NCERT Exemplar Solutions Class 11 Physics Systems of Particles and Rotational Motion — Very Short, Short and Long Answer Questions Q14–28

NCERT Exemplar Problems, Class 11 Physics — Systems of Particles and Rotational Motion, Very Short, Short and Long Answer Questions. All 15 questions solved.

The book numbers these questions 7.14 to 7.28; here they are Questions 14 to 28. The exemplar calls this Unit 7, while the rationalised NCERT textbook covers the same material in Chapter 6. Questions 14 to 18 are very short answer, 19 to 22 short answer and 23 to 28 long answer. Four ideas carry the page:

• Centre of mass by symmetry and integration. Use symmetry to fix one coordinate, then $\bar x = \int x\,dm \big/ \int dm$ for the other.
• Torque depends on the point. $\sum(\vec r_i – \vec a)\times\vec F_i = \sum\vec r_i\times\vec F_i – \vec a\times\sum\vec F_i$, so zero torque about one point means zero about every point only if the net force is zero.
• Angular momentum conservation. With no external torque about an axis, $I_1\omega_1 + I_2\omega_2$ is fixed, even though kinetic energy is lost to friction.
• Rolling. A body rolls without slipping when $v_{\text{cm}} = R\omega$; friction drives $v$ and $\omega$ towards that condition.

Key insight. In the friction problems (Questions 25, 26 and 28), write one equation for the centre of mass and one for rotation, link them by the rolling condition, and solve. The friction force is whatever those equations require.

Question 14

The centre of gravity of a body on the earth coincides with its centre of mass for a ‘small’ object whereas for an ‘extended’ object it may not. What is the qualitative meaning of ‘small’ and ‘extended’ in this regard? For which of the following the two coincide? A building, a pond, a lake, a mountain?

Solution. The centre of gravity coincides with the centre of mass when $g$ has the same value over the whole body. $g$ changes with height above the earth’s surface on the scale of the earth’s radius. So an object is ‘small’ if its vertical extent is very small compared with the earth’s radius, so that $g$ is effectively uniform over it; it is ‘extended’ if its height is large enough for $g$ to vary noticeably across it.

A building and a pond are small in this sense, so for them the two points coincide. A deep lake and a mountain are extended objects, for which they need not.

Question 15

Why does a solid sphere have smaller moment of inertia than a hollow cylinder of same mass and radius, about an axis passing through their axes of symmetry?

Solution. The moment of inertia is $I = \sum m_i r_i^2$, so it depends on how far the mass is from the axis. In a hollow cylinder every bit of mass is at the full distance $R$ from the axis, giving $I = MR^2$. In a solid sphere most of the mass lies closer to the axis than $R$, so $I = \tfrac25 MR^2$ is smaller.

Question 16

The variation of angular position $\theta$, of a point on a rotating rigid body, with time $t$ is shown in Fig. 7.7. Is the body rotating clock-wise or anti-clockwise?

Solution. The graph is a straight line with positive slope, so $\omega = \dfrac{d\theta}{dt}$ is positive and constant. Angular position is conventionally measured anticlockwise as positive, so a positive $\omega$ means the body is turning anticlockwise (at a uniform rate). That $\theta$ is negative before $t_2$ only reflects where the angle is measured from, not the sense of rotation.

Question 17

A uniform cube of mass $m$ and side $a$ is placed on a frictionless horizontal surface. A vertical force $F$ is applied to the edge as shown in Fig. 7.8. Match the following (most appropriate choice):

	Force		Result
(a)	$mg/4 < F < mg/2$	(i)	Cube will move up.
(b)	$F > mg/2$	(ii)	Cube will not exhibit motion.
(c)	$F > mg$	(iii)	Cube will begin to rotate and slip at A.
(d)	$F = mg/4$	(iv)	Normal reaction effectively at $a/3$ from A, no motion.

Solution. The force $F$ acts upwards at the edge opposite A. While the cube stays put, the floor’s normal reaction $N = mg – F$ acts at some distance $x$ from A, and torques about A balance:

$$Fa + (mg – F)x = mg\frac a2 \;\Rightarrow\; x = \frac{(mg/2 – F)\,a}{mg – F}.$$

• $F = mg/4$: $x = \dfrac{(mg/4)a}{3mg/4} = \dfrac a3$. The cube stays at rest with the normal reaction effectively at $a/3$ from A: (d)–(iv).
• $mg/4 < F < mg/2$: $x$ lies between $a/3$ and $0$, still inside the base, so there is no motion: (a)–(ii).
• $F > mg/2$: the torque of $F$ about A exceeds that of the weight, so the cube turns about A. The floor is frictionless, so A cannot grip and slips while the cube rotates: (b)–(iii).
• $F > mg$: the upward force exceeds the weight and the cube is lifted off: (c)–(i).

