expressed by the equation,

$$10\times g(45-12)-m^{\prime }g(50-45)=0$$

$$\Rightarrow m^{\prime }=\frac{10\times 33}{5}=66g$$

Therefore, the mass of the meter stick is $$66g$$.

18. A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. 

1. Will it reach the bottom with the same speed in each case? 

Ans: Mass of the sphere $$=m$$

Height of the plane $$=h$$

Velocity of the sphere at the bottom of the plane is given as $$=v$$

At the top of the plane, the total energy of the sphere i.e., Potential energy $$(P.E.)=mgh$$

At the bottom of the plane, the sphere has both translational and rotational kinetic energies which can be expressed as,

Therefore, total energy $$T.E.=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

Using the law of conservation of energy, we can state that:

$$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$$ …… (1)

For a solid sphere, the moment of inertia about its centre can be given as,

$$I=\frac{2}{5}m{r}^2$$

Therefore, equation (1) becomes:

$$\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2=mgh$$

$$\Rightarrow \frac{1}{2}{v}^2+\left(\frac{1}{5}{r}^2\right){\omega }^2=gh$$

But we have the formula,

$$v=r\omega$$

$$\frac{1}{2}{v}^2+\frac{1}{5}{v}^2=gh$$

$$\Rightarrow v^2\left(\frac{7}{10}\right)=gh$$

$$\Rightarrow v=\sqrt{\frac{10}{7}gh}$$

Therefore, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both values are constants and do not change. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

2. Will it take longer to roll down one plane than the other? 

Ans: Let us consider two inclined planes with inclinations $${\theta }_1$$and  $${\theta }_2$$ respectively related as:

$${\theta }_1<{\theta }_2$$

The acceleration generated in the sphere when it rolls down the plane inclined at $${\theta }_1$$is:

$$g\sin {\theta }_1$$

The different forces acting on the sphere are shown in the given figure.

$$R_1$$ is the normal reaction to the sphere as shown in the above figure.

Similarly, the acceleration generated in the sphere when it rolls down the plane inclined at $${\theta }_2$$ is:

$$g\sin {\theta }_2$$

The different forces that act on the sphere are shown in the given figure.

$$R_2$$ is the normal reaction to the sphere as given in the figure.

$${\theta }_1<{\theta }_2$$, $$\sin {\theta }_2>\sin {\theta }_1$$ …… (1)

$$a_2>a_1$$ …… (2)

Initial velocity of sphere, $$u=0$$

Final velocity of sphere, $$v=$$ constant

Now, by using the first equation of motion, we can obtain the time of roll as:

$$v=u+at$$

$$t\propto \frac{1}{a}$$

For inclination of angle $${\theta }_1$$:

$$t_1\propto \frac{1}{a_1}$$

For inclination of angle $${\theta }_2$$:

$$t_2\propto \frac{1}{a_2}$$ …… (3)

3. If so, which one and why?

Ans: From equations (2) and (3), we obtain:

$$t_2<t_1$$

Therefore, conclude that the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.

19. A hoop of radius $$2m$$ weighs $$100kg$$. It rolls along a horizontal floor so that its centre of mass has a speed of $$20cm/s$$. How much work has to be done to stop it?

Ans: Radius of the hoop is given as, $$r=2m$$ 

Mass of the hoop is, $$m=100kg$$

Velocity of the hoop is,

$$v=20cm/s=0.2m/s$$

Total energy of the hoop $$=$$Translational KE $$+$$Rotational KE

$$E_r=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$$

Moment of inertia of the hoop about its centre can be given as $$I=m{r}^2$$

$$E_r=\frac{1}{2}m{v}^2+\frac{1}{2}\left(m{r}^2\right){\omega }^2$$

But we have the formula,  $$v=r\omega$$

$$\Rightarrow E_1=\frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2{\omega }^2$$

$$\Rightarrow E_1=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$$

The work needed to be done for halting the hoop is the same as the total energy of the hoop.

