Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 6 - Force and Motion-II - Problems: 41



Work Step by Step

Since the cat is on a merry-go-round, we must find the centripetal acceleration. This is given by a = $v^2/r$. Since it completes one turn every 6 seconds and one turn is 2*pi radians, then we know that $\omega = 2*pi/6 = pi/3 $ rad/sec. Then to find the radial velocity of the cat we take the radius of 5.4m and use $v =\omega r =5.4 * pi/3 = 5.65 m/s$. Then we take a = $(5.65)^2/5.4 = 5.9 m/s^2$. Therefore, the centripetal force is equal to F =ma = 5.9m where m is the mass of the cat. In order to keep the cat from sliding, this force must be balanced by the force of friction. Therefore we have that Ff = 5.9m. However, we know that the force of friction is equal to the Normal Force * mu where mu is the coefficient. To find the normal force we list all the forces in the y direction and see that he is not accelerating and therefore his weight and the normal force must balance. Thus, N = mg where g = 9.81. Therefore, we have that 5.9m = mg*mu. Here we can solve for mu because the ms will cancel and g we know to be 9.81 and so mu = 0.60
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