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A cord is wound over the rim of a flywhe...

A cord is wound over the rim of a flywheel of mass 20 kg and radius 25 cm. A mass `2.5` kg attached to the cord is allowed to fall under gravity. Calculate the angular acceleration of the flywheel

A

25 rad/`sec^(2)`

B

20 rad/`sec^(2)`

C

10 rad/`sec^(2)`

D

5 rad/`sec^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Moment of inertia of flywheel about its axis
`I= (MR^(2))/(2) = (20 xx (.25)^(2))/(2)= 625 xx 10^(-3) kgm^(2)`

Torque acting on the flywheel
`tau= FR = mgR = 2.5 xx 10 xx 25 = 625 xx 10^(-2)m`
`alpha= (tau)/(I)= (625 xx 10^(-2))/(625 xx 10^(-3)) = 10 "rad"//s^(2)`
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