19 September, 03:26

# 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 2.0 kg and radius 9.90 cm which operates at 720 rev/min. When the power is shut off, you time the grindstone and find it takes 41.9 s for it to stop rotating.(a) What is the angular acceleration of the grindstone in rad/s^2? (b) What is the frictional torque exerted on the grindstone in N·m?

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1. 19 September, 03:54
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b) - 17.6 x 10ˉ³Nm

Explanation:

ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s

a) Assuming a constant angular acceleration, the formula will be

α = (ωf - ω₀) / t

As final state of the grindstone is at rest, so ωf = 0

⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²

b) Moment of inertia I for a disk about its central axis

I = ½mr²

where m=2kg and radius 'r' = 0.099m

I = ½ (2) (0.099²)

I = 9.8 x 10ˉ³ kgm²

Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:

τ = I α = > (9.8 x 10ˉ³) ( - 1.799)

τ = - 17.6 x 10ˉ³Nm

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