A stationary bicycle wheel of radius R = 0.32 m is mounted in the vertical plane

Question

A stationary bicycle wheel of radius R = 0.32 m is mounted in the vertical plane on a horizontal low-friction axle. The wheel has mass M = 1.4 kg, all concentrated in the rim (the spokes have negligible mass). A lump of clay with moss m = 0.057 kg foils and sticks to the outer edge of the wheel at the location shown, at 45 degree from the vertical, Just before the impact the clay has a speed v = 2.2 m/s. (Assume the +x axis is to the right, the + y axis is up, and the 7 axis is out.) (a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C? L^rightarrow c.i = kg middot m^2/s (b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the center C? L^rightarrow c.i = kg middot m^2/s (c) Just after the impact, what is the angular velocity of the wheel? w = radians/s (d) Qualitatively, what happens to the linear momentum of the combined system? Why? The momentum doesn't change because the surroundings don't exert forces on the system. The magnitude of the momentum decreases due to the upward force of the axle. This Is a collision so the momentum can't change.

Explanation / Answer

(A) Just before:

angular momentum of wheel = 0

of clay = m v r sin(45)

= 0.057 x 2.2 x 0.32 x sin45

= 0.02837 kg m^2 /s

(B) angular momentum of system will be conserved.

hence initial = final angular momentum

Lf = 0.02837 kg m^2 /s

(C) Lf = I w

0.02837 = ( (1.4 x 0.32^2) + ( 0.057 x 0.32^2) ) w
w = 0.19 rad/s

(d) The momentum does not change because the surroundings don't exert force on the system.

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