__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*μ*_{k}= 0.2.

__Solution__: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}g

*t*

=

*μ*

_{k}g

*t*… (

*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}

*m*g

*r*/ I .....(iii)

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

ω = ω

_{0}+ αt

= ω

_{0}+ (

*-*μ

_{k}

*m*g

*r*/ I)t ....(iv)

Rolling starts when linear velocity,

*v*=

*r*ω

∴ v = r (ω

_{0}

*-*μ

_{k}

*m*grt / I) ...(v)

Equating equations (

*ii*) and (

*v*), we get:

μ

_{k}gt = r (ω

_{0}

*-*μ

_{k}

*m*grt / I)

= rω

_{0}- μ

_{k}

*m*gr

^{2}t / I ....(vi)

For the ring:

I = mr

^{2}

∴ μ

_{k}gt = rω

_{0}- μ

_{k}

*m*gr

^{2}t / mr

^{2}

= rω

_{0}- μ

_{k}gt

2μ

_{k}gt = rω

_{0}

∴ t = rω

_{0}/ 2μ

_{k}g

= 0.1 X 10 X 3.14 / 2 X 0.2 X 9.8 = 0.80 s ....(vii)

For the disc: I = (1/2)mr

^{2}

∴ μ

_{k}gt = rω

_{0}- μ

_{k}

*m*gr

^{2}t / (1/2)mr

^{2}

= rω

_{0}- 2μ

_{k}gt

3μ

_{k}gt = rω

_{0}

∴ t = rω

_{0}/ 3μ

_{k}g

= 0.1 X 10 X 3.14 / 3 X 0.2 X 9.8 = 0.53 s .....(viii)

Since

*t*

_{d}>

*t*

_{r}, the disc will start rolling before the ring.

## No comments:

## Post a Comment