Open Query 29

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course PHY 201

11/19 12 pmIf your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

029. `query 29

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Question: `qQuery class notes #28. Explain how we can calculate the average angular velocity and the angular acceleration of an object which rotates from rest through a given angle in a given time interval, assuming constant angular acceleration.

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Your solution:

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Angular velocity is represented by omega*r where omega is the rate of change of position which can by found in we know the given angle and the time. (omega=theta/t) r is the radius from the center.

Angular acceleration will then by found by using the velocity and r. v^2/r

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Given Solution:

`a**This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration.

We have

angular acceleration = change in angular velocity / change in clock time.

The average angular velocity is change in angular position / change in clock time.

This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time.

So you can calculate the average angular velocity.

If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity.

From this information you can calculate angular acceleration. **

Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg.

Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m.

The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m.

The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position

x_cm = 1.58 kg m / (3.60 kg) = .44 meters,

placing it a bit to the left of the 1.50 kg mass.

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Self-critique (if necessary):

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Question: `qQuery problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.

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Your solution:

****

L=2lo in the center cube. Therefore, L=1/2Lo in the first cube and l=6LO in the 3rd, for a total of 8.5 LO.

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Given Solution:

`a** The mass of the second will be 2^3 = 8 times as great as the first. It takes 8 1-unit cubes to make a 2-unit cube.

The mass of the third will be 3^3 = 27 times as great as the first. It takes 27 1-unit cubes to make a 3-unit cube.

In the x direction the distance from left edge to center of first cube is 1/2 L0 (the center of the first cube).

In the y direction the distance is from lower edge to center of the first cube is 1/2 L0 (the center of the first cube).

In the x direction the distance from left edge to center of the second cube is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0.

In the y direction the distance from lower edge to center of the second cube is L0 (the center of the second cube).

In the x direction the distance from left edge to center of the third cube is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0.

In the x direction the distance from lower edge to center of the first cube is 3/2 L0 (the center of the third cube).

Moments about left edge and lower edge of first cube:

If m1 is the mass of the first cube then in the x direction you have total moment

m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0.

The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at

center of mass in x direction: 138 m1 L0 / (36 m1) = 3.83 L0.

In the y direction the moment is

m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 49 m1 L0

so the center of mass is at

center of mass in y direction: 49 m1 L0 / (36 m1) = 1.36 L0. **

STUDENT QUESTION

I don’t understand why in the Y direction the “equation” isn’t identical to that of the

X…why is X 2L0 and Y just L0…since the cubes have to be uniform this doesn’t make sense to me.

INSTRUCTOR RESPONSE

The coordinates are for the center of mass.

Each cube rests on the x axis.

The first cube extends in the vertical direction from the x axis to y = L0, so its center of mass in the y direction is at 1/2 L0.

The second cube extends in the vertical direction from the x axis to y = 2 L0, so its center of mass in the y direction is at L0.

The third cube extends in the vertical direction from the x axis to y = 3 L0, so its center of mass in the y direction is at 3/2 L0.

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Self-critique (if necessary):

I had no idea with this question and I think I understand it after reading the solution, but I still don’t quite understand why L3=9/2

The third cube 'starts' at distance 3 L0 from the origin, since the other two cubes are in the way. Its sides are 3 L0, so it 'ends' at distance 6 L0 from the origin. Halfway between its 'start' and 'end' its distance is (3 L0 + 6 L0) / 2 = 9/2 L0.

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Question: `qWhat is the mass of the second cube as a multiple of the mass of the first?

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Your solution:

****Being a cube, it will be 8 times the mass of the first because it would take 8 of the first to make the second.

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Given Solution:

`a** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. **

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Self-critique (if necessary):

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Question: `qWhat is the mass of the third cube as a multiple of the mass of the first?

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Your solution:

*****Given the solution from a previous problem, I think the mass of the third in relation to the first will be 27 times that of the first.

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Given Solution:

`a** The mass of the third cube is 3^3 = 27 times the mass of the first. **

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Question: `qHow far from the outside edge of the first cube is its center of mass?

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Your solution:

****

The outside edge is 1/2L0

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Given Solution:

`a** In the x direction the distance is 1/2 L0 (the center of the first cube).

In the y direction the distance is also 1/2 L0 (the center of the first cube). **

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Question: `qHow far from the outside edge of the first cube is the center of mass of the second cube?

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Your solution:

****

The distance will be 2 LO since it takes 1 LO to cross the first cube and another LO to reach the center.

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Given Solution:

`a** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0.

In the y direction the distance is L0 (the center of the second cube). **

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Question: `qHow far from the outside edge of the first cube is the center of mass of the third cube?

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Your solution:

****

To reach the center of the mass of the third cube, it is 2LO + 1LO (to reach the edge of the second cube)+1.5 LO to reach the center of the third, so 4.5 LO

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Given Solution:

`a** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0.

In the x direction the distance is 3/2 L0 (the center of the third cube). **

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Self-critique (if necessary):

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Question: `qHow do you use these positions and the masses of the cubes to determine the position of the center of mass of the system?

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Your solution:

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To determine the center of mass of the system, you add the 3 spots multiplied by their respective mass and divide by the sum of the mass of all

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Given Solution:

`a** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0.

In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. **

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Self-critique (if necessary):

That’s sort of what I said, I’m still not 100% clear on the center of mass.

You've got it, you just have to get used to thinking about it.

The general idea (without a lot of qualifying conditions, so these statements are not entirely complete):

The torque exerted by the weight of the object about any pivot point is the same as if its mass was all concentrated at the center of mass. The object can be balanced at its center of mass. If you strike the object at its center of mass you will not cause its state of rotation to change.

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Question: `qUniv. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible.

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Your solution:

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Given Solution:

`a** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case.

Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance

c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m

from the left end of the canoe.

A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at

c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m.

The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards.

Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. **

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Self-critique (if necessary):

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I've inserted some notes related to center of mass, and the calculation of the center of mass for that 3-cube system.

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