The moment of inertia of a uniform-density disk rotating about anaxle through its center can be shown to be (1/2)MR2. This result isobtained by using integral calculus to add up the contributions ofall the atoms in the disk (see Problem 9.3). The factor of 1/2reflects the fact that some of the atoms are near the center andsome are far from the center; the factor of 1/2 is an average ofthe square distances. A uniform-density disk whose mass is 15 kgand radius is 0.15 m makes one complete rotation every 0.3 s.(a) What is the moment of inertia of this disk?I = kgÂ·m2(b) What is its rotational kinetic energy?Krot = J(c) What is the magnitude of its rotational angular momentum?Lrot = kgÂ·m2/s

OneClass: The moment of inertia of a uniform-density disk rotating about anaxle through its center ca...

This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy Ktotal of a dumbbell of mass m when it is rotating with angular speed? and its center of mass is moving translationally with speed v. (Figure 1) Denote the dumbbell's moment of inertia about its center of mass by Icm. Note that if you approximate the spheres as point masses of mass m/2 each located a distance r from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by Icm=mr2, but this fact will not be necessary for this problem.

Find the total kinetic energy K_tot of the dumbbell.

Express your answer in terms of m, v, Icm, and ?.

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A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.

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

(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgÂ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgÂ·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 auniform disk is Â½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.

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 afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgÂ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgÂ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

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