Question

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2Question 7.30. System Of Particles And Rotational Motion

Answer 1

Radius of disc and ring (R) = 10cm = 0.1 m
initial angular speed (wo) = 10π rad/s
Coefficient of Kinetic friction (u) = 0.2

Moment of inertia of disc (Id)= mR²/2
Moment of inertia of ring (Ir)= mR²
We know, frictional force responsible for the motion of body .
So, F = fr = ma ---------(1)
Where a is linear acceleration and N is normal reaction .
umg = ma
a = ug ----------(2)
Let ā is the angular acceleration.
We know,
Torque = Iā
Also torque = R×F use this here,
R×F = Iā
Rumg = Iā
ā = umgR/I

For disc
___________
ā = umgR/mR²/2 = 2ug/R

For ring
___________
ā = umgR/mR² = ug/R

Now,use linear equation,
V = u + at
Initial velocity (u) = 0
V = at
V = ugt { from equation (2)
For rolling motion ,
V = wR
w = v/R = ugt/R -------(4)

Let t1 time taken by ring and t2 time taken by disc .
use , angular equation ,
w = wo + āt

For disc
___________
w = wo + āt2
ugt2/R = wo -2ugt2/R { from equation (4) , ā = -2ug/R frictional torque oppses the motion of the body }

t2 = Rwo/3ug
Put R , wo , u and g values
t2 = 0.1 × 10π/3×0.2 ×9.8 = 0.53 s

For ring
___,_________
w = wo + āt1
ugt1/R = wo - ugt1/R
t1 = Rwo/2ug
Put R, wo , u and g values
t1 = 0.1 × 10π/2×0.2×9.8 =0.8 s

It is clear that disc will start rolling earlier .