Important Questions for CBSE Class 11 Physics Chapter 6 - Systems of Particles and Rotational Motion

Very short answer questions (1 Mark)

1. A wheel $0.5m$ in radius is moving with a speed of $12m/s$ . Find its angular speed.

Ans: Angular speed can be given as $v=r\omega$ 

$\Rightarrow \omega ={\displaystyle \frac{v}{r}}={\displaystyle \frac{12}{0.5}}$ 

$\Rightarrow \omega =24rad/s$ 

2. State the condition for translational equilibrium of a body.

Ans: For a body to be in a translational equilibrium, the vector sum of all the forces acting on the body must be equal to zero.

3. How is angular momentum related to linear momentum?

Ans: Angular momentum can be related as $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$ 

Or $L=rp\sin \theta$ 

Where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{p}$ .

4. What is the position of the centre of mass of a uniform triangular lamina?

Ans: The centre of mass of a uniform triangular lamina is at the centroid of the triangular lamina.

5. What is the moment of inertia of a sphere of mass $20kg$ and radius ${\displaystyle \frac{1}{4}}m$ about its diameter?

Ans: $I={\displaystyle \frac{2}{5}}M{R}^2$ 

$I={\displaystyle \frac{2}{5}}\times 20\times {\left({\displaystyle \frac{1}{4}}\right)}^2$ 

$I=0.5kg{m}^2$ 

6. What are the factors on which moment of inertia of a body depends?

Ans: Moment of inertia of a body depends on: 

• Mass of the body

• Shape and size of the body

• Position of the axis of rotation

7. Two particles in an isolated system undergo head on collision. What is the acceleration of the centre of mass of the system?

Ans: Acceleration of the centre of mass of the system is zero.

8. Which component of a force does not contribute towards torque?

Ans: The component of a force that does not contribute towards torque is the radial component.

9. What is the position of centre of mass of a rectangular lamina?

Ans: The point of intersection of diagonal is the position of the centre of mass of a rectangular lamina.

10. 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., lies outside the respective body.

11. 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 of the force due to the motion of 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.

12. 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: 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 $36kW$ .

Short Answer Questions (2 Marks)

1. A planet revolves around on massive star in a highly elliptical orbit is its angular momentum constant over the entire orbit. Give reason?

Ans: When a planet revolves around the star, it takes place under the effect of gravitational force. The force is radial and does not contribute towards torque. Hence, in the absence of an external torque angular momentum of the planet remains constant.

2. Obtain the equation $\omega ={\omega }_0+\alpha t$ .

Ans: We know the relation,

$\alpha ={\displaystyle \frac{d\omega }{dt}}$ 

$d\omega =\alpha dt$ 

Integrating within the limits

${\int }_{{\omega }_0}^{\omega }d\omega ={\int }_0^t\alpha dt$ 

$\Rightarrow \omega -{\omega }_0=\alpha t$ 

$\Rightarrow \omega ={\omega }_0+\alpha t$ 

Hence, proved.

3. What is the torque of the force $\overrightarrow{F}=2\hat{i}-5\hat{j}+4\hat{k}$ acting at the point $\overrightarrow{r}=(3\hat{i}+3\hat{j}+3\hat{k})m$ about the origin?

Ans: Torque is given by

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

$\Rightarrow \overrightarrow{\tau }=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 4\\ 3& 2& 3\end{array}\right]$ 

$\Rightarrow \overrightarrow{\tau }=(17\hat{i}-6\hat{j}-13\hat{k})Nm$ 

4. What is the value of linear velocity if $\overrightarrow{\omega }=(3\hat{i}-4\hat{j}+\hat{k})$ and $\overrightarrow{r}=(5\hat{i}-6\hat{j}+6\hat{k})$ .