Question 18

A uniform sphere of mass $m$ and radius $R$ is placed on a rough horizontal surface (Fig. 7.9). The sphere is struck horizontally at a height $h$ from the floor. Match the following:

	Height		Result
(a)	$h = R/2$	(i)	Sphere rolls without slipping with a constant velocity and no loss of energy.
(b)	$h = R$	(ii)	Sphere spins clockwise, loses energy by friction.
(c)	$h = 3R/2$	(iii)	Sphere spins anti-clockwise, loses energy by friction.
(d)	$h = 7R/5$	(iv)	Sphere has only a translational motion, loses energy by friction.

Solution. Let the blow deliver an impulse $J$ to the right. It gives the centre of mass $v = J/m$, and, acting at $(h – R)$ above the centre, an angular velocity

$$\omega = \frac{J(h – R)}{\tfrac25 mR^2}.$$

For $h > R$ the blow is above the centre and the sphere spins clockwise (forward, the rolling sense); for $h < R$ it spins anticlockwise. It rolls without slipping straight away if $v = R\omega$:

$$\frac Jm = \frac{5J(h – R)}{2mR} \;\Rightarrow\; h = \frac{7R}{5}.$$

• $h = 7R/5$: rolls without slipping, no friction acts, constant velocity, no energy lost: (d)–(i).
• $h = R$: the blow passes through the centre, $\omega = 0$; the sphere slides, and friction works on it: (b)–(iv).
• $h = 3R/2$: above the centre, clockwise spin, but $R\omega = \tfrac54 v \ne v$, so it slips and loses energy to friction: (c)–(ii).
• $h = R/2$: below the centre, anticlockwise spin; it slips and loses energy: (a)–(iii).

Question 19

The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

Solution. No. Let the forces $\vec F_i$ act at $\vec r_i$ measured from O, with $\sum\vec r_i\times\vec F_i = 0$. About another point O′ at position $\vec a$,

$$\sum(\vec r_i – \vec a)\times\vec F_i = \sum\vec r_i\times\vec F_i – \vec a\times\sum\vec F_i = -\vec a\times\sum\vec F_i.$$

Since $\sum\vec F_i \ne 0$, this is non-zero for any $\vec a$ not parallel to the net force. The torque vanishes about every point only when the net force is also zero.

Question 20

A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

Solution. The centripetal force on each particle comes from the internal elastic forces of the wheel. Because the wheel is symmetric, for every particle there is one diametrically opposite with an equal and opposite acceleration; the internal forces come in action–reaction pairs and the sum of all of them is zero. The centre of mass, at the axis, does not accelerate, so no net external force is needed, and uniform rotation needs no net external torque.

A half-wheel can be set rotating in the same way, by fixing it on an axle through the centre of the full wheel and applying a torque to start it. Its centre of mass, however, is not on the axis but a distance $r_{\text{cm}}$ from it ($\tfrac{2R}{\pi}$ for a half-ring, $\tfrac{4R}{3\pi}$ for a half-disc). As it rotates, the centre of mass moves in a circle and must be given a centripetal force $M\omega^2 r_{\text{cm}}$ towards the axis. Internal forces cannot supply this, so an external force is required: the axle must keep pulling on the half-wheel, with a direction that rotates with it.

Where this differs from the printed key. The key reaches the same conclusion (external agency needed) but argues that $\mathbf L$ is not along $\boldsymbol\omega$, so an external torque is needed. For a flat half-wheel rotating about an axis perpendicular to its plane, every particle’s $\vec r\times m\vec v$ points along the axis, so $\mathbf L$ is parallel to $\boldsymbol\omega$ and no torque is needed for uniform rotation. What the half-wheel does need is the external centripetal force on its off-axis centre of mass.

Question 21

A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does its weight cause any torque about this axis? Give reason for your answer.

(Fig. 7.10 shows the door in the $xy$-plane, hinged along the vertical $y$-axis, with its weight $W$ acting vertically downwards at its centre.)

