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Chapter 6: Q96P (page 148)

A child places a picnic basket on the outer rim of a merry-go-round that has a radius of 4.6 mand revolves once every 30s. (a) What is the speed of a point on that rim? (b)What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

Short Answer

Expert verified

a) Speed of a point on rim is 0.96m/s

b)μs=0.021

Step by step solution

01

Given

Merry-go-round that has a radius of 4.6 m and revolves once every 30 s

02

Understanding the concept

The problem is based on the centrifugal force. Centrifugal force is the apparent outward force on a mass when it is rotated. It also deals with the static friction.

Formula:

Centrifugal force is given by

F=mv2R

and the maximum value of static friction is given by

fs,max=μsmg.

03

Calculate the speed of a point on that rim

(a)

The distance traveled in one revolution is

2πR=2π4.6m=29m.

The speed is

v=29m/30s=0.96m/s

04

Calculate the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride 

(b)

According to Newton’s second law

fs=mv2/R=m0.20

As N = mg in this situation, the maximum possible static friction is,

fs,max=μsmg.

Equating this withlocalid="1660973301753" fs=m0.20μs=0.20/9.8=0.021

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Most popular questions from this chapter

An airplane is flying in a horizontal circle at a speed of 480km/h(Fig. 6-41). If its wings are tilted at angleθ=40°to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.

An 8.0 kgblock of steel is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.450. A force is to be applied to the block. To three significant figures, what is the magnitude of that applied force if it puts the block on the verge of sliding when the force is directed

(a) horizontally,

(b) upward at60.0°from the horizontal, and

(c) downward at60.0°from the horizontal?

Engineering a highway curve.If a car goes through a curve too fast, the car tends to slide out of the curve. For a banked curve with friction, a frictional force acts on a fast car to oppose the tendency to slide out of the curve; the force is directed down the bank (in the direction water would drain). Consider a circular curve of radius R 200 mand bank angle u, where the coefficient of static friction between tires and pavement is. A car (without negative lift) is driven around the curve as shown in Fig. 6-11. (a) Find an expression for the car speed Vmaxthat puts the car on the verge of sliding out.

(b) On the same graph, plot Vmaxversus angle u for the range 0°to50°, first forμs=0.60(dry pavement) and then forμs=0.050(wet or icy pavement). In kilometers per hour, evaluateVmaxfor a bank angle ofθ=10°and for

(c)μs=0.60and

(d)μs=0.050. (Now you can see why accidents occur in highway curves when icy conditions are not obvious to drivers, who tend to drive at normal speeds.)

In Fig. 6-14, a block of massis held stationary on a ramp by the frictional force on it from the ramp. A force f, directed up the ramp, is then applied to the block and gradually increased in magnitude from zero. During the increase, what happens to the direction and magnitude of the frictional force on the block?

A child weighing 140Nsits at rest at the top of a playground slide that makes an angle of25°with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of0.86m/s2(down the slide, of course).

(a) What is the coefficient of kinetic friction between the child and the slide?

(b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?
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