A 7.0 -kg disk traveling at 9.0 m/s strikes a 7.0 -kg stick of length 3.0 m that

Question

A 7.0-kg disk traveling at 9.0 m/s strikes a 7.0-kg stick of length 3.0 m that is lying flat on nearly frictionless ice as shown in the overhead view in the figure below. The disk strikes at the endpoint of the stick, at a distance r = 1.5 m from the stick's center. Suppose that the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 5.25 kg

A 7.0-kg disk traveling at 9.0 m/s strikes a 7.0-kg stick of length 3.0 m that is lying flat on nearly frictionless ice as shown in the overhead view in the figure below. The disk strikes at the endpoint of the stick, at a distance r = 1.5 m from the stick's center. Suppose that the collision is perfectly inelastic so that the disk adheres to the stick at the endpoint at which it strikes. The moment of inertia of the stick about its center of mass is 5.25 kg m2. Find the velocity of the center of mass of the system. Find the angular velocity of the system after the collision.

Explanation / Answer

Given data

mass m1 = 7 kg , initial speed of disk u1 = 9 m/s

mass of the stick m2 = 7 kg , intially at rest , u2 = 0

a)

as the collision is completely inelastic

m1u1 +m2u2 =(m1+m2) Vcm

Vcm = 7(9)+0 / (7+7) = 63/14 = 4.5 m/s

the velocity of the center of mass of the system = 4.5 m/s

b)

since the collision occurs at r = 1.5 m

we know Vcm = r*w

angular speed w = Vcm /r = 4.5 / 1.5 = 3 rad/s

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