Solution. No. The torque of the weight about any point on the hinge is $\vec\tau = \vec r\times\vec W$. Here $\vec r$ lies in the plane of the door (the $xy$-plane) and $\vec W$ is along $-y$, so $\vec\tau$ is perpendicular to the $xy$-plane, along $\pm z$. Its component along the vertical ($y$) axis of rotation is zero. In other words, the weight is parallel to the axis, and a force parallel to an axis cannot turn a body about it. (The weight does try to tilt the door about a horizontal axis, which is what loads the hinges.)

Question 22

$(n – 1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$-polygon. The vacant vertex has a position vector $\mathbf a$ with respect to the centre of the polygon. Find the position vector of centre of mass.

Solution. If the vacant vertex were also filled, the $n$ equal masses would have their centre of mass at the centre of the polygon, by symmetry. Let $\mathbf b$ be the centre of mass of the $n – 1$ masses. Adding a mass $m$ at $\mathbf a$ gives the full polygon, so

$$\frac{(n – 1)m\,\mathbf b + m\,\mathbf a}{nm} = \mathbf 0 \;\Rightarrow\; \mathbf b = -\frac{\mathbf a}{n – 1}.$$

The centre of mass lies on the line through the centre and the vacant vertex, on the opposite side.

Question 23

Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

Solution. (a) Place the half-disc of radius $R$ and mass $M$ with its straight edge on the $x$-axis and centre at the origin, so that $0 \le \theta \le \pi$. The surface density is $\sigma = \dfrac{M}{\pi R^2/2} = \dfrac{2M}{\pi R^2}$, and an element has $dm = \sigma\,r\,dr\,d\theta$. By symmetry about the $y$-axis, $\bar x = 0$. Then

$$\bar y = \frac{\int_0^R\!\int_0^\pi (r\sin\theta)\,\sigma\,r\,dr\,d\theta}{\int_0^R\!\int_0^\pi \sigma\,r\,dr\,d\theta} = \frac{\tfrac{R^3}{3}\cdot 2}{\tfrac{R^2}{2}\cdot\pi} = \frac{4R}{3\pi}.$$

(b) For a quarter-disc in the first quadrant, $0 \le \theta \le \tfrac\pi2$ and $\sigma = \dfrac{4M}{\pi R^2}$:

$$\bar x = \frac{\int_0^R\!\int_0^{\pi/2} r^2\cos\theta\,dr\,d\theta}{\int_0^R\!\int_0^{\pi/2} r\,dr\,d\theta} = \frac{\tfrac{R^3}{3}\cdot1}{\tfrac{R^2}{2}\cdot\tfrac\pi2} = \frac{4R}{3\pi},$$

and by symmetry about the line $y = x$, $\bar y = \bar x$.

Question 24

Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident. (a) Does the law of conservation of angular momentum apply to the situation? Why? (b) Find the angular speed of the two-disc system. (c) Calculate the loss in kinetic energy of the system in the process. (d) Account for this loss.

Solution. (a) Yes. The external forces on the system, gravity and the reactions at the axle, act through the common axis and have no torque about it. The friction between the discs is internal. So angular momentum about the axis is conserved.

(b) The discs end up turning together at $\omega$:

$$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\,\omega \;\Rightarrow\; \omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}.$$

(c) $K_i = \tfrac12(I_1\omega_1^2 + I_2\omega_2^2)$ and $K_f = \tfrac12\dfrac{(I_1\omega_1 + I_2\omega_2)^2}{I_1 + I_2}$. Subtracting,

$$K_i – K_f = \frac{I_1 I_2}{2(I_1 + I_2)}(\omega_1 – \omega_2)^2,$$

which is positive unless $\omega_1 = \omega_2$.

(d) While the discs slip over each other, kinetic friction acts between the faces. The work done against it is turned into heat (and some sound), which is where the lost kinetic energy goes.

Question 25

A disc of radius $R$ is rotating with an angular speed $\omega_0$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $\mu_k$. (a) What was the velocity of its centre of mass before being brought in contact with the table? (b) What happens to the linear velocity of a point on its rim when placed in contact with the table? (c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table? (d) Which force is responsible for the effects in (b) and (c)? (e) What condition should be satisfied for rolling to begin? (f) Calculate the time taken for the rolling to begin.

Solution. (a) The disc was only spinning about a fixed axis through its centre, so its centre of mass was at rest: velocity zero.

(b) The lowest point of the rim moves over the table at $R\omega_0$, and kinetic friction opposes this slipping. Friction’s torque slows the spin, so the linear speed of a point on the rim (relative to the centre) decreases.

