query 29

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course phy 201

7:30

Query 29If 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:

Iwould think it is the same as the the regular except taking into account the angular aspects.

v=ds/dt

a=dv/dt

confidence rating #$&*:

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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):Ok

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Self-critique Rating:

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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:

Im not sure hot to go about this. I know with cubes that means they are cubed

So maybe 1^3=3

2^3=8

3^3=27

confidence rating #$&*:

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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):

This is difficult for me to understand, but I think I have a small grasp on it.

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Self-critique Rating:

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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:

8 timesthe first

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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):

OK

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Self-critique Rating:

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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:

27 times as great

confidence rating #$&*:

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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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Self-critique (if necessary):

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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:

1/2L0

confidence rating #$&*:

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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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Self-critique (if necessary):

In x and y direction

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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:

L0

confidence rating #$&*:

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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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Self-critique (if necessary):

Ok it is L0 in the y direction and 2 L0 in x direction

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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:

L0 +2L0 +3/2L0

confidence rating #$&*:

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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):

X direction is 3L0

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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:

Im not sure

confidence rating #$&*:

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0

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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):

I don understand this very well

@&

If you hold a mass of, say, 2 kg at the end of a 3-foot stick, using one hand, you can manage that OK. If you hold it from a stick twice as long, it's twice as hard to hold up. You have to exert twice the torque with your hand. Most people will at least have trouble holding 2 kg at the end of a 6-foot stick in this manner.

If you double that distance to 12 feet most people won't be able to hold it up at all.

To get the torque necessary to hold something up in this manner, you multiply the weight by the length of the stick.

Now these cubes not only get further from the original point but they get bigger and more massive.

If you multiply the mass of each by how far it is from the original point, you get something very much like the torque they would exert about that point, if gravity was pulling them downward. (what you get wouldn't really be a torque because you're using mass instead of force, but we can ignore that for the moment--it doesn't make a difference for this problem).

Now we ask how far all that mass would have to be from the original point, if all the masses were combined into one mass, to feel the same -- i.e., to exert the same torque. To figure that out, we divide the torque by the mass.

The result is the position of the center of mass. That's the point where the system of cubes would balance.

*@

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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:

confidence rating #$&*:

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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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Self-critique Rating:

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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:

confidence rating #$&*:

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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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Self-critique Rating:

#*&!

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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:

confidence rating #$&*:

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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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Self-critique Rating:

#*&!#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#