A stationary bicycle wheel of radius 0.7 m is mounted in the vertical plane (see

Question

A stationary bicycle wheel of radius 0.7 m is mounted in the vertical plane (see figure below). The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. The wheel has mass 4.7 kg, all concentrated in the rim (the spokes have negligible mass). A lump of clay with mass 0.5 kg falls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has speed 7 m/s, and the wheel is rotating clockwise with angular speed 0.32 rad/s.(Assume +x is to the right, +y is upward, and +z is out of the page. Assume the line connecting the center to the point of impact is at an angle of 45° from the horizontal.)

(a) Just before the impact, what is the angular momentum (magnitude) of the combined system of wheel plus clay about the center C?

(b) Just after the impact, what is the angular momentum (magnitude) of the combined system of wheel plus clay about the center C?

(c) Just after the impact, what is the angular velocity (magnitude) of the wheel?

Explanation / Answer

A) just before impact

angular momentum of the system is L1+L2

L1 = m*v*r*sin(45) = 0.5*7*0.7*sin(45) = 1.732

L2 = I*w = m*r^2*w = 4.7*0.7*0.7*0.32 = 0.73696

then angular momentum of the system is Li = 1.732-0.73696 = 0.99504 kg*m^2/s

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B) just after impact

angular momentum is = 0.99504 (this is in accordance with the law of conservation of angular momentum)

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C) angular velocity is w

then L = I*w

I is the moment of inertia = m*r^2 + M*r^2 = (4.7+0.5)*0.7*0.7 = 2.548 kg*m^2

then w = L/I = 0.99504/2.548 = 0.3905 rad/s

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