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Chapter 6: Q41P (page 143)

A cat dozes on a stationary merry-go-round in an amusement park, at a radius of5.4mfrom the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every6.0s.What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding (or the cat clinging with its claws)?

Short Answer

Expert verified

The least coefficient of static friction between the cat and the merry-go-round is0.60.

Step by step solution

01

Given

R=5.4mT=6.0s

02

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity.

Formula:

The velocity in uniform circular motion is given by,

v=2蟺RT

where, v is the velocity, R is the radius andTis the time period

03

Determining the least coefficient of static friction between the cat and the merry-go-round

As the cat moving in the circular motion, it has centripetal accelerati ac=v2R, and frictional force is also directed to the center of the circle.

By using the Newton鈥檚 2nd law along the horizontal direction,

-fs=-mv2R

渭smg=mv2R

渭s=v2gR=2蟺RT2gR=4蟺2RgT2=4脳蟺2脳5.49.81脳62=0.60

Hence, the least coefficient of static friction between the cat and the merry-go-round is 0.60.

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Most popular questions from this chapter

A toy chest and its contents have a combined weight of 180NA toy chest and its contents have a combined weight of 0.42.The child in Fig. 6-35 attempts to move the chest across the floor by pulling on an attached rope. (a) If 胃is 42掳whatis the magnitude of the force F鈫�that the child must exert on the rope to put the chest on the verge ofmoving? (b) Write an expression for the magnituderequired to put the chest on the verge of moving as a function of the angle 胃. Determine (c) the value of 胃for which Fis a minimum and (d) that minimum magnitude.

A car weighing 10.7kNand traveling at 13.4 m/swithout negative lift attempts to round an unbanked curve with a radius of 61.0 m. (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path? (b) If the coefficient of static friction between the tires and the road is 0.350, is the attempt at taking the curve successful?

A student, crazed by final exams, uses a force of magnitude 80Nand angle胃=70掳to push a 5.0 kg block across the ceiling of his room (Fig. 6-52). If the coefficient of kinetic friction between the block and the ceiling is 0.40, what is the magnitude of the block鈥檚 acceleration?

Continuation of Problem 8. Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20kgstone, which presents to the wind a vertical cross-sectional area of0.040m2and has a drag coefficient C of0.80. (a) In kilometers per hour, what wind speedValong the ground is needed to maintain the stone鈥檚 motion once it has started moving? Because winds along the ground are retarded by the ground, the wind speeds reported for storms are often measured at a height of10m. Assume wind speeds are2.00 times those along the ground. (b) For your answer to (a), what wind speed would be reported for the storm? (c) Is that value reasonable for a high-speed wind in a storm?

In Fig. 6-24, a forceacts on a block weighing 45N.The block is initially at rest on a plane inclined at angle 胃=150to the horizontal. The positive direction of the x axis is up the plane. Between block and plane, the coefficient of static friction is 渭s=0.50and the coefficient of kinetic friction is 渭k=0.34. In unit-vector notation, what is the frictional force on the block from the plane when is (a) (-5.0N)i^, (b) (-8.0N)i^, and (c) (-15N)?
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