Question

A firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 1.5 kg and radius 7.40 cm which operates at 730 rev/min. When the power is shut off, you time the grindstone and find it takes 38.8 s for it to stop rotating.
(a) What is the angular acceleration of the grindstone in rad/s^2? (Assume constant angular acceleration.)


(b) What is the frictional torque exerted on the grindstone in N·m?

Answer 1

Answer:

a) -1.97 rad/sec² b) -8.09*10⁻³ N.m

Explanation:

a) Assuming a constant angular acceleration, we can apply the definition of angular acceleration, as follows:

γ = (ωf -ω₀) / t

We know that the final state of the grindstone is at rest, so ωf =0

In order to be consistent in terms of units, we can convert ω₀ from rev/min to rad/sec:

ω₀ = 730 rev/min* (1 min/60 sec)* (2*π rad / 1 rev) = 73/3*π

⇒ γ = (0-73/3*π) / 38.8 sec = - 1.97 rad/sec²

b) In order to get the value of the frictional torque exerted on the grindstone, that caused it to stop, we can apply the rotational equivalent of the Newton's 2nd law, as follows:

τ = I * γ (1)

As the grindstone can be approximated by a solid disk, the rotational inertia I can be expressed as follows:

I = m*r² / 2, where m=1.5 kg and r = 0.074 m.

Replacing in (1) , m. r and γ (the one we calculated in a)), we get:

τ = (1.5 kg* (0.074)² m² / 2) * -1.97 rad/sec = -8.09*10⁻³ N.m

(The negative sign implies that the frictional torque opposes to the rotation of the grindstone).