Question

A wheel whose moment of inertial is 0.03kgm^(2), is accelerated from rest to 20rad//s in 5 s. When the external torque is removed, the wheel stops in 1 min. Find (a). The frictional torque. (b). The external torque.

Answer 1

The value of external torque is τ (external) = 0.12 N-m  and frictional torque is τf = 0.03 × 205 + 0.01

Explanation:

Given data:

• Moment of inertia "I" = 0.03 kgm^2

• Acceleration "a" = 20 rad/s

• Time = 5 second

Solution:

ωi = 0

ω f = 20 rad / s^2

τ = Iα

α = Δω / Δt = 20 - 0 / 5

α = 4 rad / s^2

τ (external) = 0.03 x 4 Nm

τ (external) = 0.12 N-m

Angular impulse  = change in angular momentum

τft = I x ω

or τf = Iω / t  + τf  

τf = 0.03 × 205 + 0.01

Hence the value of external torque is τ (external) = 0.12 N-m and frictional torque is τf = 0.03 × 205 + 0.01