CLASS 11 · PHYSICS · CHAPTER 6

SYSTEM OF PARTICLES AND ROTATIONAL MOTION

Simple answers to all 17 end-of-chapter questions on centre of mass, torque, angular momentum and rotation.

17 worked answersComplete chapter answers

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17 worked answers · short steps first, explanations when you need them

Chapter 6 Exercises

Textbook pages 101–103

Question 1

Where is the centre of mass of a uniform sphere, cylinder, ring and cube? Must a body’s centre of mass lie inside its material?

Steps only

Symmetry places the centre of mass at each shape’s geometric centre: the sphere’s centre, the cylinder’s midpoint on its axis, the ring’s centre, and the cube’s centre.

Answer

At each shape’s geometric centre. No; for example, a ring’s centre of mass is in its empty middle.

Question 2

In an HCl molecule, the nuclei are 1.27 Å apart. Chlorine is 35.5 times as massive as hydrogen. Find the molecule’s centre of mass, treating the nuclei as point masses.

Steps only

Put H at x=0 and Cl at x=1.27 Å. The centre of mass is X=(mH·0+mCl·1.27)/(mH+mCl).

Answer

The centre of mass is about 1.235 Å from H, or 0.035 Å from Cl.

Question 3

A child sits on a trolley moving at speed V along a smooth horizontal floor. The child stands and runs around on the trolley. What is the speed of the trolley-child system’s centre of mass?

Steps only

There is no net external horizontal force on the trolley-child system, so its total horizontal momentum and centre-of-mass velocity stay constant.

Answer

V; the centre-of-mass speed does not change.

Question 4

Show that the area of a triangle formed by vectors a and b is half the magnitude of a×b.

Steps only

The parallelogram made by a and b has base |a| and perpendicular height |b|sinθ, so its area is |a||b|sinθ.

Answer

Atriangle=½|a×b|.

Question 5

Explain why |a·(b×c)| gives the volume of the parallelepiped formed by vectors a, b and c.

Steps only

The magnitude |b×c| is the area of the base parallelogram. The component of a perpendicular to that base is the height.

Answer

Volume=|a·(b×c)|.

Question 6

A particle has position r=(x,y,z) and momentum p=(px,py,pz). Find the components of angular momentum and show that for motion in the xy-plane only the z-component remains.

Steps only

Angular momentum is L=r×p. Expanding the cross product gives Lx=ypz−zpy, Ly=zpx−xpz and Lz=xpy−ypx.

Answer

L=(ypz−zpy, zpx−xpz, xpy−ypx). In the xy-plane, only Lz can be non-zero.

Question 7

Two particles of mass m and speed v move in opposite directions along parallel lines separated by distance d. Show that the total angular momentum is the same about any chosen point.

Steps only

Their momenta are equal and opposite, so total momentum is zero. Shifting the origin by vector a changes total angular momentum by −a×Ptotal.

Answer

The angular momentum is independent of the chosen point; its magnitude is mvd.

Question 8

A 2 m non-uniform bar hangs at rest from two strings. The strings make 36.9° and 53.1° with the vertical at the left and right ends. Find the centre of gravity’s distance d from the left end.

Steps only

Let the left and right tensions be TL and TR. Horizontal balance gives 0.6TL=0.8TR, so TL=4TR/3. Vertical balance gives 0.8TL+0.6TR=W, hence TR=0.6W.

Answer

The centre of gravity is 0.72 m from the left end.

Question 9

A car has mass 1800 kg, axle separation 1.8 m and centre of gravity 1.05 m behind the front axle. Find the ground’s upward force on each front wheel and on each back wheel. Use g=9.8 m/s².

Steps only

Let total upward reactions at the front and back axles be Ff and Fb. Vertical balance gives Ff+Fb=mg. Taking moments about the front axle: Fb×1.8=mg×1.05.

Answer

Each front wheel: about 3.68×10³ N. Each back wheel: about 5.15×10³ N.

Question 10

Equal torques act for equal times on a hollow cylinder and a solid sphere of the same mass and radius. Which gains greater angular speed? The cylinder rotates about its symmetry axis; the sphere about a diameter.

Steps only

For the same torque, angular acceleration α=τ/I. The hollow cylinder has I=MR²; the solid sphere has I=2MR²/5.

Answer

The solid sphere gains greater angular speed.

Question 11

A 20 kg solid cylinder of radius 0.25 m spins at 100 rad/s. Find its rotational kinetic energy and angular momentum about its axis.

Steps only

For a solid cylinder, I=½MR²=½×20×0.25²=0.625 kg·m².

Answer

Rotational kinetic energy: 3125 J. Angular momentum: 62.5 kg·m²/s.

Question 12

A child spins on a frictionless turntable at 40 rev/min with arms out. Pulling the arms in reduces the child’s moment of inertia to 2/5 of its original value. (a) Find the new angular speed. (b) Why does rotational kinetic energy increase?

Steps only

With no external torque, angular momentum Iω is conserved. Therefore ωf=(Ii/If)ωi=(5/2)×40=100 rev/min.

