A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad-s^-2 produced in it due to the torque. Then, moment of inertia of the wheel is (g =10 ms^-2 )

2 kg-m²

1 kg-m²

4 kg-m²

8 kg-m²

Solution

2 kg-m²

Torque is defined as the moment of force or turning effect of force or turning effect of force about the given axis or point. It is measured as the cross product of position and force vector while as angular momentum is the rotational analogue of linear momentum.

Given:-

r = 0.4 m

Angular acceleration $α = 8 rad s^{- 2}$

m = 4 kg

l = ?

Torque $τ = I α$

But we know Τ = Fr

and The equation for the force of gravity is ( F = mg ) where g is the acceleration due to gravity.

mgr = $I α$

4 × 10 × 0.4 = I × 8

⇒ $I = \frac{16}{8}$

I = 2 kg.m²

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