Suggested Solutions for Tutorial 7

SUGGESTED SOLUTIONS FOR TUTORIAL 7

(KINEMATICS OF ROTATION OF RIGID BODIES)

1. A flywheel of radius 20 cm starts from rest, and has a constant angular acceleration of 60 rad/s2. Find :

a. the magnitude of the net linear acceleration of a point on the rim after 0.15 s.

(Ans : a = 20.2 ms-2 )

b. the number of the revolutions completed in 0.25 s.

(Ans : 0.3 rev)

Sol.1.a) The tangential acceleration is constant and given by

at = r a

= ( 0.2 m ) ( 60 rad s-2 )

= 12 m s-2

To find the angular velocity at the given time,

w = wo + a t

= 0 + (60 rad s-2 ) ( 0.15 s )

= 9 rad s-1

radial acceleration ;

ar = w2 r

= ( 9 )2 ( 0.2 )

= 16.2 m s-2

The magnitude of the net linear acceleration is

a = √( ar2 + at2 )

= √( 16.22 + 122 )

= 20.2 m s-2

b) q = wot + 1 a t2 , wo = o

2

q = 1 a t2

2

= 1 ( 60 rad s-2) ( 0.25 s )2

2

= 1.88 rad

This corresponds to ;

= ( 1.88 rad ) ( 1 rev )

2p rad

= 0.3 rev

Figure 1

2.  The outer diameter of the tires on a bicycle is 70cm. When the bike is moving 3 m/s, what is the velocity of points P, Q and R in figure 1 with respect to

a.  center of mass of wheel (Ans : vP = 3ms-1, vQ = 0ms-1, vR = 3 m s-1 )

b.  the ground (Ans : vP = 6ms-1, vQ = 3ms-1, vR = 0 m s-1)

Sol.2

a) respect to the centre of mass

Vp = rw , VQ = 0 m s-1, VR = 3 m s-1

= 3 m s-1

b) respect to the ground

VP = 2 Vbycle , VQ = Vbycle , VR = 0 m s-1

= 2 ( 3 ) = 3 m s-1

= 6 m s-1

3. For the system of masses as shown in figure below, find the centre of mass.

2.0 kg

3 m

4 m

1.0 kg 3.0 kg

(Ans : 3.33 , 1.00)

Sol.3.

( 4,3 )

m3 = 2.0 kg

( 0 , 0 )

( 4,0 )

m1 = 1.0 kg m2 = 3.0 kg

Centre of mass

( Xcm , Ycm ) = ( å m¡ x¡ , å m¡ y¡ )

å m¡ å m¡

å m¡ x¡ = ( 1 ) ( 0 ) + ( 3 ) ( 4 ) + 2 ( 4 ) = (12 + 8)/ 6 = 20/6

å m¡ 1.0 + 3.0 + 2.0

= 3.33 m

å m¡ y¡ = ( 1 ) ( 0 ) + 3 ( 0 ) + 2 ( 3 )

å m¡ 1.0 + 3.0 + 2.0

= 6

6

= 1.00 m

Centre of mass = ( 3.33 , 1.00 )

4. A flywheel on a motor increases its rate of rotation uniformly from 240 rev min-1 to 360 rev min-1 in 20 s. Calculate :

a. its angular acceleration. ( Ans : 0.2p rad s-2 )

b. its angular displacements in this time. ( Ans : 200p rad )

Sol.4.

a) wo = 240 x 2p rad min-1 = 480p rad min-1 = 8p rad s-1

w = 360 x 2p rad min-1 = 720p rad min = 12 p rad s-1

we have t = 20 s, so angular acceleration is

a = change of angular velocity

time taken

= w − wo

t

= 12p − 8p

20

= 0.2p rad s-2

b) Angular displacement q is given by

q = Average angular velocity X time

= 1 ( w + wo ) x t

2

= 1 ( 12p + 8p ) x 20

2

= 200p rad

5.  A wheel starts from rest and attains a rotational velocity of 240 rev/s is 2.0 min, what is its average angular acceleration?

(Ans : a = 2.00 rev s-1)

Sol.5

We know that,

wo = 0 , wƒ = 240 rev s-1 , t = 2.0 min

= 120 s

from the definition of angular acceleration

a = wƒ − wo = ( 240 − 0 )rev s-1 = 2.00 rev s-1

t 120 s

6. Two particles of mass 2.5 kg and 6.0 kg are mounted 4.0 m apart on a light rod (whose mass can be ignored) as shown in figure below.

4.0 m 2.5 kg 6.0 kg

Calculate the moment of inertia of the system,

a. when rotated about an axis passing halfway between the masses.

b. when the system rotates about an axis located 0.5 m to the left of 2.5 kg mass.

(Ans : a. 34 kg m2 , b. 122.13 kg m2 )

Sol.6

6.

(a). 2.0 m 2.0 m

2.5 kg 6.0 kg

axis

Both particles are the same distance 2.0 m, from the axis rotation,

I = å m r2 = 2.5 ( 2.0 )2 + ( 6.0 ) ( 2.0 )2

= 34 kg m2

(b).

0.5 m

4.0 m

2.5 kg 6.0 kg

axis

The 2.5 kg mass is now 0.5 m from the axis and the 6.0 kg is 4.5 m from the axis.

