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).
