A wheel of radius (8 ~cm ) is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of (0.4 ~kg ) attached to its free end. When the mass is released, it descends through (1 ~m ) in 10 seconds, then its moment of inertia is (Acceleration due to gravity, (g=10 ~ms ⁻² ))

Step-by-step solution

Given, radius of wheel, (R=8 10⁻² ~m ), mass of weight, (m=0.4 ~kg ), descending length, (L=1 ~m ) and time, (t=10 ~s ) As, torque ( =F . R=m g R ) putting the given values, we get (=0.4 10 8 10⁻² ) So, ( =0.32 Nm ) Let the wheel is rotated by an angle ( ), ( = arc radius = 1 0.08 =12.5 rad ) So, from the equation of rotational motion, ( aligned & = ₀ t+ t^2 2 & = 2 t^2 = 2 12.5 10 10 =0.25 rad s ⁻² aligned ) ( [ ₀=0 ] ) Now, as Torque, ( =I ) ( ) Moment of inertia of a wheel, (I= = 0.32 0.25 =1.28 ~kg - m ^2 ) Hen