Lab 16

#$&*

course Phy 201

Effect of loading on angular acceleration of strap

You loaded a strap with either dominoes or magnets and obtained data to determine its angular acceleration. You then repositioned the load and repeated. You might have repositioned the load once more.

Insert a copy of your data here, along with any previously submitted work you wish to include:

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2 seconds in 1/4 turn

6 seconds per rotation

.9cm dominoes

strap is equal to 20cm

Dom on each side 1/8 turn in 1 second

8 seconds per rotation

2.5 seconds 3/16 rotation = 13.25 seconds per rotation

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As best you can determine, what was the average angular acceleration of the system for each of the systems you observed?

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2 sec 1.57 radians

2 sec 1.2 radians

1st attempt

.79 rad/sec is d'omega

2nd attempt

.6 rad/sec is d'omega

aplha = d'omega/d't

1st attempt

alpha = .4 rad/s^2

2nd attempt

alpha = .3 rad/s^2

@&
It looks like you divided average angular velocity by the time interval.

You need to divide the change in angular velocity by the time interval.
*@

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Assuming each domino to have mass 16 grams, the strap to have mass 70 grams and each magnet to have mass 40 grams determine the moment of inertia of each system.

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I= SUM mr^2

Domino 2*16 = 32 grams + 70 grams for the strap

I= SUM 102grams(10cm^2)

I= 10,200 gcm^2 for total system

@&
The dominoes are pretty much 10 cm from the axis of rotation, but the strap itself is spread out from 0 to 15 cm relative to the axis.

So you can do 32 gram * (10 cm)^2.

But you have to figure the moment of inertia of the strap separately. What's the formula for a thin strap or a rod rotating about its center of mass?
*@

On each end

I= 16 g (25 cm^2)

I= 400 gcm^2

@&
You would have to square that 25 cm. You would have 16 g * (25 cm)^2, closer to 10 000 g cm^2.
*@

Tnet= alphaI

Tnet= (.4 rad/sec)(-10,200 gcm^2)

Tnet= -4080

@&
Angular acceleration is in rad / sec^2, but your idea is correct. If you use the units in your calculation, you'll get the units of torque.
*@

Tnet = (.3rad/sec) (-10,200gcm^2)

Tnet= -3060

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Determine the average frictional torque that brought each system to rest.

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@&
This is pretty much what you've just done above.
*@

@&
You should take a few minutes and make a couple of corrections, mostly for the sake of getting ready for the test. You want to be sure you're doing these calculations correctly. You're on the right track.
*@

Lab 16

@& It looks like you divided average angular velocity by the time interval. You need to divide the change in angular velocity by the time interval. *@

@& You would have to square that 25 cm. You would have 16 g * (25 cm)^2, closer to 10 000 g cm^2. *@

@& Angular acceleration is in rad / sec^2, but your idea is correct. If you use the units in your calculation, you'll get the units of torque. *@

@& This is pretty much what you've just done above. *@

@& You should take a few minutes and make a couple of corrections, mostly for the sake of getting ready for the test. You want to be sure you're doing these calculations correctly. You're on the right track. *@

@&
It looks like you divided average angular velocity by the time interval.

You need to divide the change in angular velocity by the time interval.
*@

@&
You would have to square that 25 cm. You would have 16 g * (25 cm)^2, closer to 10 000 g cm^2.
*@

@&
Angular acceleration is in rad / sec^2, but your idea is correct. If you use the units in your calculation, you'll get the units of torque.
*@

@&
This is pretty much what you've just done above.
*@

@&
You should take a few minutes and make a couple of corrections, mostly for the sake of getting ready for the test. You want to be sure you're doing these calculations correctly. You're on the right track.
*@