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 4.2 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 8 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 and direction) of the combined system of wheel plus clay about the center C?

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

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

(d) Qualitatively, what happens to the linear momentum of the combined system? Why?

Some of the linear momentum is changed into angular momentum.There is no change because linear momentum is always conserved.    Some of the linear momentum is changed into energy.The downward linear momentum decreases because the axle exerts an upward force.

Explanation / Answer

Just before impact

angular momentum of the system is Li = (m*v*r*sin(45)) - (I*w) = (0.5*8*0.7*sin(45)) - (M*r^2*w)

negative sign before I*w shows that motion is in clockwise direction

Li = (0.5*8*0.7*sin(45)) - (M*r^2*w)

Li = (0.5*8*0.7*sin(45)) - (4.2*0.7^2*0.32)

Li = 1.32 kg-m^2/sec along +z-axis

b) Using law of conservation of angular momentum ,

angular momentum of the system before impact = angular momentum of the system after impact

1.32 = Lf along -Z-axis

C) after impact both wheel and clay rotating together

so Lf = (I1+I2)*w = 1.32

w= 1.32/(I1+I2) = 1.32/(Mr^2 + m*r^2) = 1.32/((M+m)*r^2)

w = 1.32/((4.2+0.5)*0.7^2)

w = 0.573 rad/sec along -Z-axis

d) since linear momentum is conserved,hence there is no change because linear momentum is always conserved

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