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

NCERT Class 11 Physics — Systems of Particles and Rotational Motion, Chapter 6 Exercises. All 17 questions solved.

The exercises at the end of Chapter 6 fall into four groups. Questions 6.1 to 6.3 and 6.15 locate the centre of mass and use the fact that internal forces cannot change its motion. Questions 6.4 to 6.7 practise the vector product, first on areas and volumes and then on angular momentum. Questions 6.8, 6.9 and 6.16 are balance problems, solved with the two conditions for equilibrium. Questions 6.10 to 6.14 and 6.17 are rotation about a fixed axis: moment of inertia, torque, rotational kinetic energy, angular momentum and power. Almost everything runs on three sets of results:

• Centre of mass — $\vec R = \dfrac{\sum m_i\vec r_i}{\sum m_i}$, and $M\vec a_{\text{cm}} = \vec F_{\text{ext}}$, so with no external force the centre of mass keeps a constant velocity.
• Equilibrium of a rigid body — $\sum\vec F = 0$ and $\sum\vec\tau = 0$, where the torque of a force is $\vec\tau = \vec r\times\vec F$; once the forces balance, the torques may be taken about any point.
• Rotation about a fixed axis — $\tau = I\alpha$, $L = I\omega$, $K = \tfrac12 I\omega^2$ and $P = \tau\omega$, with $I = MR^2$ for a ring or thin hollow cylinder, $\tfrac12 MR^2$ for a solid cylinder or disc and $\tfrac25 MR^2$ for a solid sphere, each about its axis of symmetry.

Where a weight has to be found from a mass, $g = 9.8\ \text{m s}^{-2}$.

Key insight. Rotation about a fixed axis is linear motion with the symbols renamed — $I$ for $m$, $\omega$ for $v$, $\alpha$ for $a$, $\tau$ for $F$. So $\tau = I\alpha$ is Newton’s second law (6.10, 6.13), $\tfrac12I\omega^2$ is kinetic energy (6.11, 6.17), $P = \tau\omega$ is $P = Fv$ (6.14), and with no external torque $I\omega$ is conserved just as momentum is (6.12). The one thing with no linear counterpart is the choice of axis: in the balance problems (6.8, 6.9, 6.16) take moments about the point where an unknown force acts, and that force drops out of the equation.

Question 6.1

Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Solution. All four bodies are symmetric about a central point: for every small piece of mass at position $\vec r$ measured from that point, there is an equal piece at $-\vec r$. In the sum $\sum m_i\vec r_i$ the pieces cancel in pairs, so the sum is zero and the centre of mass is that central point, the geometrical centre. For the sphere it is the centre of the sphere, for the cylinder the midpoint of its axis, for the ring the centre of the circle, and for the cube the point where its body diagonals cross.

The ring answers the second part. Its centre of mass is at the centre of the circle, where there is no material at all. The same is true of a hollow sphere, a hollow cylinder, a hollow cube or a bangle, so the centre of mass need not lie inside the body.

Question 6.2

In the HCl molecule, the separation between the nuclei of the two atoms is about $1.27\ \text{Å}$ ($1\ \text{Å} = 10^{-10}\ \text{m}$). 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.

Solution. Since nearly all the mass sits in the nuclei, the molecule is two point masses $1.27\ \text{Å}$ apart. Put the hydrogen nucleus at the origin and the chlorine nucleus at $x = 1.27\ \text{Å}$, with masses $m$ and $35.5m$:

$$X = \frac{m(0) + 35.5m(1.27)}{m + 35.5m} = \frac{35.5\times1.27}{36.5}\ \text{Å} = 1.24\ \text{Å}$$

The mass $m$ cancels, which is why only the ratio of the masses was needed. The centre of mass lies on the line joining the nuclei, very close to the heavy chlorine nucleus — only about $0.035\ \text{Å}$ from it.

Question 6.3

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?

Solution. Treat the trolley and the child as one system. The forces on it from outside are the weights and the normal reaction of the floor, which are vertical and balance, and, since the floor is smooth, no horizontal force at all. The pushes between the child’s feet and the trolley when the child runs are internal forces: they come in equal and opposite pairs and cancel in the total. With no net external force, $M\vec a_{\text{cm}} = 0$, so the centre of mass keeps the velocity it had before the child moved. The trolley itself may speed up or slow down as the child runs one way or the other, but the centre of mass of the pair does not.

