Asst_29_query

course Phy 121

??????_??????assignment #029????n???????????Physics I

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07-23-2006

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10:29:56

Query 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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RESPONSE -->

to get the angular velocity you take the radius and mulyiply by the change in angular positio.

the angular acceleration is calculated by divideing the torque by the moment of inertia.

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10:31:24

**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. **

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RESPONSE -->

change in angukar position divided by the change in clock time = average angular velocity.

calculate acceleration from this.

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10:33:07

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.

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RESPONSE -->

1kg(1) + 1.5kg(.50) + 1.10kg(.25) / .75

4.05m

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10:34:17

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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RESPONSE -->

So the first one is at 0 not 1. This changes everything.

The x coordinates depend on the position about which you calculate the torques. That position is your choice, but all torques have to be figured with respect to that reference position.

If you calculate torques about a point 1 meter to the right of the 1 kg mass, then you will in fact get the numbers pretty much as you indicate. However the masses would exert negative torques (clockwise torques) relative to this point.

The correct expression would then be

( -(1 kg * 1.00 m) - (1.5 kg * .50 m) - (1.10 kg * .25 m) ) / (1 kg + 1.5 kg + 1.1 kg)

= - 2.025 kg * m / (3.6 kg) = -.56 m.

This would put the center of mass .56 m to the left of your reference position.

This in fact coincides with the position obtained in the given solution, which is .44 m to the right of the 1 kg mass (i.e., .56 m to the left of the point which is 1 meter to the right of the 1 kg mass is 1 m - .56 m = .44 m to the right of that mass).

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10:35:21

Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.

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RESPONSE -->

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