(a) What is the angular momentum of a 2.7 kg uniform cylindrical grinding wheel of radius 18 cm when rotating?

Question

(a) What is the angular momentum of a 2.7 kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1300 rpm?

---kg·m2/s

(b) What magnitude of torque is required to stop it in 6.0 s?

---N·m

Answer 1

Let:

I be the moment of inertia,

w be the angular velocity,

L be the angular momentum,

T be the torque,

a be the angular acceleration,

t be the time taken to stop.

(a)

I = 2.7 * 0.18^2 / 2 kg m^2

w = 1300 * 2pi / 60 rad / s.

L = Iw

= 5.95 kg m^2 / s.

(b)

T = L / t

= L / 6.0

= 0.992 Nm.