Ans: Linear velocity is given by

$\overrightarrow{\upsilon }=\overrightarrow{\omega }\times \overrightarrow{r}$ 

$\Rightarrow \overrightarrow{\upsilon }=(3\hat{i}-4\hat{j}+\hat{k})\times (5\hat{i}-6\hat{j}+6\hat{k})$ 

$\Rightarrow \overrightarrow{\upsilon }=-18\hat{i}-13\hat{j}+2\hat{k}$ 

5. Establish the third equation of rotational motion ${\omega }^2-{{\omega }_0}^2=2\alpha \theta$ .

Ans: We have the relation, $\alpha ={\displaystyle \frac{d\omega }{dt}}$ 

Multiply and divide by $d\theta$ 

$\alpha ={\displaystyle \frac{d\omega }{dt}}\times {\displaystyle \frac{d\theta }{d\theta }}$ 

$\Rightarrow \alpha ={\displaystyle \frac{d\omega }{d\theta }}\times {\displaystyle \frac{d\theta }{dt}}$ 

$\Rightarrow \alpha d\theta =\omega d\omega$ 

Integrating we get

${\int }_0^{\theta }\alpha d\theta ={\int }_{{\omega }_0}^{\omega }\omega d\omega$ 

$\Rightarrow \alpha {\left[\theta \right]}_0^{\theta }={\left[{\displaystyle \frac{{\omega }^2}{2}}\right]}_{{\omega }_0}^{\omega }$ 

$\Rightarrow \alpha \theta ={\displaystyle \frac{{\omega }^2}{2}}-{\displaystyle \frac{{{\omega }_0}^2}{2}}$ 

$\Rightarrow {\omega }^2-{{\omega }_0}^2=2\alpha \theta$ 

Hence, proved.

6. Find the expression for radius of gyration of a solid sphere about one of its diameters.

Ans: Moment of inertia of a solid sphere about its diameter can be given as,

$={\displaystyle \frac{2}{5}}M{R}^2$ 

$K=$ Radius of Gyration

$I=M{K}^2={\displaystyle \frac{2}{5}}M{R}^2$ 

$\Rightarrow K^2={\displaystyle \frac{2}{5}}{R}^2$ 

$\Rightarrow K=\sqrt{{\displaystyle \frac{2}{5}}}R$ 

7. Prove that the centre of mass of two particles divides the line joining the particles in the inverse ratio of their masses?

Ans: We can express the relation as ${\overrightarrow{r}}_{cm}={\displaystyle \frac{m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2}{m_1+m_2}}$ 

If centre of mass is at the origin,

$\Rightarrow {\overrightarrow{r}}_{cm}=0$ 

$\Rightarrow m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2=0$ 

$m_1{\overrightarrow{r}}_1=-m_2{\overrightarrow{r}}_2$ 

In terms of magnitude, it can be expressed as,

$m_1\left|{\overrightarrow{r}}_1\right|=m_2\left|{\overrightarrow{r}}_2\right|$ 

$\Rightarrow {\displaystyle \frac{m_1}{m_2}}={\displaystyle \frac{r_2}{r_1}}$ 

8. Show that cross product of two parallel vectors is zero.

Ans: $\overrightarrow{A}\times \overrightarrow{B}=AB\sin \theta \hat{x}$ 

 If $\overrightarrow{A}$ and $\overrightarrow{B}$ are parallel to each other,

Then, $\theta =0^{\circ }$ 

$\Rightarrow \overrightarrow{A}\times \overrightarrow{B}=0$ 

9. Prove the relation $\overrightarrow{\tau }={\displaystyle \frac{d\overrightarrow{L}}{dt}}$ .

Ans: We are aware of the relation, $\overrightarrow{L}=I\overrightarrow{\omega }$ 

Differentiating on both sides with respect to time

${\displaystyle \frac{d\overrightarrow{L}}{dt}}={\displaystyle \frac{d}{dt}}\left(I\overrightarrow{\omega }\right)=I{\displaystyle \frac{d\overrightarrow{\omega }}{dt}}=I\overrightarrow{\alpha }$ …… (1)

Where, $\overrightarrow{L}=$ Angular Momentum

$\overrightarrow{\tau }=$ Torque

And ${\displaystyle \frac{d\overrightarrow{\omega }}{dt}}=\overrightarrow{\alpha }$ 

$\Rightarrow \overrightarrow{\tau }=I\overrightarrow{\alpha }$ …… (2)

From (1) and (2)

$\overrightarrow{\tau }={\displaystyle \frac{d\overrightarrow{L}}{dt}}$ 

Hence, proved.

