A cord of negligible mass is wound around the rim of a wheel of radius 20 cm. An object of mass 600 g is attached to the end of the cord and is allowed to fall from rest. It took 5 s for the object to fall by 2 m. Calculate the angular acceleration and moment of inertia of the wheel. Assume that the axis of wheel is horizontal.

Options

A. 25 rad/`sec^(2)`
B. 20 rad/`sec^(2)`
C. 10 rad/`sec^(2)`
D. 5 rad/`sec^(2)`

Answer

C

Explanation

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)`