Question

# what torque required stopping awheel of moment of inertia 6 × 10^-3kgm2 from speed of 40rad/s in 20 sec.

Answer

0.12 Nm

- Identify the known variables. A: moment of inertia = 6 × 10^-3kgm2, initial angular velocity = 40 rad/s, time = 20 sec.
- Use the kinematic equation, ωf = ωi + αt, where ωf = 0, α = -τ/I, and solve for torque (τ). A: τ = -I(ωf - ωi)/t = -6 × 10^-3 kgm2(0 - 40 rad/s)/20 s = 0.12 Nm.