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

22 views7 pages

copperlizard489

14 Mar 2018

School

Western University

Department

Kinesiology

Course

Kinesiology 2241A/B

Professor

Thomas Richard Jenkyn

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Linear momentum is the mass of an object multiplied by it"s velocity. "how hard it will be to slow something down" Isaac newton called this the "quantity of motion" that an object had. It is a measure of how hard it will be to stop or slow down. A slow moving train has a lot of momentum, since m is large. A thrown baseball has a lot of momentum, since v is large. Time rate of change is related to the amount of force required. Just like a force, linear momentum is a vector. Point of application is the body center of mass. Line of action is always tangent to the pathway of the body. Velocity is always the tangent to the line. If you suddenly stop steering the car, linear momentum will tell you where you"re going to go.

Related Questions

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
6
7
8

puceworm982

A ball of mass 1.5 kg is thrown upward. It leaves the thrower's hand with a velocity of 10 m/s. The following questions refer to the motion after the ball leaves the thrower's hand. Assume that the upward direction is positive.
A. How long before the ball returns to the throwers hand?
B. What is final velocity of the ball just before it reaches the hand?
C. What is the change in momentum of the ball?
D. What is the impulse calculated from the change of momentum?
E. from the impulse, find the average force acting on the ball. What causes the force?

chocolatefrog316

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.

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

Document Summary

Get access

Related Documents

Kinesiology 2241A/B Lecture Notes - Lecture 18: Momentum, Mattress, Lincoln Near-Earth Asteroid Research

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

Kinesiology 2241A/B Lecture Notes - Lecture 17: Momentum, Mattress, Memory Stick

Related Questions