Answer

(a) 100 rev/min. (b) Kinetic energy rises to 2.5 times its initial value because the child’s muscles do work.

Question 13

A 3 kg hollow cylinder of radius 0.40 m has a light rope wound around it. The rope is pulled with 30 N without slipping. Find the cylinder’s angular acceleration and the rope’s linear acceleration.

Steps only

For a hollow cylinder, I=MR²=3×0.40²=0.48 kg·m². The torque is τ=FR=30×0.40=12 N·m.

Answer

Angular acceleration: 25 rad/s². Rope acceleration: 10 m/s².

Question 14

An engine applies torque 180 N·m to keep a rotor turning at 200 rad/s. Find the power required.

Steps only

Rotational power is P=τω.

Answer

36 kW.

Question 15

A uniform disk of radius R has a circular hole of radius R/2 removed. The hole’s centre is R/2 from the disk’s centre. Locate the centre of gravity of the remaining disk.

Steps only

Treat the removed hole as negative mass. If the complete disk has mass M, the hole has mass M/4. Put the original centre at x=0 and the hole centre at x=R/2.

Answer

The centre of gravity is R/6 from the original centre, on the side opposite the hole.

Question 16

A metre stick balances at its 50 cm mark. Two 5 g coins are placed at 12 cm, and the balance point moves to 45 cm. Find the stick’s mass.

Steps only

Take moments about the new balance point, 45 cm. The stick’s weight acts at 50 cm, 5 cm to the right; the 10 g of coins act 33 cm to the left.

Answer

66 g.

Question 17

An oxygen molecule has mass 5.30×10⁻²⁶ kg and moment of inertia 1.94×10⁻⁴⁶ kg·m². Its mean translational speed is 500 m/s. If rotational kinetic energy is 2/3 of translational kinetic energy, find its average angular speed.

Steps only

Translational energy is ½mv². Rotational energy is (2/3)(½mv²)=⅓mv², and it also equals ½Iω².

Answer

About 6.75×10¹² rad/s.

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SYSTEM OF PARTICLES AND ROTATIONAL MOTION

Chapter 6 Exercises

Where is the centre of mass of a uniform sphere, cylinder, ring and cube? Must a body’s centre of mass lie inside its material?

Answer

In an HCl molecule, the nuclei are 1.27 Å apart. Chlorine is 35.5 times as massive as hydrogen. Find the molecule’s centre of mass, treating the nuclei as point masses.

Answer

A child sits on a trolley moving at speed V along a smooth horizontal floor. The child stands and runs around on the trolley. What is the speed of the trolley-child system’s centre of mass?

Answer

Show that the area of a triangle formed by vectors a and b is half the magnitude of a×b.

Answer

Explain why |a·(b×c)| gives the volume of the parallelepiped formed by vectors a, b and c.

Answer

A particle has position r=(x,y,z) and momentum p=(px,py,pz). Find the components of angular momentum and show that for motion in the xy-plane only the z-component remains.

Answer

Two particles of mass m and speed v move in opposite directions along parallel lines separated by distance d. Show that the total angular momentum is the same about any chosen point.

Answer

A 2 m non-uniform bar hangs at rest from two strings. The strings make 36.9° and 53.1° with the vertical at the left and right ends. Find the centre of gravity’s distance d from the left end.

Answer

A car has mass 1800 kg, axle separation 1.8 m and centre of gravity 1.05 m behind the front axle. Find the ground’s upward force on each front wheel and on each back wheel. Use g=9.8 m/s².

Answer

Equal torques act for equal times on a hollow cylinder and a solid sphere of the same mass and radius. Which gains greater angular speed? The cylinder rotates about its symmetry axis; the sphere about a diameter.

Answer

A 20 kg solid cylinder of radius 0.25 m spins at 100 rad/s. Find its rotational kinetic energy and angular momentum about its axis.

Answer

A child spins on a frictionless turntable at 40 rev/min with arms out. Pulling the arms in reduces the child’s moment of inertia to 2/5 of its original value. (a) Find the new angular speed. (b) Why does rotational kinetic energy increase?

Answer

A 3 kg hollow cylinder of radius 0.40 m has a light rope wound around it. The rope is pulled with 30 N without slipping. Find the cylinder’s angular acceleration and the rope’s linear acceleration.

Answer

An engine applies torque 180 N·m to keep a rotor turning at 200 rad/s. Find the power required.

Answer

A uniform disk of radius R has a circular hole of radius R/2 removed. The hole’s centre is R/2 from the disk’s centre. Locate the centre of gravity of the remaining disk.

Answer

A metre stick balances at its 50 cm mark. Two 5 g coins are placed at 12 cm, and the balance point moves to 45 cm. Find the stick’s mass.

Answer

An oxygen molecule has mass 5.30×10⁻²⁶ kg and moment of inertia 1.94×10⁻⁴⁶ kg·m². Its mean translational speed is 500 m/s. If rotational kinetic energy is 2/3 of translational kinetic energy, find its average angular speed.

Answer

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