﻿ 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. from Physics System of Particles and Rotational Motion Class 11 Manipur Board

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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.

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω=10 π rad s–1

Coefficient of kinetic friction, μk = 0.2

Initial velocity of both the objects, u = 0

The frictional force causes motion between two objects.

As per Newton’s second law of motion,

Frictional force,
f =ma

μkmg= ma

where,

a = Acceleration produced in the objects

= Mass

Therefore,

a = μkg                                           … (i)

As per the first equation of motion,

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= – I α

where,

α = Angular acceleration

μkmgr = –I α

α =                                    ...(iii)

Using the first equation of rotational motion to obtain the final angular speed,

ω = ω0 + α

= ω0 + (-μkmgr / I)t                          ...(iv)

Rolling starts when linear velocity,

v = rω

Therefore,

v = r (ω0 μkmgrt / I)                         ...(v)

Equating equations (ii) and (v), we get

μkgt = r (

= rω0 -                           ...(vi)

For the ring,

I = mr

∴ μkgt = rω0 -
= rω0 - μkgt

2μkgt = rω

∴    t =

=

=0.80 s                                       ...(vii)

For the disc:

I = mr2

∴    μkgt = rω0 - μkmgr2t / mr

= rω0 - 2μkgt

3μkgt = rω

Therefore,

t =

=       ...(viii)

Since td > tr, the disc will start rolling before the ring.
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