A 2.50-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m. P

Question

A 2.50-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m.

Part A

What constant torque will bring it from rest to an angular speed of 1200 rev/min in2.5 s?

Express your answer with the appropriate units.

Part B

Through what angle has it turned during that time?

Part C

Use equation W=z(21)=z to calculate the work done by the torque.

Express your answer with the appropriate units.

Part D

What is the grinding wheel’s kinetic energy when it is rotating at 1200 rev/min?

Express your answer with the appropriate units.

Part E

Compare your answer in part (D) to the result in part (C).

Explanation / Answer

I = ½mr² = ½ * 2.5kg * (0.1m)² = 0.0125 kg·m²
= 1200rev/min * 1min/60s * 2rads/rev = 125.66 rad/s
= - / t = -125.66rad/s / 2.5s = 50.264 rad/s²

A) = I = -0.0125kg·m² * 50.264rad/s² = -0.6283 N·m (that is, against the initial velocity)

B) = ½t² = ½ * 50.264rad/s² * (2.5s)² = 157.075 rads

C) W = = -0.6283N·m * 157.075rads = -98.69 J

D) KE = ½I² = ½ * 0.0125kg·m² * (125.66rad/s)² = 98.69 J

E) They compare well.

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