Hence, required work to be done can be given as,

$$W=m{v}^2=100\times {\left(0.2\right)}^2=4J$$

20. The oxygen molecule has a mass of $$5.30\times {10}^{26}kg$$ and a moment of inertia of $$1.94\times {10}^{-46}kg{m}^2$$about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $$500m/s$$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Ans: Given that,

Mass of an oxygen molecule is given as, $$m=5.30\times {10}^{26}kg$$

Moment of inertia of oxygen molecule is given as, $$I=1.94\times {10}^{-46}kg{m}^2$$

Velocity of the oxygen molecule is given as, $$v=500m/s$$

Let the separation between the two atoms of the oxygen molecule be $$2r$$

Mass of each oxygen atom in the oxygen molecule $$=\frac{m}{2}$$

Therefore, moment of inertia $$I$$, can be calculated as:

$$\left(\frac{m}{2}\right)r^2+\left(\frac{m}{2}\right)r^2=m{r}^2$$

$$r=\sqrt{\frac{l}{m}}$$

$$\Rightarrow \sqrt{\frac{1.94\times {10}^{-46}}{5.36\times {10}^{26}}}=0.60\times {10}^{-10}m$$

It is provided that:

$$K{E}_{rot}=\frac{2}{3}K{E}_{trans}$$

$$\Rightarrow \frac{1}{2}I{\omega }^2=\frac{2}{3}\times \frac{1}{2}m{v}^2$$

$$\Rightarrow m{r}^2{\omega }^2=\frac{2}{3}m{v}^2$$

$$\Rightarrow \omega =\sqrt{\frac{2}{3}}\times \frac{v}{r}$$

$$\Rightarrow \omega =\sqrt{\frac{2}{3}}\times \frac{500}{0.6\times {10}^{-10}}$$

$$\Rightarrow \omega =6.80\times {10}^{12}rad/s$$, which is the required average angular velocity.

21. A solid cylinder rolls up an inclined plane of angle of inclination

$$30{}^{\circ }$$. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of $$5m/s$$. How far will the cylinder go up the plane?

How long will it take to return to the bottom?

Ans:  A solid cylinder rolling up an inclination is pictured in the following figure.

Initial velocity of the solid cylinder on the inclined plane, $$v=5m/s$$

Angle of inclination is $$30{}^{\circ }$$

Height reached by the cylinder on the inclined plane $$=h$$

Energy of the cylinder on the inclined plane at point $$A$$:

$$K{E}_{rot}=K{E}_{trans}$$

$$\Rightarrow \frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2$$

Energy of the cylinder at point $$B$$ $$=mgh$$

Let us use the law of conservation of energy, we can express:

$$\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2=mgh$$

Moment of inertia of the solid cylinder is $$I=\frac{1}{2}m{r}^2$$

$$\Rightarrow \frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\omega }^2+\frac{1}{2}m{v}^2=mgh$$

$$\Rightarrow \frac{1}{4}m{r}^2{\omega }^2+\frac{1}{2}m{v}^2=mgh$$

$$\Rightarrow \frac{1}{4}{r}^2{\omega }^2+\frac{1}{2}{v}^2=gh$$

But we have the expression, $$v=r\omega$$

$$\Rightarrow \frac{1}{4}{v}^2+\frac{1}{2}{v}^2=gh$$

$$\Rightarrow \frac{3}{4}{v}^2=gh$$

$$\Rightarrow h=\frac{3}{4}\times \frac{v^2}{g}$$

$$\Rightarrow h=\frac{3}{4}\times \frac{5\times 5}{9.8}=1.91m$$

In $$\mathrm{\Delta }ABC$$, 

$$\sin \theta =\frac{BC}{AB}$$

$$\Rightarrow \sin 30{}^{\circ }=\frac{h}{AB}$$

$$AB=\frac{1.91}{0.5}=3.82m$$

Therefore, the cylinder will move $$3.82m$$ up the inclined plane.

For radius of gyration $$K$$, the velocity of the cylinder at the instance when it rolls back to the bottom is given by the formula:

$$v={\left(\frac{2gh}{1+\frac{K^2}{R^2}}\right)}^{\frac{1}{2}}$$

$$\Rightarrow v={\left(\frac{2gAB\sin \theta }{1+\frac{K^2}{R^2}}\right)}^{\frac{1}{2}}$$

For the solid cylinder we can write $$K^2=\frac{R^2}{2}$$ 

$$\Rightarrow v={\left(\frac{2gAB\sin \theta }{1+\frac{1}{2}}\right)}^{\frac{1}{2}}$$