Question 6.4

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

Solution. Draw $\vec a$ and $\vec b$ from a common point $O$, with angle $\theta$ between them; the line joining their tips completes the triangle. Take $\vec a$ as the base, of length $|\vec a|$. The height is the perpendicular distance from the tip of $\vec b$ to the line of $\vec a$, which is $|\vec b|\sin\theta$. So

$$\text{area} = \tfrac12\times\text{base}\times\text{height} = \tfrac12\,|\vec a|\,|\vec b|\sin\theta$$

But $|\vec a|\,|\vec b|\sin\theta$ is exactly the magnitude of the vector product $\vec a\times\vec b$. Hence the area is $\tfrac12|\vec a\times\vec b|$. The whole parallelogram on $\vec a$ and $\vec b$ has area $|\vec a\times\vec b|$, and the triangle is half of it.

Question 6.5

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

Solution. Take the face formed by $\vec b$ and $\vec c$ as the base. It is a parallelogram of area $|\vec b\times\vec c|$ (twice the triangle of Question 6.4), and the vector $\vec b\times\vec c$ points along the normal to that face. The volume of a parallelepiped is base area times perpendicular height, and the height is the component of the third edge $\vec a$ along that normal, $|\vec a|\,|\cos\phi|$, where $\phi$ is the angle between $\vec a$ and $\vec b\times\vec c$. So

$$V = |\vec b\times\vec c|\;|\vec a|\,|\cos\phi| = |\vec a\cdot(\vec b\times\vec c)|$$

because $\vec a\cdot(\vec b\times\vec c) = |\vec a|\,|\vec b\times\vec c|\cos\phi$ by the definition of the scalar product. The sign of $\vec a\cdot(\vec b\times\vec c)$ depends only on whether $\vec a$ lies on the same side of the base as $\vec b\times\vec c$ or on the opposite side, which is why the result holds in magnitude.

Question 6.6

Find the components along the $x$, $y$, $z$ axes of the angular momentum $\vec l$ of a particle, whose position vector is $\vec r$ with components $x$, $y$, $z$ and momentum is $\vec p$ 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.

Solution. Angular momentum is $\vec l = \vec r\times\vec p$, and a vector product in components is a determinant:

$$\vec l = \begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ x & y & z\\ p_x & p_y & p_z\end{vmatrix}$$

Expanding along the top row,

$$l_x = yp_z – zp_y,\qquad l_y = zp_x – xp_z,\qquad l_z = xp_y – yp_x$$

If the particle moves only in the $x$-$y$ plane, it is always at $z = 0$ and never moves in the $z$ direction, so $p_z = 0$ as well. Both terms of $l_x$ and of $l_y$ then vanish, leaving $\vec l = (xp_y – yp_x)\,\hat k$ along the $z$-axis — perpendicular to the plane of motion, as the vector product of two vectors lying in that plane must be.

Question 6.7

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

Solution. The angular momentum of a particle about a point has magnitude $mv$ times the perpendicular distance from the point to its line of motion, and its sense follows from $\vec r\times\vec p$.

Take any point $P$ between the lines, a distance $y$ from the line of $A$ and $d – y$ from the line of $B$. About $P$, $A$ has angular momentum $mvy$ and $B$ has $mv(d – y)$. The particles are on opposite sides of $P$ and moving in opposite directions, so both turn the same way about $P$ — clockwise in the figure — and the two add:

$$L = mvy + mv(d – y) = mvd$$

Now take a point $Q$ outside the lines, a distance $y’$ beyond the line of $A$ and so $d + y’$ from the line of $B$. Both particles are now on the same side of $Q$, moving in opposite directions, so they turn in opposite senses, and $B$, being farther away, wins:

$$L = mv(d + y’) – mvy’ = mvd$$

in the same clockwise sense as before. In both cases the position of the point has cancelled out, and the direction is perpendicular to the plane containing the two lines. The distances do not change as the particles move, so the result is also the same at every instant.

The same result follows in one line of vectors. About any origin, $\vec L = \vec r_A\times m\vec v + \vec r_B\times(-m\vec v) = m(\vec r_A – \vec r_B)\times\vec v$, and $\vec r_A – \vec r_B$, the position of $A$ relative to $B$, does not depend on where the origin is.

