A disk with a hole cut out in the middle has a c value of 0.54 and its radius (o

Question

A disk with a hole cut out in the middle has a c value of 0.54 and its radius (outer) is 0.52 meters. It is moving on a horizontal surface. Initially, it is spinning at 57.1 rad/s counter-clockwise about its center and its center of mass is moving at 23 m/s in the +x direction. Hence, it is skidding and it is not rolling without slipping. If the coefficient of kinetic friction is 0.72, how far does the disk skid, in meters, before it rolls without slipping?

Hint: The kinetic friction force must be included now since it is slipping. By applying both Newton's second law and summing the torques created by all forces about the center one can find the amount of the acceleration of the center of mass and the angular acceleration. This will then change both the velocity of the center of mass (through acceleration) and the angular velocity (through angular acceleration). Find the time it takes for the center of mass to be moving at the speed needed to roll without slipping and then use a constant acceleration equation to determine how far the object moves when this criteria is met.

Explanation / Answer

Frictional Torque T = Fk * R2

also Torque T = I*Alpha

where ALpha is angualr accleration

I is moment of inertia m, here MOI = M/2 * (R1^2 + R2^2)

frictional force fk = - m a

i.e u m g = - m a

a = u g

a = 0.72 * 9.81

accelration a = 7.06 m/s^2

thus (m/2) *(R1^2 + R2^2) ALPHA = uk m g R2

solving for ALpha

((0.53/2)^2 + 0.52^2) * Alpha = 2 * 0.72 * 9.81 * 0.52

Alpha = 21.56 rad/s^2

Condition for Rolling with out slipping is V = R W 2

Vo- at = R2*(Wo - At)

Sovling for time

t = (Vo - R2W)/(a- Ralpha)

t = ((23 - (0.52 * 57.1))/((7.06 - (0.52 * 21.56))

t = 1.612 secs

DIstance travelled S = Uo t - 0.5 at^2

S = 23 * 1.61 - (0.5* 7.06 * 1.61^2)

S = 27.87 m is the distance travelled.

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