A playground ride consists of a disk of mass m 46 kg and


A playground ride consists of a disk of mass M = 46 kg and radius R = 2.3 m mounted on a low-friction axle. A child of mass m = 20 kg runs at speed v = 2.5 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.

ANGULAR MOMENTUM

(a) Consider the system consisting of the child and the disk, but not including the axle. Which of the following statements are true, from just before to just after the collision?
The axle exerts a force on the system but nearly zero torque.
The momentum of the system changes.
The momentum of the system doesn't change.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardly changes.
The torque exerted by the axle is zero because the force exerted by the axle is very small.
The torque exerted by the axle is nearly zero even though the force is large, because I i I is nearly zero.

(b) Relative to the axle, what was the magnitude of the angular momentum of the child before the collision?

I icl = kg•m2/s

(c) Relative to the axle, what was the angular momentum of the system of child plus disk just after the collision?

1rd = kgA•m2/s

(c) Relative to the axle, what was the angular momentum of the system of child plus disk Just after the collision? I LCI = kgA•m2/s

(d) If the disk was initially at rest, now how fast is it rotating? That is, what is its angular speed? (The moment of inertia of a uniform disk is AV2MR2.) = radians/s

(e) How long does it take for the disk to go around once?
Time to go around once =
ENERGY

(f) If you were to do a lot of algebra to calculate the kinetic energies before and after the collision, you would find that the total kinetic energy just after the collision is less than the total kinetic energy just before the collision. Where has most of this energy gone?
Increased chemical energy in the child.
Increased translational kinetic energy of the disk. Increased thermal energy of the disk and child.
MOMENTUM

(g) What was the speed of the child just after the collision? v= m/s

(h) What was the speed of the center of mass of the disk just after the collision? vcm = m/s

(I) What was the magnitude of the linear momentum of the disk Just after the collision?
IPI = kgA•m/s

(j) Calculate the change in linear momentum of the system consisting of the child plus the disk (but not including the axle), from just before to just after impact, due to the impulse applied by the axle. Take the x axis to be in the direction of the initial velocity of the child.
APx = Px, Px,i = kgA•m/s

ANGULAR MOMENTUM

(k) The child on the disk walks inward on the disk and ends up standing at a new location a distance R/2 = 1.15 m from the axle. Now what is the angular speed? (It helps to do this analysis algebraically and plug in numbers at the end.)

w = radians/s

ANGULAR MOMENTUM

(k) The child on the disk walks inward on the disk and ends up standing at a new location a distance R/2 = 1.15 m from the axle. Now what is the angular speed? (It helps to do this analysis algebraically and plug in numbers at the end.)
w = radians/s
ENERGY

(I) If you were to do a lot of algebra to calculate the kinetic energies In part (k), you would find that the total kinetic energy after the move is greater than the total kinetic energy before the move. Where has this energy come from?
Decreased chemical energy In the child.
Decreased thermal energy of the disk and child. Decreased momentum of the disk.

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Physics: A playground ride consists of a disk of mass m 46 kg and
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