A stationary bicycle wheel of radius 0.8 m is mounted in the vertical plane (see

Question

A stationary bicycle wheel of radius 0.8 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 4.5 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 6 m/s, and the wheel is rotating clockwise with angular speed 0.29 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 degree from the horizontal.) Just before the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? Just after the impact, what is the angular momentum (magnitude and direction) of the combined system of wheel plus clay about the center C? Just after the impact, what is the angular velocity (magnitude and direction) of the wheel?

Explanation / Answer

a) I_wheel = M*R^2

= 4.4*0.8^2

= 2.816 kg.m^2

angular momentum of wheel = -2.816*0.26

= -0.73216 kg.m^2/s

angular momentum of caly = m*v*R*sin(135)

= 0.5*8*0.8*sin(135)

= 2.26 kg.m^2/s

Total angular momentum = 2.26 -0.73216

= 1.5278 kg.m^2/s

Direction : -z

b) here angular momentum is conserved.

so, final angular momentum = initial angular momentum

= 1.5278 kg.m^2/s

directio : -z axis

c) final angular momentum = (I_wheel + I_caly)*w

1.5278 = ( M*R^2 + m*R^2)*w

1.5278 = (2.816 + 0.5*0.8^2)*w

w = 1.5278/3.136

= 0.4872 rad/s

direction : -z

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