Q.323. Rotational Motion

Question 323.
A 2.05 kg mass is attached to a light cord that is wrapped around a pulley of radius 4.80 cm, which turns with negligible friction. The mass falls at a constant acceleration of 2.40 m/s2. Find the moment of inertia of the pulley.

Answer 323.
Let T = tension in the cord
=> mg - T = ma
=> T
= m(g - a)
= (2.05) * (9.81 - 2.40) N
= 15.1905 N

Torque on the pulley
= T * R
= (15.1905) * (0.048) meter-newton

Angular acceleration of the pulley
= a/R
= (2.40) / (0.048) rad/s^2

Moment of inertia of the pulley
= Torque / angular acceleration
= [(15.1905) * (0.048)] / (2.40) / (0.048)]
= 0.01458 kgm^2.

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