10. Show that for an isolated system the centre of mass moves with uniform velocity along a straight line path.

Ans: For the given conditions let $\overrightarrow{M}$ be the total mass concentrated at centre of mass whose position vector is $\overrightarrow{r}$ .

$\overrightarrow{F}={\displaystyle \frac{M{d}^2\overrightarrow{r}}{d{t}^2}}$ 

$\Rightarrow \overrightarrow{F}={\displaystyle \frac{Md}{dt}}\left({\displaystyle \frac{d\overrightarrow{r}}{dt}}\right)={\displaystyle \frac{Md}{dt}}\left({\overrightarrow{V}}_{cm}\right)$ 

For an isolated system it can be given as, $\overrightarrow{F}=0$ 

$\Rightarrow {\displaystyle \frac{Md}{dt}}\left({\overrightarrow{V}}_{cm}\right)=0$ 

Or ${\displaystyle \frac{d}{dt}}\left({\overrightarrow{V}}_{cm}\right)=0$ since $M\ne 0$ 

$\Rightarrow \left({\overrightarrow{V}}_{cm}\right)=constant$ 

11. The angle $\theta$ covered by a body in rotational motion is given by the equation $\theta =6t+5t^2+2t^3$ . Determine the value of instantaneous angular velocity and angular acceleration.

at time $t=2s$ .

Ans: In the question it is given that $\theta =6t+5t^2+2t^3$ .

Now, Angular velocity $\omega ={\displaystyle \frac{d\theta }{dt}}=6+10t+6t^2$ 

At $t=2s$ 

$\omega =6+10\left(2\right)+6{\left(2\right)}^2$ 

$\Rightarrow \omega =50rad/\sec$ 

Again, angular acceleration can be given as,

$\alpha ={\displaystyle \frac{d\omega }{dt}}={\displaystyle \frac{d^2\theta }{d{t}^2}}=10+12t$ 

At $t=2s$ 

$\alpha =10+12\left(2\right)$ 

$\alpha =34rad/s^2$ 

12. A solid cylinder of mass $20kg$ rotates about its axis with angular speed $100rad/\sec$ . 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 as, $m=20kg$ 

Angular speed is given as, $\omega =100rad/s$ 

Radius of the cylinder, $r=0.25m$ 

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

$I={\displaystyle \frac{m{r}^2}{2}}$ 

Now,

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

$\Rightarrow I=0.625kg{m}^2$ 

Kinetic energy can be given as: $K.E.={\displaystyle \frac{1}{2}}I{\omega }^2$ 

$\Rightarrow K.E.={\displaystyle \frac{1}{2}}\times 6.25\times {\left(100\right)}^2=3125J$ 

Angular momentum can be given as, 

$L=I\omega$ 

$\Rightarrow L=6.25\times 100$ 

$\Rightarrow L=62.5Js$ 

13. 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:

$\tau =I\alpha$ 

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

$\Rightarrow \alpha =25rad{s}^{-2}$ 

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

14. A bullet of mass $10g$ and speed $500m/s$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0m$ wide and weighs $12kg$ . It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ${\displaystyle \frac{M{L}^2}{3}}$ .) 

Ans: Given that,

Mass of the bullet is given as, $m=10g=10\times {10}^{-3}kg$ 

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

Width of the door, $L=1.0m$ 

Radius of the door, $r={\displaystyle \frac{1}{2}}m$ 

Mass of the door is given, $M=12kg$ 

Angular momentum transmitted by the bullet on the door:

$\alpha =mvr$ 

$\Rightarrow \alpha =(100\times {10}^{-3})\times \left(500\right)\times {\displaystyle \frac{1}{2}}=2.5kg{m}^2{s}^{-1}$ 

Moment of inertia of the door can be given as:

$I={\displaystyle \frac{M{L}^2}{3}}$ 

$\Rightarrow I={\displaystyle \frac{1}{3}}\times 12\times {\left(1\right)}^2=4kg{m}^2$ 

But we have the relation,

$\alpha =I\omega$ 

$\Rightarrow \omega ={\displaystyle \frac{\alpha }{I}}={\displaystyle \frac{2.5}{4}}=0.625rad{s}^{-1}$ , which is the required angular speed.

15. Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

(a). Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins. 

Ans: To roll the given disc, some torque is necessary. As per the definition of torque, the rotating force must be tangential to the disc. Since the frictional force at point $B$ is along the tangential force at point $A$ , a frictional force is necessary for making the disc roll.

Force of friction will act in the opposite direction to the direction of velocity at point $B$ . The direction of linear velocity at point $B$ can be pointed tangentially leftward. Therefore, frictional force will act tangentially rightward. The frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.

(b). What is the force of friction after perfect rolling begins?

Ans: Since frictional force will act opposite to the direction of velocity at point $B$ , perfect rolling will start when the velocity at that point becomes equal to zero. This will make the frictional force that acts on the disc as zero.

Long Answer Questions (3 Marks)

1. The moment of inertia of a solid sphere about a tangent is ${\displaystyle \frac{7}{5}}M{R}^2$ . Find the moment of inertia about a diameter.

Ans: From the diagram we can infer that a tangent $KCL$ is drawn at point $C$ of a solid sphere of mass $M$ and radius $R$ . Now, draw a diameter $AOB\parallel KCL$ .

Then according to Theorem of parallel axis, $I_1=I+M{\left(OC\right)}^2$ 

M.I about the tangent can be given as $I_1={\displaystyle \frac{7}{5}}M{R}^2$ 

$I=I_1-M{\left(OC\right)}^2$ 

$\Rightarrow I={\displaystyle \frac{7}{5}}M{R}^2-M{R}^2$ 

$\Rightarrow I={\displaystyle \frac{2}{5}}M{R}^2$ , which is the moment of inertia about a diameter.

2. Four particles of mass $1kg$ , $2kg$ , $3kg$ and $4kg$ are placed at the four vertices $A$ , $B$ , $C$ and $D$ of square of side $1m$ . Find the position of centre of mass of the particle.

Ans: From the given data we can infer,

$m_1=1kg$ at $(x_1,y_1)=(0,0)$ 

$m_2=2kg$ at $(x_2,y_2)=(1,0)$ 

$m_3=3kg$ at $(x_3,y_3)=(1,1)$ 

$m_4=4kg$ at $(x_4,y_4)=(0,1)$ 

Now, 

$X_{cm}={\displaystyle \frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}}$ 

$\Rightarrow X_{cm}=0.5m$ 

$Y_{cm}={\displaystyle \frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}}$ 

$\Rightarrow Y_{cm}=0.7m$ 

Therefore, the centre of mass is located at $(0.5m,0.7m)$ .

3. A circular ring of diameter $40cm$ and mass $1kg$ is rotating about an axis normal to its plane and passing through the centre with a frequency of $10$ rotations per second. Calculate the angular momentum about its axis of rotation.

Ans: From the question we can write,

$R={\displaystyle \frac{40}{2}}cm=20cm=0.2m$ 

$M=1kg$ 

$v=10rotations/\sec$ 

Now, Moment of inertia can be calculated as,

$M.I.=M{R}^2=1\times {\left(0.2\right)}^2=0.04kg{m}^2$ 

$\omega =2\pi v=2\pi \times 10=20\pi rad/s$ 

Angular momentum can be calculated as,

$\Rightarrow L=0.04\times 20\pi$ 

$\Rightarrow L=2.51kg{m}^2/s$ 

4. 

(a). Which physical quantities are represented by the

(i) Rate of change of angular momentum?

Ans: Torque i.e., $\tau ={\displaystyle \frac{d\overrightarrow{L}}{dt}}$ 

(ii) Product of $I$ and $\overrightarrow{\omega }$ ?

Ans: Angular momentum i.e., $L=I\omega$ 

(b). Show that angular momentum of a satellite of mass $M_s$ revolving around the earth having mass $M_e$ in an orbit of radius $r$ is equal to

Ans: Given that

Mass of satellite is given as $=M_s$ 

Mass of earth is given as $=M_e$ 

Radius of satellite is given as $=r$ 

Required centripetal force can be given as $={\displaystyle \frac{M_s{v}^2}{r}}$ …… (1)

Where $v$ is the orbital velocity with which the satellite revolves around the earth.

