A cyclotron designed to accelerate protons has an outer radius of0.330 m. The protons are emitted nearly at rest from a source at the center and are accelerated through 600 V each time they cross the gap between the dees. There's are between the poles of an electromagnet where the field is 0.550 T. (a) Find the cyclotron frequency for the protons in this cyclotron. (b) Find the speed at which protons exit the cyclotron. (c) Find their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron? (e) For what time interval does one proton accelerate?

OneClass: A cyclotron designed to accelerate protons has an outer radius of0.330 m. The protons are e...

You are working as a medical assistant at a proton beam facility, where high-speed protons are used to bombard cancer cells. The protons are accelerated with a cyclotron, which you find very interesting because of your background in physics. You are explaining this to a patient who has some familiarity with cyclotrons. She asks, “How many revolutions does a proton make in the cyclotron before it reaches its exit kinetic energy?” You are taken aback, both by the high quality of her question and the fact that you never thought of such a question before. You tell her you will try to get her an answer before she finishes her treatment today. When you are finished preparing her for treatment, you go into the cyclotron room and look at the machine. Only three numbers are available on the machine labeling:

The exit energy K = 250 MeV, the radius at which the protons exit, r = 0.850 m, and the accelerating potential difference between the dees, ΔV = 800 V. You go back to the patient prepared to give her a total number of times the protons go around the cyclotron before exiting.

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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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