29 Open query

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course PHY 201

11/29/2011 8:38 PM

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:

You should be able to use the same reasoning

for angular velocity as you use for linear.

Initial velocity is zero, and if acceleration

is constant, average velocity will be 1/2 vf.

Knowing the angle of travel and the time

interval you can find the final velocity with

the second equation of uniform motion. By

converting the angle to radians you can

determine the velocity in radians per second.

Knowing the vf now, and using the fourth

equation of uniform motion you should be able

to solve for acceleration in radians per second

per second.

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):

Was this supposed to be a seperate question?

If so I understand finding the moment of a

simple system.

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Self-critique Rating:3

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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:

What is L0? Okay, finished reading through the

solution.

The edge of the first cube is at (0,0), and

each consecutive cube touches the first.

If each cube is measured in units based on the

first cube, Cube 2L0 is eight times more

massive than cube LO, and Cube 3L0 is 27 times

more massive than cube L0.

The distance from zero in the x direction to

the center of each cube will be:

1/2 L0 to the center of Cube L0,

L0 + L0 = 2L0 to the center of Cube 2L0,

L0 + L0 + L0 + 3L0/2 = 9L0/2 to the center of

Cube 3L0.

The distance from zero in the y direction to

the center of each cube will be:

1/2 L0 to the center of Cube L0,

L0 to the center of Cube 2L0,

3L0/2 to the center of Cube 3L0.

The moment in x will be:

(L0/2 * m1) + (2L0 * 8m1) + (9L0/2 * 27m1) =

((L0*m1)/2) + ((16L0*m1)/2) + ((243L0*m1)/2) =

276/2 L0*m1 = 138L0*m1

M_x/Mass = 138L0*m1 / 36m1 = 3.83L0 to the

center of mass in the x-axis.

The moment in y will be:

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

((L0*m1)/2) + ((16L0*m1)/2) + ((81L0*m1)/2) =

49L0*m1

M_y/Mass = 49L0*m1 / 36m1 = 1.36L0 to the

center of mass in the y-axis.

Center of mass for the system is at (3.83L0,

1.36L0)

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):

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Self-critique Rating:OK

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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:

8m1 because the volume of the second cube is

2L0^3 it would take eight of the L0 cubes to

fill the 2L0 cube.

confidence rating #$&*:

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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:

27m1

Volume of cube 3L0 is 3L0^3 so it would take 27

of the first cube to fill the 3L0 cube.

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:

L0/2

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):

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Self-critique Rating:OK

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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:

2L0

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):

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Self-critique Rating:OK

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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:

9L0/2

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):

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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:

Find the moment in x and y for the system.

Divide this moment by the total mass of the

system to obtain the distance in x and y from

zero to the center of mass for the system.

The moment in x will be:

(L0/2 * m1) + (2L0 * 8m1) + (9L0/2 * 27m1) =

((L0*m1)/2) + ((16L0*m1)/2) + ((243L0*m1)/2) =

276/2 L0*m1 = 138L0*m1

M_x/Mass = 138L0*m1 / 36m1 = 3.83L0 to the

center of mass in the x-axis.

The moment in y will be:

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

((L0*m1)/2) + ((16L0*m1)/2) + ((81L0*m1)/2) =

49L0*m1

M_y/Mass = 49L0*m1 / 36m1 = 1.36L0 to the

center of mass in the y-axis.

Center of mass for the system is at (3.83L0,

1.36L0)

confidence rating #$&*:

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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):

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Self-critique Rating:OK

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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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&#Very good responses. Let me know if you have questions. &#