Question

a grinding wheel of mass 5.0kg and diameter 0.4 m is rotating with an angular speed of 2 rev S-1,calculate the torque which will increase its angular speed to 8 revolution per second in 2s.

Answer 1

torque=moment of inertia* alpha

alpha= 8-2/2=3rev/s

moment of inertia(if considered ring)=mr^2=5*0.2^2=0.2 units

torque=0.2*3=0.6 Nm

Answer 2

Hello dear,

◆ Answer-
τ = 251.33 Nm

◆ Explanation-
# Given-
m = 5 kg
r = 4/2 = 2 m
f1 = 2 rps
f2 = 8 rps
t = 2 s

# Solution-
Rotational acceleration is-
α = (ω2-ω1) / t
α = 2π(f2-f1) / t
α = 2π(8-2) / 2
α = 4π rad/s^2

Moment of inertia is -
I = mr^2
I = 5 × 2^2
I = 20 kgm^2

Rotational torque is calculated by-
τ = Iα
τ = 20 × 4π
τ = 251.33 Nm

Therefore, torque required is 251.33 Nm.

Hope this helps you...