An bullet of mass 8 grams is fired into a block of mass 250 grams that is initially at rest at the edge of a table of height of 1metre. The bullet remains in the block, and after the impact the block lands d. 2 metres from the bottom of the table. Determine the initial speed of the bullet

You need to (a) determine the velocity of the bullet+block that would cause it to land on the floor 2 meters from the table, then (b) determine the momentum of the bullet + block at that velocity, and finally (c) determine the initial velocity of the bullet needed to equal this momentum. Post back with what you get for an answer.

Hello!
Would you help me with this one please?
A centrifuge in a medical laboratory rotates at an angular speed of 3600 revolution per minute. When switched off, it rotates through 50 revolutions before coming to rest. Find the constant angular acceleration (in rad/seconds square) of the centrifuge.

I find it helpful to think about the anaolgy between equations of linear motion and equations of rotational motion. If you wanted to determine acceleration of a mass that is initially moving at velocity v that comes to a stop over distance d you would use the equation:



Well in this problem you use an analogous equation, but with rotation components:



where = final rotational velocity, which is 0 rad/s; = initial radial velocity in radians/second; = rotational acceleration, and = the angle the mass rotates through, in radians. You'll have to covert the data you were given - 3600 RPM to radians/second, and 50 rotations to radians - then plug and chug.

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Originally Posted by ebaines

I find it helpful to think about the anaolgy between equations of linear motion and equations of rotational motion. If you wanted to determine acceleration of a mass that is intially moving at velocity v that comes to a stop over distance d you would use the equation:



Well in this problem you use an analogous equation, but with rotation components:



where = final rotational velocity, which is 0 rad/s; = initial radial velocity in radians/second; = rotational acceleration, and = the angle the mass rotates through, in radians. You'll have to covert the data you were given - 3600 RPM to radians/second, and 50 rotations to radians - then plug and chug.

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Hello again!
what about this one
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 revolutions per second in 8 seconds. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12 seconds. Through how many revolutions does the tub turn during the entire 20 seconds interval? Assume constant angular acceleration while it is starting and stopping.

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Originally Posted by rubenita03

Hello again!
what about this one
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 revolutions per second in 8 seconds. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12 seconds. Through how many revolutions does the tub turn during the entire 20 seconds interval? Assume constant angular acceleration while it is starting and stopping.

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Perhaps you can apply the principles I describEd previously to this problem. Post back with your answer and we'll check it.

Quote:

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Originally Posted by ebaines

Perhaps you can apply the principles I descrivebd previously to this problem. Post back with your answer and we'll check it.

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Hello!

I have the answer 65 revolution in 20 seconds
I did this way
I came up with the angular acceleration which is w/t and it = 0.625 rev/sec2
after this I look for the angular displacement in 2 parts the first one in the first 8 seconds with the formula A0= Wot+1/2@t2
since in the first the Wo=0 and Wf = 5rev/sec
and in the second Wo= 5rev/sec and Wf=0
is it OK
help me please
thank you
Monica

Quote:

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Originally Posted by rubenita03

I have the answer 65 revolution in 20 seconds
I did this way
I came up with the angular acceleration which is w/t and it = 0.625 rev/sec2p

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So far so good.

Quote:

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Originally Posted by rubenita03

after this I look for the angular displacement in 2 parts the first one in the first 8 seconds with the formula A0= Wot+1/2@t2
since in the first the Wo=0 and Wf = 5rev/sec

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What did you get for the angle cobvered in this first phase? I get 20 revolutions.

[QUOTE=rubenita03;3324783] and in the second Wo= 5rev/sec and Wf=0[/quopte]

Don't forget that the angular acceleration is different in the second part, since it takes 12 seconds to slow from its initial speed of 5 rev/sec. I calculate that it takes 30 revolutions to slow to 0, so total number of revolutions = 50.