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8.01 Lecture Notes - Lecture 1: Rotation Around A Fixed Axis, Angular Momentum, Angular Acceleration

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School
MIT
Department
Physics
Course
8.01
Professor
Peter Dourmashkin

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Document Summary

Translational motion: the gravitational external force acts on the center of mass. Rotational motion: object rotates about center of mass. Example: for straight line motion, the bicycle wheels rotate about a xed direction and the center of mass translates: s > x slipping, s = x rolling without slipping, s < x skidding. S = x > r 0 = v t > v = r > v = rw. The sign depends on the choice on linear coordinate system and rotational coordinate system. Translational motion of center of mass: newton"s second and third laws. The momentum of a system remains constant unless the system is acted on by an external force in which case the acceleration of center of mass satis es. Fixed axis rotation about the center of mass of a rigid body. Angular momentum for two dimensional rotation and translation. The angular momentum for a translating and rotating object is given by.

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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Related Questions

Just a few questions from a review packet i'm stuck on, anyhelpwould be appreciated!

1) The moment of inertia about its axis is given as .5 mr^2. Ifthiscylinder rolls without slipping the ratio of it's rotationalkineticenergy to it's translational kinetic energy is
a. 2:1
b. 1:1
c. 1:2
d. 1:3

2) An object’s angular momentum changes by 20 kg,m2/s in 4.0s. Whatmagnitude average torque acted on this object?
A. 2.5 N’m
B. 5.0 N-m
C. 40 N-m
D. 80 N’m

3) A solid cylinder of mass 10 kg is pivoted about africtionlessaxis thought the center O. A rope wrapped around theouter radiusR1 = 1.0 m, exerts a force F1 = 5.0 N to the right. Asecond ropewrapped around another section of radius R2 = 0.50 mexerts a forceF2 = 6.0 N downward. (See Fig. 8-1.) What is theangularacceleration of the cylinder?
A) 1.0 rad/s2
B) 0.60 rad/s2 >my guess<
C) 0.40 rad/s2
D) 0.80 rad/s2

4) A solid sphere of mass 1.0 kg and radius 0.010 m starts fromrestand rolls without slipping down a 1.0-m high inclined plane.What isthe speed of the sphere when it reaches the bottom of theinclinedplane?
A) 3.7 m/s >My guess<
B) 4.4 m/s
C) 5.6 m/s
D) 6.3 m/s

5)A railroad car with a mass of 200 kg is going at 10 m/s. A 70kgstunt man drops directly down on it from a distance of 4 mabove.What is the speed of the railroad car after?
a. 4.7 m/s
b. 7.4 m/s
c. 2.8 m/s
d. 10 m/s

6) A projectile is launched with an initial velocity of 60 m/s atanagnle of 30 degrees above the horizontal. How far will itgo?
a. 318 m
b. 160 m
c. 152 m
d. 184 m

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.