Question 6.8

A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. The angles made by the strings with the vertical are $36.9^\circ$ and $53.1^\circ$ respectively. The bar is $2\ \text{m}$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

Solution. The bar is non-uniform, so its centre of gravity is not at its midpoint, and the tensions $T_1$ (left string) and $T_2$ (right string) are unknown. Three conditions are available — no net horizontal force, no net vertical force and no net torque — which is exactly enough for $T_1$, $T_2$ and $d$.

The angles are those of a 3-4-5 triangle: $\sin36.9^\circ = 0.6 = \cos53.1^\circ$ and $\cos36.9^\circ = 0.8 = \sin53.1^\circ$. Each angle is measured from the vertical, so a string’s vertical component uses the cosine and its horizontal component the sine.

Horizontal forces. The strings pull the bar sideways in opposite directions, and the bar does not move sideways:

$$T_1\sin36.9^\circ = T_2\sin53.1^\circ \;\Rightarrow\; 0.6\,T_1 = 0.8\,T_2 \;\Rightarrow\; T_1 = \tfrac43\,T_2$$

Vertical forces. Together the strings hold up the weight:

$$W = T_1\cos36.9^\circ + T_2\cos53.1^\circ = 0.8\left(\tfrac43T_2\right) + 0.6\,T_2 = \tfrac53\,T_2$$

Torques about the left end. $T_1$ acts at the left end, so it has no moment about that point and drops out. The horizontal part of $T_2$ acts along the bar and has no moment either. What is left is the weight, at $d$ from the left end, turning the bar one way, and the vertical part of $T_2$, at $2\ \text{m}$, turning it the other:

$$W d = (T_2\cos53.1^\circ)(2) = 1.2\,T_2 \;\Rightarrow\; d = \frac{1.2\,T_2}{\tfrac53\,T_2} = 0.72\ \text{m}$$

Question 6.9

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

Solution. “Weighs $1800\ \text{kg}$” gives the mass; the weight is $W = mg = 1800\times9.8 = 17640\ \text{N}$, acting at the centre of gravity $G$. Let $R_F$ be the total upward force of the ground on the two front wheels and $R_B$ that on the two back wheels. The car is in equilibrium, so both the forces and the torques balance.

Forces: $R_F + R_B = 17640\ \text{N}$.

Torques about $G$, where the weight acts and so drops out: the front axle is $1.05\ \text{m}$ in front of $G$ and the back axle $1.8 – 1.05 = 0.75\ \text{m}$ behind it, so

$$R_F(1.05) = R_B(0.75)$$

Solving the two equations together,

$$R_F = \frac{0.75}{1.8}\times17640 = 7350\ \text{N},\qquad R_B = \frac{1.05}{1.8}\times17640 = 10290\ \text{N}$$

Each axle has two wheels, which share its load equally: $7350/2 = 3675\ \text{N}$ on each front wheel and $10290/2 = 5145\ \text{N}$ on each back wheel. The back wheels carry more because the centre of gravity is nearer the back axle.

Question 6.10

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?

Solution. For rotation about a fixed axis $\tau = I\alpha$, so the same torque gives the body with the smaller moment of inertia the larger angular acceleration, and starting from rest the angular speed after a time $t$ is $\omega = \alpha t$. In the hollow cylinder all the mass is at distance $R$ from the axis, so $I_{\text{cyl}} = MR^2$. In the solid sphere much of the mass is close to the axis, and $I_{\text{sph}} = \tfrac25MR^2$. Then

$$\frac{\omega_{\text{sph}}}{\omega_{\text{cyl}}} = \frac{\alpha_{\text{sph}}}{\alpha_{\text{cyl}}} = \frac{I_{\text{cyl}}}{I_{\text{sph}}} = \frac{MR^2}{\tfrac25MR^2} = 2.5$$

Question 6.11

A solid cylinder of mass $20\ \text{kg}$ rotates about its axis with angular speed $100\ \text{rad s}^{-1}$. The radius of the cylinder is $0.25\ \text{m}$. 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?

Solution. For a solid cylinder about its own axis,

$$I = \tfrac12MR^2 = \tfrac12(20)(0.25)^2 = 0.625\ \text{kg m}^2$$

The kinetic energy of rotation is the rotational form of $\tfrac12mv^2$:

$$K = \tfrac12I\omega^2 = \tfrac12(0.625)(100)^2 = 3125\ \text{J}$$

and the angular momentum is the rotational form of $mv$:

$$L = I\omega = (0.625)(100) = 62.5\ \text{J s}$$

(A joule second is the same unit as $\text{kg m}^2\,\text{s}^{-1}$.)