Now, Gravitational force between the satellite and the earth can be given as:

$={\displaystyle \frac{G{M}_e{M}_s}{r^2}}$ …… (2)

Equating equations (1) and (2)

${\displaystyle \frac{M_s{v}^2}{r}}={\displaystyle \frac{G{M}_e{M}_s}{r^2}}$ 

$\Rightarrow v=\sqrt{{\displaystyle \frac{G{M}_e}{r}}}$ 

Now angular momentum of the satellite can be given as,

$L=M_{s}vr$ 

$\Rightarrow L=M_s\times \sqrt{{\displaystyle \frac{G{M}_e}{r}}}\times r$ 

$\Rightarrow L=\sqrt{G{M}_e{M_s}^{2}r}$ 

Hence, proved

5. In the $HCl$ molecule, the separation between the nuclei of the two atoms is about $1.27\stackrel{o}{A}$ $20kg$ . 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 $=1.27\stackrel{o}{A}$ 

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:

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

$\Rightarrow m(1.27-x)+35.5mx=0$ 

$\Rightarrow 1.27-x=-35.5x$ 

$\Rightarrow x={\displaystyle \frac{-1.27}{(35.5-1)}}=-0.037\stackrel{o}{A}$ 

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 $0.037\stackrel{o}{A}$ from the $Cl$ atom.

6. 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:

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

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

Now, 

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

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

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

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

7. 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,

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

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

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

Long Answer Questions (4 Marks)

1. 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 $n$ be a unit vector perpendicular to both $b$ and $c$ . Therefore,

$n$ and $a$ have the same direction.

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

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

$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=bcn$ 

Now,

$\overrightarrow{a}(\overrightarrow{b}\times \overrightarrow{c})=a\cdot \left(bcn\right)$ 

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

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

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

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

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

2. 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: Given that,

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

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

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

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

But we have the formula, $v=r\omega$ 

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

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

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

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

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

3. 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 $={\displaystyle \frac{m}{2}}$ 

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

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

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

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

It is provided that:

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

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

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

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

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

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

4. A man stands on a rotating platform, with his arms stretched horizontally holding a $5kg$ weight in each hand. The angular speed of the platform is $30$ revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from $90cm$ to $20cm$ . The moment of inertia of the man together with the platform may be taken to be constant and equal to $7.6kg{m}^2$ .

(a). What is his new angular speed? (Neglect friction.)

Ans: Moment of inertia of the man-platform system is given as, 

$7.6kg{m}^2$ 

Moment of inertia when the man stretches his hands to a distance of 90 cm:

$M{I}_{HandsStretched}=2\times m{r}^2$ 

$\Rightarrow M{I}_{HandsStretched}=2\times 5\times {\left(0.9\right)}^2$ 

$\Rightarrow M{I}_{HandsStretched}=8.1kg{m}^2$ 

Initial moment of inertia of the system can be given as,

$I_i=7.6+8.1=15.7kg{m}^2$ 

Angular speed can be expressed as,

${\omega }_1=300rev/min$ 

Angular momentum can be given as,

$L_i=I_i{\omega }_i=15.7\times 30$ …… (1)

Moment of inertia when the man folds his hands to a distance of 20 cm becomes:

$M{I}_{Handsat20cm}=2m{r}^2$ 

$\Rightarrow M{I}_{Handsat20cm}=2\times 5{\left(0.2\right)}^2=0.4kg{m}^2$ 

Final moment of inertia is given as,

$I_f=7.6+0.4=8kg{m}^2$ 

Final angular speed can be given as, ${\omega }_f$ 

Final angular momentum can be expressed as,

$L_f=I_f{\omega }_f=0.79\omega$ …… (2)

From the conservation of angular momentum, we can write:

$I_i{\omega }_i=I_f{\omega }_f$ 

${\omega }_f={\displaystyle \frac{15.7\times 30}{8}}=58.88rev/min$ 

(b). Is kinetic energy conserved in the process? If not, from where does the change come about?

Ans: Kinetic energy is not conserved in the mentioned process. With the decrease in the moment of inertia, there is an increase in kinetic energy. The additional kinetic energy is generated from the work done by the man to fold his hands toward himself.

