A 2.30 kg grinding wheel is in the form of a solid cylinder of radius 0.200 m .

Question

A 2.30 kg grinding wheel is in the form of a solid cylinder of radius 0.200 m .

a-What constant torque will bring it from rest to an angular speed of 1700 rev/min in 2.80 s ?

b-Through what angle has it turned during that time?

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

d-What is the grinding wheel's kinetic energy when it is rotating at 1700 rev/min ?

e-

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

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

Explanation / Answer

I = ½mr² = ½ * 2.30 kg * (0.200m)² = 0.046 kg·m²
= 1700 rev/min = 178.023 rad/sec
= - / t = -178.023 rad/s / 2.80 sec = 63.58 rad/s²

A) = I = -0.046kg·m² * 63.58 rad/s² = -2.9246 N·m (that is, against the initial velocity)

B) = ½t² = ½ * 63.58 rad/s² * (2.8s)² = 249.23 rads

C) W = = -2.9246N·m * 249.23rads = -728.90 J

D) KE = ½I² = ½ * 0.046 kg·m² * (178.023rad/s)² = 728.92 J

E) They compare well.

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