Q. 305.0( 4 Votes )

# A solid disc and

Answer :

Given,

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

Angular velocity of the disc and ring, ω_{0} = 10 π rad/s

Initial velocity of both the objects, u = 0 m/s

Coefficient of kinetic friction, μ_{k} = 0.2

Motion of the objects start due to frictional force,

f = ma

Where f = frictional force ,

⇒ f = μ_{k}mg

m = mass of the body

a = acceleration of the body

∴ μ_{k}mg = ma

⇒ a = μ_{k}g ------ > (1)

We have final velocity as per first equation of motion,

v = u + at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

∴ v = μ_{k}g ------- > (2)

The torque generated by friction initially causes reduction in angular velocity. Thus,

T = -Iα

Where,

I = moment of inertia of the body

α = Angular acceleration

and, T = fr

Where,

f = frictional force

r = radius

∴ T = -μ_{k}mgr

⇒ ------- > (3)

As per the frist equation of rotational motion, final angular velocity,

ω = ω_{0} + αt

Where,

ω_{0} = intial angular velocity

α = angular acceleration

t = time

∴ ------ > (4)

Rolling starts when linear velocity, v = rω

∴ -------- > (5)

From the equations 2 & 5 we get,

⇒ ------- > (6)

We know that,

For the ring, I = mr^{2}

∴

⇒ μ_{k}gt = rω_{0} – μ_{k}gt

⇒ 2μ_{k}gt = rω_{0}

∴

⇒ t_{ring}

⇒ t_{ring} = 0.8s

For the disc, I = 0.5mr^{2}

∴

⇒

⇒ 3μ_{k}gt = rω_{0}

∴

⇒ t_{disc}

Since the t_{disc} < t_{ring.}

Therefore, the disc will start rolling before the ring.

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