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A uniform solid disk is set into rotation with an angular speed X about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and released.

(a) What is the angular speed of the disk once pure rolling takes place?

(b) Find the fractional loss in kinetic energy from the time the disk is released until pure rolling occurs.

(c) Furthermore, assume that the coefficient of friction between disk and surface is u. Show that the time interval before pure rolling motion occurs is RX/3ug

(d) Show that the distance the disk travels before pure rolling occurs is (RX)^2 / (18ug)

For part (a), I assumed a constant frictional force acting at the point of contact until pure rolling occurs. Say the force = F, final angular velocity = Y, and t = duration of the force. Moment of inertia of the disk = 0.5MR^2

Using Newton's Second Law (both linear and rotational),

Ft = Mv = MRY ................ (1)

FRt = (0.5MR^2)(X-Y)

Equating the two equations,

0.5X - 0.5Y = Y

1.5Y = 0.5X

Y = 1/3 X

(b) Have calculated the fractional loss as 2/3.

(c) Using equation (1), uMgt = MR(1/3 X)

t = RX/3ug

(d) I couldn't solve this part. I tried to apply conservation of energy for the loss of energy due to friction.

Let the distance traveled be s and F be the frictional force.

Fs = uMgs = loss of energy = (2/3)(0.5 x 0.5MR^2 x X^2) = (1/6 MR^2X^2)

s = (1/6 MR^2X^2) / uMg = (1/6 R^2X^2) / ug

My answers are correct for part a,b,c

Couldn't get the coefficient as 1/18 as required by question :(

Someone please tell me what's going wrong :(