I = å m r2

= 2.5 ( 0.5 )2 + ( 6.0 ) ( 4.5 )2 = 0.625 + 121.5

= 122.13 kg m2

7. Calculate the kinetic energy of a cylinder of mass 10kg and radius 0.10m if it is rolling along a plane with a translation velocity of 0.20ms-1 . The moment of inertia of the cylinder is 0.24kg m2 .

(Ans : 0.68J)

Sol.7

Refer to Fig 7.5.1. For there to be no sliding between point P on the wheel and the plane it touches, P must be stationary at the instant of contact. This means the translational velocity, v (forwards ) must be cancelled by speed rw in the opposite direction due to rotation of the wheel,

i.e.

v = rw

so, w = v

r

Now ;

( Total KE ) = Translational KE + Rotational KE

= 1 m v2 + 1 Iw2

2 2

we have , m = 10 kg , v = 0.20 m s-1 , I = 0.24 kg m2

r = 0.10 m

\ w = v = 0.2

r 0.10

= 2.0 rad s-1

Hence ;

Total KE = 1 ( 10 ) ( 0.2 )2 + 1 ( 0.24 ) ( 2.0 )2

2 2

= 0.68 J

8. The larger wheel in Fig. 7.5.2 has a mass of 60 kg and a radius r of 30 cm. It is driven by a belt as shown.

Figure 7.5.2

The tension in the upper part of the belt is 6.0 N, and that in the lower part is essentially zero.

a. How long does it takes for the belt to accelerate the larger wheel from rest to a speed of 3.0 rev / s ?

( Ans : 28.13 s )

b. How far does the wheel turn in this time ? ( Ans : 265.1 rad )

c.  What is the rotational kinetic energy of the wheel ? ( Ans : 79.9 J )

d. Calculate the power involved in the motion. ( Ans : 16.96 w )

Sol.8. First let us change the final speed of the wheel to radians per second :

wƒ = 3.0 rev s-1 = 6p rad s-1

(a) We use t = I a to find the acceleration of the wheel.

Because the applied force is 6.0 N with a lever arm of 0.30 m, the applied torque is

t = force x lever arm

= ( 6.0 N ) ( 0.30 )

= 1.8 N.m

The moment of inertia of a disk is

I disk = 1 m r2

2

= 1 ( 60 kg ) ( 0.30 m )2

2

= 2.7 kg m2

Then, from

t = I a

\ a = t

I

= 1.8 N m

2.7 kg m2

= 0.67 rad s-2

Now we can use ;

wƒ = wo + a t

\ t = wƒ − wo

a

= ( 6 p − 0 ) rad s-1

( 0.67 ) rad s-2

= 28.13 s

(b) The angular distance turned in this time is given by ;

q = w t

= 1 (wƒ + wo ) t

2

= 1 ( 6 p + 0 ) ( 28.13 )

2

= 265.1 rad

(c) We know that ;

KEr = 1 I w2

2

= 1 ( 2.7 kg m2 ) (6 p rad s-1 )2

2

= 79.9 J

or ;

using work energy theorem ;

w = ( KEr )ƒ − ( KEr )¡

= 79.9 − 0

= 79.9 J

(d) We know that ;

P = t

= t ( wo + wƒ )

2

= 1.8 ( 0 + 6 p )

2

= 16.96 W

9.  A turbine fan in a jet engine has a moment of inertia of 2.5 kg m2 about its axis of rotation. As the turbine is starting up, its angular velocity as a function of time is,

w = (400 rad/s3 ) t2

Find the fan’s angular momentum as a function of time and find its value at time t = 3.0 s.

(Ans : 9000 kg m2 s-1)

Sol.9. L = I w

= ( 2.5 kg m2 ) ( 400 rad / s3 ) t2

= ( 1000 kg m2 / s3 ) t2

At time, t = 3.0 s

L = ( 1000 kg m2 / s3 ( 3.0 s )2

= 9000 kg m2 s-1

10.  A merry-go-round of radius, R = 2.0 m has a moment of inertia, I = 250 kg m2 and is rotating at 10 rev/min. A 25 kg child jumps onto the edge of the merry-go-round. What is the new angular speed of merry-go-round?

(Ans : 0.75 rad/s)

Sol.10.

R = 2.0 m ; wo = 2 p x 10 = p rad s-1

I = 250 kg m2 60 3

m = 25 kg

å L¡ = å Lƒ

I wo = ( I + Ic ) w1 = ( I + MR2 ) w1

w1 = I wo

I + MR2

w1 = = 0.75 rad s-1

11. A roulette wheel turning at 3.0 rev / s coast to rest uniformly is 18.0 s. What is its deceleration? How many revolutions does it turn through?

(Ans : θ = 27 rev )

Sol.11

wo = 3.0 rev / s, w¦ = 0, t = 18.0 s, a = ?, q = ?

from the defining equation for a,

a = w¦ - wo = (0 - 3.0) rev / s

t 18.0 s

= - 0.167 rev s-2

q = wo t + 1 a t 2

2

= (3) (18) + 1 (- 0.167) (18) 2 = 27 rev

2