(c) The same friction force acts forwards on the disc, so the centre of mass, initially at rest, speeds up: its linear speed increases.

(d) Kinetic friction between the disc and the table.

(e) Rolling begins when the contact point stops slipping, that is when $v_{\text{cm}} = R\omega$.

(f) Friction $\mu_k mg$ gives $a_{\text{cm}} = \mu_k g$, so $v_{\text{cm}} = \mu_k g\,t$. Its torque $\mu_k mgR$ about the centre gives, with $I = \tfrac12 mR^2$, an angular deceleration $\alpha = \dfrac{\mu_k mgR}{I} = \dfrac{2\mu_k g}{R}$, so $\omega = \omega_0 – \dfrac{2\mu_k g}{R}t$. Rolling begins when $v_{\text{cm}} = R\omega$:

$$\mu_k g\,t = R\omega_0 – 2\mu_k g\,t \;\Rightarrow\; t = \frac{R\omega_0}{3\mu_k g}.$$

In general form, $t = \dfrac{R\omega_0}{\mu_k g\left(1 + \dfrac{mR^2}{I}\right)}$, which gives the same result for a disc.

Question 26

Two cylindrical hollow drums of radii $R$ and $2R$, and of a common height $h$, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise), respectively. Their axes, fixed, are parallel and in a horizontal plane separated by $(3R + \delta)$. They are now brought in contact $(\delta \to 0)$. (a) Show the frictional forces just after contact. (b) Identify forces and torques external to the system just after contact. (c) What would be the ratio of final angular velocities when friction ceases?

Solution. Take the small drum (radius $R$) on the left and the large drum (radius $2R$) on the right, touching at a point between them.

(a) At the contact point the small drum, turning anticlockwise, moves upwards at speed $R\omega$. The large drum, turning clockwise, also moves upwards there, at $2R\omega$. The large drum’s surface is faster, so it drags the small drum’s surface upwards and is itself held back. The friction force $F$ on the left (small) drum is upwards, and the equal and opposite force $F$ on the right (large) drum is downwards.

(b) The frictional forces are internal to the two-drum system. The external forces are those from the fixed axles, which hold each drum in place against the friction: a downward force $F’ = F$ at the small drum’s axle and an upward force $F” = F$ at the large drum’s axle. The net external force is zero, but these two form a couple with arm $3R$, the distance between the axes, so the external torque is $F\times3R$ (anticlockwise). That is why angular momentum of the pair is not conserved here.

(c) Friction stops when there is no relative slipping at the contact, so the surface speeds are equal. Both drums keep their original senses (the small one speeds up anticlockwise, the large one slows down clockwise), and

$$R\omega_1 = 2R\omega_2 \;\Rightarrow\; \frac{\omega_1}{\omega_2} = 2.$$

Question 27

A uniform square plate S (side $c$) and a uniform rectangular plate R (sides $b$, $a$) have identical areas and masses (Fig. 7.11). Show that (i) $I_{xR}/I_{xS} < 1$; (ii) $I_{yR}/I_{yS} > 1$; (iii) $I_{zR}/I_{zS} > 1$.

(In Fig. 7.11 the $x$- and $y$-axes pass through the centre of each plate, parallel to its sides; the rectangle has its longer side $a$ along $x$ and shorter side $b$ along $y$, so $a > b$.)

Solution. Equal areas give $c^2 = ab$, with $b < c < a$. For a uniform plate, the moment of inertia about an axis through the centre parallel to one side is $\tfrac{M}{12}$ times the square of the other side.

(i) $I_{xR} = \dfrac{Mb^2}{12}$ and $I_{xS} = \dfrac{Mc^2}{12}$, so $\dfrac{I_{xR}}{I_{xS}} = \dfrac{b^2}{c^2} < 1$ since $b < c$.

(ii) $I_{yR} = \dfrac{Ma^2}{12}$ and $I_{yS} = \dfrac{Mc^2}{12}$, so $\dfrac{I_{yR}}{I_{yS}} = \dfrac{a^2}{c^2} > 1$ since $a > c$. (Note that the two ratios multiply to $\dfrac{a^2b^2}{c^4} = 1$, so one is less than 1 exactly when the other is greater.)

