Question

A flywheel of moment of inertia 10^7 gm cm^2 is rotating at a speed of 120 rotations per minute

Answer 1

min = 120 rev
5/2 rev = 60/120x5/2=1.25sec
t=1.25sec,
T=Iαα
ωω=ωωo+ααt
0=2ππ(120)/60+1.25αα
αα=10.05 rad/sec2
T=Iαα
=10x10.05=100.53N-m

Answer 2

Hey dear,
Posted question is incomplete.

◆ Complete question is -
A flywheel of moment of inertia 10^7 g cm? is rotating at a speed of 120 rotations per minute. Find the constant breaking torque required to stop the wheel in 5 roations.

◆ Answer-
τ = -2.514 Nm

◆ Explaination-
# Given-
I = 10^7 gcm^2 = 1 kgm^2
f1 = 120 rpm
f2 = 0
θ = 5 rotations.

# Solution-
Angular velocity is calculated by -
ω1 = 2πf = 2 × 3.142 × 2
ω1 = 12.57 rad/s

Using kinematic eqn for rotational motion-
ω2 = ω1 + αθ
0 = 12.57 + 5×α
α = -2.514 rad/s^2

Torque required is calculated by -
τ = I×α
τ = 1 × (-2.514)
τ = -2.514 Nm

Torque of 2.514 Nm in opposite direction is required to stop the wheel.

Hope this helps you...