Question

A wheel of mass 40 kg and radius of gyration 0.5
E m comes to rest from as speed of 1800 revolutions
per minute in 30 s. The retardation torque
(in N-m) is
(1) -10 TI -- .--. .... (2) 20 .... .....
(3) 30
(4) 40
​

Answer 1

Answer: option (2). 20π N-m

Explanation:

Mass of the wheel, m = 40 kg

Radius of gyration, k = 0.5 m

No of revolutions = 1800 per minute

Time, t = 30 s

Let the initial angular acceleration be “ω₀” and the final angular acceleration be “ω” which is zero as the wheel comes to rest finally.

So, the instantaneous angular acceleration be “α” is given as

α = dω / dt = [ω- ω0] / t = [0 – {1800/60}*2π] / 30 = - [30*2π]/30 = - 2π rad/s²

Now, using Newton’s second law of Rotational motion,  

Retardation torque, τ = (moment of Inertia) * (instantaneous angular acceleration)

τ = I * α = m * k² * α = 40 * (0.5)² * (-2π) = - 20π N-m

Thus, the retardation torque is 20π N-m.