(iii) By the perpendicular-axis theorem, $I_z = I_x + I_y$ for a lamina, so

$$I_{zR} – I_{zS} = \frac{M}{12}\left(a^2 + b^2 – 2c^2\right) = \frac{M}{12}\left(a^2 + b^2 – 2ab\right) = \frac{M}{12}(a – b)^2 > 0,$$

hence $\dfrac{I_{zR}}{I_{zS}} > 1$.

Question 28

A uniform disc of radius $R$, is resting on a table on its rim. The coefficient of friction between disc and table is $\mu$ (Fig. 7.12). Now the disc is pulled with a force $\mathbf F$ as shown in the figure. What is the maximum value of $\mathbf F$ for which the disc rolls without slipping?

(In Fig. 7.12 the horizontal force $\mathbf F$ is applied at the centre of the disc.)

Solution. Let the disc have mass $M$ and the friction $f$ act backwards at the contact point. For the centre of mass and for rotation about it:

$$Ma = F – f, \qquad fR = \tfrac12 MR^2\,\alpha.$$

Rolling without slipping requires $\alpha = a/R$, so $f = \tfrac12 Ma$. Then $F = Ma + \tfrac12 Ma = \tfrac32 Ma$, giving $a = \dfrac{2F}{3M}$ and $f = \dfrac F3$. Static friction can supply at most $\mu Mg$, so

$$\frac F3 \le \mu Mg \;\Rightarrow\; F \le 3\mu Mg.$$

Common mistakes

• Calling a lake or mountain ‘small’ in Question 14. The test is height compared with the earth’s radius, not area.
• Reading the sense of rotation from the sign of $\theta$ in Question 16. It is the slope $d\theta/dt$ that matters.
• Forgetting the frictionless floor in Question 17. Once $F > mg/2$ the cube turns, and A slips because nothing holds it.
• Assuming a blow through the centre makes a sphere roll in Question 18. It needs $h = 7R/5$; at $h = R$ there is no spin at all.
• Assuming torque balance about one point is enough in Question 19. With a net force, the torque changes from point to point.
• Using angular momentum conservation for the drums in Question 26. The axles exert an external couple; the condition that ends the slipping is equal surface speeds.
• Taking $f = \mu Mg$ as the friction actually acting in Question 28. The friction is $F/3$; $\mu Mg$ is only its upper limit.

Practise next

• Clutch plates. A flywheel with $I_1 = 0.1$ kg m² spinning at $100$ rad/s is coupled to a second, initially stationary flywheel with $I_2 = 0.3$ kg m² on the same axis. Find the common angular speed and the kinetic energy lost.

Show answer

$25$ rad/s; $375$ J lost. $\omega = \tfrac{0.1\times100}{0.4} = 25$ rad/s. The kinetic energy falls from $\tfrac12(0.1)(100)^2 = 500$ J to $\tfrac12(0.4)(25)^2 = 125$ J, a loss of $375$ J, which matches $\tfrac{I_1I_2}{2(I_1+I_2)}\omega_1^2$.

• Spinning ring set down. A thin ring of radius $0.5$ m spinning at $20$ rad/s about its axis is gently placed on a floor with $\mu_k = 0.2$ ($g = 10$ m/s²). How long until it rolls without slipping, and at what speed?

Show answer

$2.5$ s, at $5$ m/s. For a ring $I = mR^2$, so $t = \tfrac{R\omega_0}{\mu_k g(1 + mR^2/I)} = \tfrac{0.5\times20}{0.2\times10\times2} = 2.5$ s. Then $v = \mu_k g t = 5$ m/s, and $\omega = 20 – 4\times2.5 = 10$ rad/s, so $R\omega = 5$ m/s as required.

• Pulling a solid sphere. A horizontal force $F$ is applied at the centre of a solid sphere of mass $M$ on a floor with coefficient of friction $\mu$. What is the largest $F$ for which it rolls without slipping?

Show answer

$F_{\max} = \tfrac72\mu Mg$. With $I = \tfrac25MR^2$, $fR = \tfrac25MR^2(a/R)$ gives $f = \tfrac25Ma$, and $F – f = Ma$ gives $f = \tfrac27F$. Setting $\tfrac27F \le \mu Mg$ gives $F \le 3.5\,\mu Mg$.

• Exemplar Systems of Particles and Rotational Motion: Multiple Choice Questions — Questions 1 to 13, on centre of mass, angular momentum and moments of inertia.
• NCERT textbook Chapter 6 — Systems of Particles and Rotational Motion for the theory of centre of mass, torque, angular momentum and moment of inertia.
• Exemplar Gravitation — the next unit.