5. Read each statement below carefully, and state, with reasons, if it is true or false;

(a). During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

Ans: False

Frictional force acts in the opposite direction of motion of the centre of mass of a body. In the case of rolling, the direct point of motion of the centre of mass is in backward direction. Therefore, frictional force acts in the forward direction.

(b). The instantaneous speed of the point of contact during rolling is zero.

Ans: True

Rolling can be considered as the rotation of a body about an axis that passes through the point of contact of the body with the ground. Therefore, its instantaneous speed is zero.

(c). The instantaneous acceleration of the point of contact during rolling is zero.

Ans: False

When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.

(d). For perfect rolling motion, work done against friction is zero.

Ans: True

When perfect rolling begins, the frictional force that acts at the lowermost point becomes zero. Therefore, the work done against friction is also zero.

(e). A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Ans: True.

The rolling of a body occurs when a frictional force will act between the body and the surface. This frictional force will give the torque necessary for rolling. When the frictional force is not present, the body slips from the inclined plane under the effect of its own weight.

6. 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 on the point about which it is taken.

Very Long Answer Questions (5 Marks)

1.

(a). Why is moment of inertia called rotational inertia?

Ans: Rotational inertia measures the moment of inertia during its rotational motion. Clearly, the moment of inertia is called rotational inertia.

(b). Calculate M.I of a uniform circular disc of mass $500gm$ and radius $10cm$ about 

(i) Diameter 

Ans: Moment of inertia can be given as:

$I_d={\displaystyle \frac{1}{4}}M{R}^2$ 

$\Rightarrow I_d={\displaystyle \frac{1}{4}}\times 500\times {10}^2=12500gmc{m}^2$ 

(ii) axis tangent to the disc and parallel to diameter

Ans: Moment of inertia can be given as:

${I_d}^{\prime }=I_d+M{R}^2$ 

$\Rightarrow {I_d}^{\prime }={\displaystyle \frac{1}{4}}M{R}^2+M{R}^2$ 

$\Rightarrow {I_d}^{\prime }={\displaystyle \frac{5}{4}}M{R}^2$ 

$\Rightarrow {I_d}^{\prime }={\displaystyle \frac{5}{4}}\times 500\times {10}^2$ 

$\Rightarrow {I_d}^{\prime }=62500gmc{m}^2$ 

(c). Axis passing through centre and perpendicular to its plane?

Ans: Moment of inertia can be calculated:

$I={\displaystyle \frac{1}{2}}M{R}^2$ 

$\Rightarrow I={\displaystyle \frac{1}{2}}\times 500\times {\left(10\right)}^2$ 

$\Rightarrow I=25000gmc{m}^2$ 

2.

(a). A cat is able to land on its feet after a fall. Why?

Ans: When cat lands to the ground, it stretches its tail as a result of which, the moment of inertia increases.

Also, $I\omega =cons\tan t$ 

Angular speed will be small because increase in moment of inertia and thus, the cat is able to land on its feet without any harm.

(b). If angular momentum moment of inertia is decreased, will its rotational be also conserved? Explain.

Ans: The moment of inertia of a system decreases from $I$ to $I^{\prime }$ .

Then the angular speed will increase from $\omega$ to ${\omega }^{\prime }$ .

We can write,

$\Rightarrow I\omega =I^{\prime }{\omega }^{\prime }$ 

Since, $(I\omega =constant)$ 

${\omega }^{\prime }={\displaystyle \frac{I\omega }{I^{\prime }}}$ 

K.E. of rotation of the system can be given as

$K.E.={\displaystyle \frac{1}{2}}{I}^{\prime }{{\omega }^{\prime }}^2$ 

$\Rightarrow K.E.={\displaystyle \frac{1}{2}}{I}^{\prime }{\left({\displaystyle \frac{I\omega }{I^{\prime }}}\right)}^2$ 

$\Rightarrow K.E.={\displaystyle \frac{1}{2}}{\displaystyle \frac{I^2{\omega }^2}{I^{\prime }}}$ 

Therefore, because $I^{\prime }<I$ the K.E of the system will increase. Hence, the kinetic energy will not be conserved.

3. 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,

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

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

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

Now, 

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

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. 

4. 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.