System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.30

Question 7.30:
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 = 0.2.
Solution:
Disc
Radii of the ring and the disc, r = 10 cm = 0.1 m
Initial angular speed, ω0 =10 π rad s–1
Coefficient of kinetic friction, μk = 0.2
Initial velocity of both the objects, u = 0
Motion of the two objects is caused by frictional force. As per Newton’s second law of motion, we have frictional force, f = ma
μkmg= ma
Where,
a = Acceleration produced in the objects
m = Mass
∴ a = μkg … (i)
As per the first equation of motion, the final velocity of the objects can be obtained as:
v = u + at
= 0 + μkgt
= μkgt … (ii)
The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.
Torque, τ= –
α = Angular acceleration
μkmgr = –
α = -μkmgr / I     .....(iii)
Using the first equation of rotational motion to obtain the final angular speed:
ω = ω0 + αt
= ω0 + (-μkmgr / I)t    ....(iv)
Rolling starts when linear velocity, v = rω
∴ v = r (ω0 - μkmgrt / I)    ...(v)
Equating equations (ii) and (v), we get:
μkgt = r (ω0 - μkmgrt / I)
= rω0 - μkmgr2t / I    ....(vi)
For the ring:
I = mr2
μkgt = rω0 - μkmgr2t / mr2
= rω0 - μkgt
2μkgt = rω0
∴ t = rω0 / 2μkg
= 0.1 X 10 X 3.14 / 2 X 0.2 X 9.8  =  0.80 s    ....(vii)
For the disc: I = (1/2)mr2
μkgt = rω0 - μkmgr2t / (1/2)mr2
= rω0 - 2μkgt
3μkgt = rω0
∴ t = rω0 / 3μkg
= 0.1 X 10 X 3.14 / 3 X 0.2 X 9.8  =  0.53 s   .....(viii)
Since td > tr, the disc will start rolling before the ring.

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