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Kinesiology 2241A/B Lecture Notes - Lecture 18: Momentum, Body Force, Mattress

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amethystgoat421
30 Oct 2018
School
Western University
Department
Kinesiology
Course
Kinesiology 2241A/B
Professor
Thomas Richard Jenkyn

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Document Summary

If you are trying to understand how much motion does one have. It come from how far you are going and mass you have. Momentum is mass times velocity, just like force. The more momentum it has= harder it is to stop it. Eg baseball, which is very fast but not too much mass would still have a lot of momentum. He defined it as time rate of change of momentum. Time rate of change of linear momentum is the amount of things you need to change. Linear momentum is body force so p of a is always center of mass. Last equation of linear momentum is a vector equation. To slow the ball down to zero momentum, she need to apply an impulse in the opposite direction of 10 kgm/s. Since impulse is f time dt, then the force can be large and the dt small, or the force can be small and dt large.

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Related Questions

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| = kg·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kg·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 ½MR2.)
= 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| = kg·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 = kg·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.

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| = kg·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kg·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 ½MR2.)
= 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| = kg·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 = kg·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.

) Ion Rocket Engine. A special type of rocket engine useselectricity for propulsion: the ion engine. A soft metal, such ascesium, is vaporized and then electrified (ionized) with highvoltage. These cesium ions are the accelerated by strong electricfields to produce a current of atoms at high velocity. Outside theengine, the ions are then recombined with electrons to restoreelectrical neutrality.

A particular ion engine has an exhaust velocity of 20,000 m/sec.The mass flow (dm/dt) is just 1 milligram per second.

a) What is the thrust produced by this engine?

b) The engine propels a 200 kg spacecraft. What velocity change(delta-v) is achievable by running the engine continuously for onesolid year? (For simplicity, assume a constant mass for the system.One year = 31.56 million seconds.)
---------------------------
) Potential Energy. How much work is required to lift one metricton (1000 kg) of water to an altitude of 1000 meters? (The Sun isdoing this all this time to the Earth’s weather.)
--------------------------------

q) Kinetic Energy 1. A car has a mass of 1200 kg. How much kineticenergy does it have when traveling at a speed of 30 m/sec?

-----------------------
Kinetic Energy 2. The Space Shuttle orbits the Earth at a speed ofsome 7750 meters per second (which is about 260 times that of atypical highway driving speed). With a typical payload, the Shuttlehas a mass of 120,000 kg.

a) What is the kinetic energy of the Shuttle in orbit?
b) The Shuttle reaches orbit in some 500 seconds. At what rate ofwork (power) did the rocket engines perform on the Shuttle to placethe craft and its payload into orbit? Compare this to the maximumpower output of a typical nuclear electric generatingstation.
-----------------------

Solar Power. In space, outside the atmosphere, some 1360 Watts ofradiant power pass through every square meter (at normalincidence).

a) What total amount of power does the Earth intercept from theSun? (The projected area of the Earth is ?RE2, where RE is theradius of the Earth, 6.38 million meters.)

b) Compare this figure to the total amount of power produced by allof human civilization, currently some 40 trillion (1012) Watts.What conclusions about solar power can you draw from thiscomparison?



q














-) Angular Momentum. Assuming a uniform distribution of mass insidethe Earth, calculate:

a) the Earth’s moment of inertia, I;
b) its angular momentum, L = omega(omega=2pi/T) the angularmomentum of the Moon in its orbit around the Earth (L = mvr).Compare these two values. (The sidereal period of the Moon is 27.3days; its mass is 1/81 of that of the Earth; its average orbitalradius is 383,000 km.)

------------------------------------

Q0 Water Pressure. Calculate the value of the water pressure at theChallenger Deep, the deepest known place on the surface of theEarth. The currently accepted depth of the water there is 10,911meters. (This is deep enough to cover Mount Everest, and still havemore than one mile of water above the peak. On January 23, 1960,Swiss explorer Jacques Piccard and US Navy Lieutenant Don Walshbecame the first, and so far, the only people ever to visit thisremote place, descending in the Italian-built bathyscaphTrieste.)



q-8
Mass of Earth’s Atmosphere. The pressure of the Earth’s atmosphereat sea level is 101,300 Pascals (Newtons per square meter.) Thismeans that every square meter of Earth’s surface supports a weightof 101,300 Newtons. From this, determine the mass of Earth’satmosphere. What fraction of the Earth’s total mass is this?