1.90-kg grinding wheel is in the form of a solid cylinder of radius 0.200 m. Wha

Question

1.90-kg grinding wheel is in the form of a solid cylinder of radius 0.200 m. What constant torque will bring it from rest to an angular speed of 1050 rev/min in 2.5 s? Through what angle has it turned during that time? rad (Use Eq. (10.21) to calculate the work done by the torque.W = Tz(Theta2 - Theta1) = Tz Delta Theta (work done by a constant torque) (10.21) What is the grinding wheel's kinetic energy when it is rotating at 1050 rev/min? Compare your answer to the result in part (c). The answer to part (d) is greater. The answer to part (c) is greater. The two answers are the same.

Explanation / Answer

I = ½mr² = ½ * 1.9kg * (0.2m)² = 0.038 kg·m²
= 1050 rev/min * 1min/60s * 2 rads/rev = 109.96 rad/s
= - / t = -109.96 rad/s / 2.5 s = 43.98 rad/s²

A) = I = -0.038 kg·m² * 43.98 rad/s² = - 1.67 N·m (that is, against the initial velocity)

B) = ½t² = ½ * 43.98rad/s² * (2.5s)² = 137.44 rads

C) W = = -1.67N·m * 137.44 rads = -229.52 J

D) KE = ½I² = ½ * 0.038kg·m² * (109.96 rad/s)² = 229.73 J

E) They compare well.

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