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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-7 System Of Particles And Rotational Motion



Q.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.

 

Answer:-



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 = r0 μkmgrt / I)    …(v)
Equating equations (ii) and (v), we get:
μkgt = r0 μkmgrt / I)
= rω0 – μkmgr2t / I    ….(vi)
For the ring:
I = mr2
∴ μkgt = rω0 – μkmgr2t / mr2
= rω0 – μkgt
kgt = rω0
t = rω0 / 2μkg
= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8  =  0.80 s    ….(vii)
For the disc: I = (1/2)mr2
∴ μkgt = rω0 – μkmgr2t / (1/2)mr2
= rω0 – 2μkgt
kgt = rω0
t = rω0 / 3μkg
= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8  =  0.53 s   …..(viii)
Since td > tr, the disc will start rolling before the ring.