Question

10.4-5-6)

A)

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 2.10 m/s². The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.

________

(Hint: You are not given a value of the radius of the track. Think through the problem using the symbol R for this value and see if a value is needed for the final solution.)

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B)

A 250-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s? (State the magnitude of the force.)

_____N

(Hint: Determine the moment of inertia of the merry-go-round and use the relation between torque and angular velocity to find the force needed to accelerate the merry-go-round.)

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C)

A potter's wheel—a thick stone disk of radius 0.500 m and mass 110 kg—is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.00 s by pressing a wet rag against the rim and exerting a radially inward force of 73.0 N. Find the effective coefficient of kinetic friction between the wheel and rag.

_______

(Hint: Can you determine the force that must be applied to the outside rim in order to stop the wheel in 6.00 s?)

Answer 1

First question

(A) Given that -

Tangential acceleration of the car, a = 2.10 m/s^2

Now,

In order for the car to stay within the circular track, the static friction between the tires of the car and the road must provide the centripetal acceleration.

Just when the car begins to skid:

F(friction,static) = F(centripetal)

=> u(s)mg = mv^2/R

=> v^2 = u(s)Rg.

The car skids once it travels a distance of (1/4)(2πR) = πR/2, where R is the radius of the circle.

At a constant acceleration a, we see that the speed of the car after it has traveled a distance of (1/2)πR is:

v^2 = 2ad = 2a(πR/2) = πaR.

Equating the values of v^2 gives:

πaR = u(s)Rg

=> u(s) = πa/g

Plug-in the value of the variables -

u(s) = (3.14)(2.10 m/s^2)/(9.8 m/s^2)

= 0.673

Therefore, coefficient of static friction between the car and track = 0.673 (Answer)