Q.88 In FIGURE CP12.88, a 200 g toy c... [FREE SOLUTION]

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鈥檚 switch is turned on, it soon reaches a steady speed of 0.75 m/s relative to the track. What then is the track鈥檚 angular velocity, in rpm?

Short Answer

Expert verified

Angular velocity of the track is 3.974 rev /min

Step by step solution

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

Explanation

First find the angular speed

$蝇 = \frac{v}{r} 鈬� 蝇 = \frac{0.75 (m / s)}{(0.3 m)} = 2.5 rad / \sec 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� (1)$

From The law of conservation of angular momentum is constant

L = L^'

Momentum = I x angular velocity

$I 蝇 = I^{'} 蝇^{'}$

$鈬� m r^{2} 蝇 = (M r^{2} + m r^{2}) 蝇^{'} 鈬� 蝇^{'} = \frac{m r^{2} 蝇}{(M + m) r^{2}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Now substitute the given values

$鈬� 蝇^{'} = \frac{(0.2 k g) (2.5 \frac{rad}{s})}{(1 kg + 0.2 kg)} 鈬� 蝇^{'} = 0.416 \frac{rad}{s} 鈬� 蝇^{'} = \frac{0.416}{2 蟺} 脳 60 鈬� 蝇^{'} = 3.974 revolution / \min$