Class 11 Physics

# Rotational Motion

## NCERT Solution

### Part 1

Question 1: Give the location of the center of mass of a (i) sphere, (ii) cylinder, (iii) ring and (iv) cube, each of uniform mass density. Does the center of mass of a body necessarily lie inside the body?

Answer: Center of mass of these objects is at their geometric centers because they have uniform mass density. It is not necessary to have center of mass inside a body. For example; the center of mass of a circular ring is at the center of the ring where there is no mass.

Question 2: In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 ÔB; (1 ÔB; = 10-10 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.

Answer: Let us assume that the nucleus of hydrogen atom is the origin.

Mass of hydrogen atom m1 = 1 unit

Mass of chlorine atom m2 = 35.5 unit

x_1=0 and x_2=1.27xx10^(-10) m

Center of mass of HCl molecule from the origin can be calculated as follows:

X=(m_1x_1+m_2x_2)/(m_1+m_2)

=(1xx0+35.5xx.127xx10^(-10))/(1+35.5) m

= (35.5xx1.27)/(36.5)xx10^(-10)m

= 1.235 × 10-10 m = 1.235 ÔB;

Question 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 trollley in any manner what is the speed of the CM of the (trolley + child) system?

Answer: The child and the trolley are working like a single system. So, there will be no change in velocity of the system because no external force is applied on the system.

Question 4: Show that the area of the triangle contained between the vectors a an b is one half of the magnitude of a × b.

Answer: Let us draw a triangle represented by a and b.

a is represented by OP and b is represented by OQ.

∠POQ = θ

Let us complete the parallelogram OPRQ and then join PQ.

Draw QN ⊥OP

In Δ OQN: sin θ = (QN)/(OQ) = (QN)/b

Or, QN = b sin θ

By definition: ab sin θ = a × b = (OP) (QN)

=2xx1/2xx\OP\xxQN

= 2 × area of Δ OPQ

So, area of Δ OPQ = 1/2 a × b proved

Question 5: Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a,b andc.

Answer: Let us make a parallelepiped on three vectors, so that OA = a, OB = b and OC = c

We have, b × c = bc sin 90° n = bc n

Where n is unit vector along OA perpendicular to the plane containing b and c.

Now, a.(b × c) = a . bc n

= (a) (bc) cos 0° = abc

This is equal in magnitude to the volume of parallelepiped.

Question 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 and z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Answer: Angular momentum l of a particle with position vector r and momentum p is given as follows:

l = r × p

But r = [xi + yj + zk]

And p = [pxi + pyj + pzk]

So, l = r × p

= [xi + yj + zk] × [pxi + pyj + pzk]

= (ypz - zpy)i + (zpx - xpz)j + (xpy = ypxk

From above equation, we can make following conclusions:

l_x=yp_z-zp_y

l_y=zp_x-xp_z

l_z=xp_y-yp_x

If the given particle moves only in x-y plane, then z = 0 and pz = 0

So, l = (xpy = ypz) k

This is only the z-component of l

This shows that for a particle moving only in x-y plane, angular momentum has only z-component.

Question 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.

Answer: Let us assume that two particles are at points P and Q as shown in this figure, and distance between them = d

LP = mv × 0 + mv × d = mvd ………. (1)

LQ = mv × d + mv × 0 = mvd ……….. (2)

Let us consider a third point R which is at a distance of y from Q

So, QR = y and PR = d – y

LR = mv × y + mv × (d – y) = mvd ………..(3)

This shows that: LP = LQ = LR Proved

Question 8: A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in this figure. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from its left end.

Answer: The following diagram shows tensions on the bar in various directions.

Length of bar l=2 m

T1 and T2 are the tensions respectively in the left and right strings.

At translational equilibrium of the bar:

T1 sin 36.9° = T2 sin 53.1°

Or, (T_1)/(T_2)=4/3

Or, T_1=4/3T_2

For rotational equilibrium (on taking the torque about center of gravity), we have

T1 cos 36.9° × d = T2 cos 53.1° (2 – d)

Or, T1× 0.8 d = T2 × 0.6 (2 – d)

Or, 4/3 × T2 × 0.8d = T2 (0.6 × 2 – 0.6d)

(3.2)/3d=1.2-0.6d

Or, 3.2d = 3.6 – 1.8d

Or, 3.2d + 1.8d = 3.6

Or, 5d = 3.6

Or, d=(3.6)/5=7.2 m

So, the center of gravity is at 7.2 m from left.

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

Answer: Given, mass m = 1800 kg

Distance d = 1.8 m

Distance of front axle from center of gravity = 1.05 m

So, distance of back axle from center of gravity = 1.8 – 1.05 = 0.75 m

Let us assume that Rf and Rb are the forces exerted by ground respectively on front and back axles.

At translational equilibrium:

R_f+R_b=mg

= 1800 × 9.8 =17640 N ………..(1)

For rotational equilibrium (torque about center of gravity)

R_f(1.05)=R_b(0.75)

Or, (R_b)/(R_f)=(1.05)/(0.75)=7/5=1.4

Or, R_b=1.4R_f ……………..(2)

Substituting this value in equation (1), we get:

1.4R_f+R_f=17640

Or, 2.4R_f=17640

Or, R_f=(17640)/(2.4)=7350 N

And, Rb = 17640 – 7350 = 10290 N

So, force exerted on each front wheel =(7350)/2=3675 N

Force exerted on each back wheel =(10290)/2=5145 N

Question 10:

(a) 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 diameter to be (2MR^2)/5, where M is the mass of the sphere and R is the radius of the sphere.

Answer: Given, moment of inertia of a sphere about its diameter =(2MR^2)/5

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

So, moment of inertia about a tangent of the sphere

=(2MR^2)/5+MR^2=(7MR^2)5

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be (MR^2)/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer: Moment of inertia of a disc about its diameter =(MR^2)/4

According to the theorem of perpendicular axis, moment of inertia of a lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

So, moment of inertia of disc about its center

=(MR^2)/2+(MR^2)/4=(MR^2)/2

This figure shows the situation.

On applying the theorem of parallel axes:

MI =(MR^2)/2+MR^2=(3MR^2)/2