Question

A wheel has moment of inertia 5x 10-3 kg-m² and is making 20 revolutions per second. The magnitude of torque needed to stop it in 10 s is:

A 4.5xx 10-2 N-m
B.2.5xx 10-2 N-m
C. 2x 10-2 N-m
D. 4xx 10-2 N-m

Answer 1

Answer:

F = M a Newton's second law

Γ = I α Corresponding law for circular motion

α = 20 * 2 π rad / sec / 10 sec = 4 π rad / sec^2

α = 12.6 /sec^2 (rad not an actual unit)

Γ = 5E-3 * 12.6E1 = .63 kg-m^2 / sec^2

(5E-3 / 12.6 = 4E-4 = 4xxE-2 seems like a mistake made in arithmetic)

Note: kg-m^2 * 1 / sec^2 = kg-m^2 / sec^2 = N-m

One needs to "multiply" inertia * rad/sec^2 to get required units of torque