A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with an initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The coefficient of kinetic friction is µk = 0.2.

Asked by Abhisek | 1 year ago |  211

##### Solution :-

Given,

Radii of the ring and the disc, r = 10 cm  = 0.10 m
Initial angular speed, ω=10 π rad s–1
Coefficient of kinetic friction, μk = 0.2

The motion of the two objects is caused by the force of friction. According to Newton’s second,

the force of friction, f = ma
μkmg= ma
Where,
a = Acceleration produced in the disc and the ring
m = Mass
∴ a = μkg    . . . . .  .. . . . . .  ( 1 )

Using the first equation of motion :
v = u + at
= 0 + μkgt
= μkgt       . . . . . . . . . . . . . .   (2)

The frictional force applies a torque in perpendicularly outward direction and reduces the initial angular speed.
Torque, T= –Iα
Where, α = Angular acceleration
μkmgr = –Iα
∴ α = -μkmgr / I     . . .  . . . . . . (3)

According to the first equation of rotational motion, we have :
ω = ω0 + αt
= ω0 + (-μkmgr / I )t    . . . . . . . .(4)
Rolling starts when linear velocity, v = rω
∴ v = r (ω0 – μkmgrt / I )    …(5)

Using  equation ( 2 ) and equation ( 5 ), we have:
μkgt = r (ω0 – μkmgrt / I )
= rω0 – μkmgr2t / I    . . . . . . . . (6)

For the ring:
I = mr2
∴ μkgt = rω0 – μkmgr2t / mr2
= rω0 – μkgt
kgt = rω0
∴ t = rω0 / 2μkg
= ( 0.1 × 10 × 3.14) / (2 × 0.2 × 10 )  =  0.80 s    . . . . (7)

For the disc: I = ($$\dfrac{1}{2}$$)mr2
∴ μkgt = rω0 – μkmgr2t / ($$\dfrac{1}{2}$$)mr2
= rω0 – 2μkgt
kgt = rω0
∴ t = rω0 / 3μkg
= ( 0.1 x 10 × 3.14) / (3 × 0.2 × 9.8 )  =  0.53  s   …..(8)

Since tD > tR, the disc will start rolling before the ring.

Answered by Pragya Singh | 1 year ago

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