Question 6.12

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

Question 6.12 (a)

Solution. The turntable is frictionless, so nothing outside the child exerts a torque about the axis; folding his arms involves only forces inside his own body. Angular momentum about the axis is therefore conserved, $I_1\omega_1 = I_2\omega_2$, taking the moment of inertia to be the child’s as the question does. With $I_2 = \tfrac25I_1$:

$$\omega_2 = \omega_1\,\frac{I_1}{I_2} = 40\times\frac52 = 100\ \text{rev/min}$$

Only a ratio of angular speeds is involved, so there is no need to convert to $\text{rad s}^{-1}$.

Question 6.12 (b)

Solution. Kinetic energy of rotation can be written in terms of the angular momentum as $K = \tfrac12I\omega^2 = \dfrac{L^2}{2I}$. $L$ is the same before and after, so

$$\frac{K_2}{K_1} = \frac{I_1}{I_2} = \frac52 = 2.5$$

The same follows from the angular speeds: $\dfrac{K_2}{K_1} = \dfrac{I_2}{I_1}\left(\dfrac{\omega_2}{\omega_1}\right)^2 = \dfrac25\times\left(\dfrac52\right)^2 = 2.5$. So the kinetic energy increases, even though angular momentum is conserved.

The extra energy comes from the child. His outstretched arms, going round in circles, tend to fly outwards, and to pull them in he has to exert an inward force through a distance, doing work. That work, drawn from the chemical energy stored in his muscles, is what raises the kinetic energy of rotation.

Question 6.13

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

Solution. The rope leaves the cylinder along a tangent, so its pull acts at right angles to a radius and its lever arm is the full radius, $0.40\ \text{m}$:

$$\tau = FR = 30\times0.40 = 12\ \text{N m}$$

In a thin-walled hollow cylinder all the mass is at distance $R$ from the axis, so

$$I = MR^2 = 3\times(0.40)^2 = 0.48\ \text{kg m}^2,\qquad \alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25\ \text{rad s}^{-2}$$

With no slipping, the rope unwinds exactly as fast as the rim moves, so its acceleration equals the tangential acceleration of the rim:

$$a = R\alpha = 0.40\times25 = 10\ \text{m s}^{-2}$$

The radius drops out in the end, since $a = R\cdot\dfrac{FR}{MR^2} = \dfrac FM = \dfrac{30}{3}$.

Question 6.14

To maintain a rotor at a uniform angular speed of $200\ \text{rad s}^{-1}$, an engine needs to transmit a torque of $180\ \text{N m}$. 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.

Solution. When a torque $\tau$ turns a body through a small angle $d\theta$ it does work $\tau\,d\theta$, so the rate of doing work is $P = \tau\,d\theta/dt = \tau\omega$, the rotational form of $P = Fv$. Here the engine’s torque exactly balances the frictional torque, which is why the speed stays constant, and all the work it does goes into overcoming friction:

$$P = \tau\omega = 180\times200 = 3.6\times10^4\ \text{W}$$

An engine that is $100\%$ efficient needs exactly this much power.

Question 6.15

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

Solution. Put the hole back. The original full disc is the remaining body plus the piece that was cut out, and the centre of mass of the full disc is its centre $O$. For a uniform sheet, mass is proportional to area. The cut-out piece has half the radius and so a quarter of the area, and a quarter of the mass, $M/4$; the remaining body has mass $3M/4$.

Take $O$ as the origin and the $x$-axis through $C$, the centre of the hole, so that $C$ is at $x = R/2$. If the centre of mass of the remaining body is at $x$, then the full disc’s centre of mass is the weighted average of the two parts:

$$M(0) = \tfrac34M\,x + \tfrac14M\left(\tfrac R2\right) \;\Rightarrow\; x = -\frac{R/8}{3/4} = -\frac R6$$

The negative sign puts it on the opposite side of $O$ from the hole, which makes sense: removing material from one side shifts the balance point to the other. Gravity is uniform over a body this small, so the centre of gravity coincides with the centre of mass.

Question 6.16

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5\ \text{g}$ are put one on top of the other at the $12.0\ \text{cm}$ mark, the stick is found to be balanced at $45.0\ \text{cm}$. What is the mass of the metre stick?

