### 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 *μ*_{k} = 0.2.

**Answer:-**Radii of the ring and the disc, *r* = 10 cm = 0.1 m

Initial angular speed, ω_{0 }=10 π rad s^{–1}

Coefficient of kinetic friction, *μ*_{k} = 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*

μ_{k}*m*g= *ma*

Where,

*a* = Acceleration produced in the objects

*m *= Mass

∴ *a* = *μ*_{k}*g* … **(i)**

As per the first equation of motion, the final velocity of the objects can be obtained as:

*v* = *u* + *at*

= 0 + *μ*_{k}*gt*

= *μ*_{k}*gt* … **(ii)**

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed.

Torque, τ= –*Iα*

α = Angular acceleration

μ_{k}*m*g*r* = –*Iα*

∴ *α = –*μ_{k}*mgr* / *I* …..**(iii)**

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

ω = ω_{0} + α*t*

= ω_{0} + (*–*μ_{k}*m*g*r* / *I*)*t* ….**(iv)**

Rolling starts when linear velocity, *v* = *r*ω

∴ *v* = *r* (ω_{0} *– *μ_{k}*mgrt* / *I*) …**(v)**

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

μ_{k}*gt* = *r* (ω_{0} *– *μ_{k}*mgrt* / *I*)

= rω_{0} – μ_{k}*mgr*^{2}*t* / *I* ….**(vi)**

For the ring:

*I* = *mr*^{2}

∴ μ_{k}*gt* = *r*ω_{0} – μ_{k}*m**gr*^{2}t / *mr*^{2}

= *r*ω_{0} – μ_{k}gt

2μ_{k}*gt* = *r*ω_{0}

∴ *t* = rω_{0} / 2μ_{k}*g*

= 0.1 × 10 × 3.14 / 2 × 0.2 × 9.8 = 0.80 s ….**(vii)**

For the disc: I = (1/2)*mr*^{2}

∴ μ_{k}*gt* = *r*ω_{0} – μ_{k}*m**gr*^{2}*t* / (1/2)*mr*^{2}

= *r*ω_{0} – 2μ_{k}*gt*

3μ_{k}*gt* = rω_{0}

∴ *t* = *r*ω_{0} / 3μ_{k}*g*

= 0.1 × 10 × 3.14 / 3 × 0.2 × 9.8 = 0.53 s …..**(viii)**

Since *t*_{d} > *t*_{r}, the disc will start rolling before the ring.