9. A 1.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m
ID: 1501039 • Letter: 9
Question
9. A 1.80-kg grinding wheel is in the form of a solid cylinder of radius 0.100 m.
(a) What constant torque will bring it from rest to an angular speed of 1450 rev/min in 3.0 s?
N · m
(b) Through what angle has it turned during that time?
rad
(c) Use Eq. (10.21) to calculate the work done by the torque.
W = z(2 - 1) = z (work done by a constant torque) (10.21)
J
(d) What is the grinding wheel's kinetic energy when it is rotating at 1450 rev/min?
J
Compare your answer to the result in part (c).
The answer to part (c) is greater.The two answers are the same. The answer to part (d) is greater.
Explanation / Answer
I = ½mr² = ½ * 1.8kg * (0.1m)² = 0.009 kg·m²
= 1450rev/min * 1min/60s * 2rads/rev = 151.76 rad/s
= - / t = -151.76 rad/s /3.0s = 50.58 rad/s²
A) = I = -0.009kg·m² *50.58rad/s² = -0.455 N·m (that is, against the initial velocity)
B) = ½t² = ½ * 50.58rad/s² * (3s)² = 227.61 rads
C) W = = -0.455N·m *227.61rads = -103.56 J
D) KE = ½I² = ½ * 0.009kg·m² * (151.76rad/s)² = 103.63 J
The answer to part (d) is greater
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