Solution. The stick balances on its own at the $50.0\ \text{cm}$ mark, which tells us its centre of gravity is there. With the coins added, the balance point moves to $45.0\ \text{cm}$. Take moments about this new knife edge, so that the knife edge’s upward push has no moment and drops out. The coins, $10\ \text{g}$ in all, are $45.0 – 12.0 = 33.0\ \text{cm}$ to one side; the stick’s weight $mg$ acts at $50.0\ \text{cm}$, $5.0\ \text{cm}$ to the other side. For balance the two moments are equal:

$$(10\ \text{g})\,g\times33.0 = m\,g\times5.0 \;\Rightarrow\; m = \frac{10\times33.0}{5.0} = 66.0\ \text{g}$$

The $g$ cancels, so the masses can stay in grams and the distances in centimetres.

Question 6.17

The oxygen molecule has a mass of $5.30\times10^{-26}\ \text{kg}$ and a moment of inertia of $1.94\times10^{-46}\ \text{kg m}^2$ 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 $500\ \text{m/s}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Solution. The kinetic energy of translation uses the mass and the speed:

$$K_{\text{trans}} = \tfrac12mv^2 = \tfrac12(5.30\times10^{-26})(500)^2 = 6.63\times10^{-21}\ \text{J}$$

Two thirds of this is the kinetic energy of rotation, $K_{\text{rot}} = 4.42\times10^{-21}\ \text{J}$. Setting it equal to $\tfrac12I\omega^2$ and solving for $\omega$:

$$\omega = \sqrt{\frac{2K_{\text{rot}}}{I}} = \sqrt{\frac{2\times4.42\times10^{-21}}{1.94\times10^{-46}}} = \sqrt{4.55\times10^{25}} = 6.75\times10^{12}\ \text{rad s}^{-1}$$

In one line, $\tfrac12I\omega^2 = \tfrac23\cdot\tfrac12mv^2$ gives $\omega = v\sqrt{2m/3I}$, which is the same number.

Common mistakes

• Question 6.1: saying the centre of mass must lie inside the body. For a solid sphere or cube it does, which makes it feel like a rule. The ring shows it is not: its centre of mass is at the centre of the circle, where there is no material.
• Question 6.2: putting the centre of mass near the hydrogen. Students who divide $1.27$ by $35.5$, or expect the centre of mass to be roughly midway, place it near the light atom. It is a weighted average, so it sits close to the heavy chlorine nucleus: $1.24\ \text{Å}$ from the hydrogen, only $0.035\ \text{Å}$ from the chlorine.
• Question 6.8: using the sine for the vertical components. The angles are measured from the vertical, so the vertical components are $T\cos\theta$. Swapping sine and cosine gives $d = 1.28\ \text{m}$, which is the distance from the wrong end.
• Question 6.9: forgetting that each axle has two wheels, or loading the front more. The equations give the load on each axle, which is shared by two wheels. And it is the axle nearer the centre of gravity — the back one here — that carries more, not less.
• Question 6.12(b): assuming kinetic energy is conserved as well. Angular momentum is conserved because there is no external torque, but the child does work in pulling his arms in, so the kinetic energy rises by a factor of $2.5$.
• Question 6.13: using $\tfrac12MR^2$ for the hollow cylinder. That is the solid cylinder’s moment of inertia. In a thin hollow cylinder all the mass is at distance $R$, so $I = MR^2$; the wrong formula doubles the answers to $50\ \text{rad s}^{-2}$ and $20\ \text{m s}^{-2}$.
• Question 6.15: taking the hole’s mass as half the disc’s. Halving the radius quarters the area, so the hole has a quarter of the mass. Using $M/2$ puts the centre of gravity at $R/2$, on the rim, instead of $R/6$.
• Question 6.16: leaving out the stick’s own weight, or placing it at the knife edge. The stick’s weight acts at its centre of gravity, the $50.0\ \text{cm}$ mark found from the first balance, and its $5.0\ \text{cm}$ lever arm about the new knife edge is what balances the coins.

Practise next

• Chapter 7, Gravitation — the next chapter; the conservation of angular momentum used in 6.12 reappears as Kepler’s second law.
• Chapter 4, Laws of Motion — the force balances behind 6.8 and 6.9, and the $F = ma$ on which $\tau = I\alpha$ is modelled.
• Chapter 5, Work, Energy and Power — kinetic energy and power, which this chapter extends to $\tfrac12I\omega^2$ and $\tau\omega$.