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Chapter 12: Q. 88 (page 335)

In FIGURE CP12.88, a 200 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg track is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car’s switch is turned on, it soon reaches a steady speed of 0.75 m/s relative to the track. What then is the track’s angular velocity, in rpm?

Short Answer

Expert verified

Angular velocity is 3.974 rev /min

Step by step solution

01

Given information

Mass of the toy =200 g =0.2 kg
Diameter of the track =60 cm =0.6 m
So radius of track =0.3 m
Mass of the frictionless track= 1 kg
Speed relative to track (v)=0.75m/s

02

Explanation

First find the angular speed

ω=vr⇒ω=0.75(m/s)(0.3m)=2.5rad/sec......................(1)

From The law of conservation of angular momentum is constant

L = L'

Momentum = I x angular velocity

Iω=I'ω'⇒Ir2ω=Mr2+mr2ω'⇒ω'=Ir2ω(M+m)r2.....................................(2)

Now substitute the given values

⇒ω'=(0.2)(2.5m/s)(1kg+0.2kg)⇒ω'=0.416rad/s⇒ω'=0.4162π×60⇒ω'=3.974revolution/min

Angular speed is 3.974 rev/min.

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