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PHYS 101 Lecture 13: Momentum,​ ​Energy,​ ​and​ ​Work

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School
McGill University
Department
Physics
Course
PHYS 101
Professor
Kenneth J Ragan

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PHYS 101 Full Course Notes
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PHYS 101 Full Course Notes
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Document Summary

One of the significant things about momentum and energy is that they are conserved in many situations. Scalar direction, but it can have a sign. Joules . we sometimes also measure it in calories: The mks unit of work is n-m or calorie = 4. 184 j. Example 3: you push an object across a table with a force of 10 n through a distance of 2 m. Example 4: you tug at a heavy box (50 kg) with a rope, making an angle of 30 o to the horizontal. the force you . = - mg sin( ) d = - 50 kg x 9. 8 m/s 2 . For example, if you were to carry the box up the same height at a constant velocity (ie, no net force), your work would be w grav (ie, a positive quantity) and the total work done would be zero again.

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Physics: Principles with Applications
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ISBN: 9780321625922

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Related Questions

The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.
20061122120126329809921253700004765.png
(a) What is F_t, the magnitude of the tangential force that acts on the pole due to the tension in the rope?

Express your answer in terms of T and theta.

When using the tangential force method, you calculate the torque using the equation

\tau = f_{\rm t} d,

where d is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque).

(b) What is the magnitude of the torque tau on the pole, about point A, due to the tension in the rope?
Express your answer in terms ofT, L, and theta.

The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm.

For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.
20061122121536329809931322450004365.png
(c) Find R_m, the length of the moment arm of the force.
Express your answer in terms ofL and theta.
2006112212386329809938811512508683.png

(d)Now consider a woman standing on the ball of her foot as shown. A normal force of magnitude N acts upward on the ball of her foot. Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a forceF. This small bone has a length x, and the angle between this bone and the Achilles' tendon is phi. The horizontal displacement between the ball of the foot and the point P is D.
D oll
Find τp, the torque about point P due to the force applied by the Achilles' tendon. Express your answer in terms of F, ϕ, and x.

1. An object is dropped from a height.Once it is moving,which ofthe following statements are true,at least at one point?
a)its velocity is more than its accelerations
b)its velocity is less than its acceleration
c)its velocity is the same as its acceleration
d)its velocity is never equal to its acceleration

2. Which type of quantity is characterized by both magnitude anddirection?
a)scalar
b)vector
c)trigonometric
d)algebraic variable

3. A hiker walks 200m west and then walks 100m north. In whatdirection is her resulting displacement?
a)north
b)west
c)northwest
d)none of the answer is correct

4. Vector 1 is 7 units long and is at 70 degree. Vector 2 is 5units long and is at 225 degree. Vector 3 is a 3 units long and isat 150 degree. Which vector has equal magnitude components?
a)Vector 1
b)vector2
c)vector 3
d)none of the vectors has equal magnitude components??????

5. Which of the following is an example of the type of force thatacts at a distance?
a)gravitational
b)magnetic
c)electrical
d)all of the above

6. A crate weight W is being pushed across a horizontal surface bya force directed at an angle of 20 degrees below the horizontal.What is the magnitude of the normal force on the crate?
a)it is less then W
b)it equals W
c)it is more than W
d)none of the above since the coefficient of kinetic friction isnot given

7. The net force on an object is in the positive x-direction.Consider the following statements.
(i) The object can be moving in the negative x-direction.
(ii) The object can be speeding up.
(iii) The object can be slowing down.
(iv) The object can be moving in the positive y-direction.

Which of the statements are true?
a. (i) and (ii)
b. (ii) and (iii)
c. (iii) and (iv)
d. Choose this answer if all the statements are true.

8.The work done by static friction can be:
a. positive.
b. negative.
c. zero.
d. Any of the above.

9.The rate at which work is done is equivalent to which of thefollowing?
a. increase in potential energy
b. thermal energy
c. potential energy
d. power

10.The dimensional equivalent of the quantity impulse in terms ofthe fundamental quantities (mass, length,
time) is which of the following?
a. MLT-1
b. ML2T-2
c. MLT
d. MLT-2

11. The dimensional equivalence of the quantity "momentum" in termsof the fundamental quantities (mass,
length, time) is:
a. MLT-1.
b. ML2T-2.
c. MLT.
d. MLT-2.

12.In order for a satellite to be geosynchronous, its orbitmust:
a. go over the North and South Poles.
b. be over the equator.
c. be over a single longitude.
d. emit television signals.

13. The quantity "angular momentum" (in terms of the fundamentalquantities of mass, length, and time) is equivalent
to:
a. MLT-2.
b. ML2T-1.
c. ML2T-3.
d. ML3T.
d. MLT-2.

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.