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.

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.)

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.

Answer 1

In order to find the coefficient of static friction, we must examine the force of friction at the slipping point where Ff = µsFN = µsmg. We don’t know the mass of the car, but hopefully it will cancel out somewhere along the way. The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions. First the tangential direction

∑Ft = Fft = mat

And then in the centerward direction

∑Fc = Ff c = mac = mV²/r

Going back to our constant acceleration equations we see that

vt²= vti² +2at∆x = 2atπr/ 2

So going backwards and plugging in

Ff c = m 2atπr /2r = πmat

Ff =Fft² +Ffc²= mat1+²

  µs = Ff/ mg = at 1+π² /g= 1.70 m/s2/ 9.80 m/s2 p( 1+π² )= 0.572

C   

KE = (1/2) * ((1/2) * m * r^2) * w0^2

KE = (1/4) * m * r^2 * w0^2 [EQ 2]

Wf = Friction force (Ff) * Total Distance that the rag rubs on the edge of the wheel during that 6 sec (X).

Wf = Ff * X

Ff = u * N

Wf = u * N * X [EQ 3]

We must assume the acceleration is constant. Now we need one of the rotational motion equations, the one without acceleration.

x = x0 + (1/2) * (w0 + w1) * t

x - x0 = the number of radians that the disk rotates

X = R * (x-x0)

(x-x0) = X / R

w1 =0

So

X/R = (1/2) * w0 * t

X = R * (1/2) * w0 * t

X = (1/2) * R * w0 * t [EQ 4]

Now shove eq 4 into eq 3

Wf = u * N * X

Wf = u * N * (1/2) * R * w0 * t [EQ 5]

Now substitute eq 2 and eq 5 into eq 1

KE = Wf

(1/4) * m * r^2 * w0^2 = u * N * (1/2) * R * w0 * t

Now we just solve for u, LOL

u = ((1/4) * m * R^2 * w0^2) / ( N * (1/2) * R * w0 * t)

u = ((1/2) * m * R * w0) / (N * t)

m = 100 kg

R = 0.500 m

w0 = 50 rev/min * (1 min/60 sec) * (2 pi radians/1 rev)

w0 = 5.23 radians/sec

N = 74.0 N

t = 6 sec

u = ((1/2) * 100 kg * 0.500 m * 5.23 radians/sec) / (74.0 N * 6 sec)

u = 0.294