Pulley Problem: Speed after falling a known distance

  • #1
A 1.2·kg block is tied to a string that is wrapped around the rim of a pulley of radius 7·cm. The block is released from rest.

Assuming the pulley is a uniform disk with a mass of 0.37·kg, find the speed of the block after it has fallen through a height of 0.43·m.




2. F=mrω^2 and possibly F=mg and v=r*omega



3. I wasn't sure how to do this problem at all. I looked around online for help and got a little bit but I'm still unsure. I used F=mg to determine the F then used that to determine ω^2. Is that correct? Obviously there would be more to the problem but this is as far as I got and I'm not even sure if it's correct.

So I got F=11.76 by multiplying 1.2kg * 9.8m/s^2 which I then put in the equation for 11.76=(.37kg)(.07m)w^2 and then got 21.3 for omega. This seems high and is probably due to the conversion from CM to M which I'm unsure is correct. Assuming it is I moved on and put that omega value 21.3 into v=wr for v=21.3*.07 and got 1.49 which is incorrect.
 
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Answers and Replies

  • #2
adjacent
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That depends on the friction of the pulley.
 
  • #3
That depends on the friction of the pulley.
I believe it's assumed frictionless because friction isn't mentioned
 
  • #4
adjacent
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Conversion from meters to cm is wrong.(thats all I know) 7/1000 is 0.007m
 
  • #5
Conversion from meters to cm is wrong.(thats all I know) 7/1000 is 0.007m
Isn't a centimeter 1/100th of meter?
 
  • #6
adjacent
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Ah.Sorry,I was mistaken.You are right.
 
  • #7
adjacent
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It seems you didn't read this part(find the speed of the block after it has fallen through a height of 0.43·m. Hope it helps.
 
  • #8
haruspex
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You need to consider the pulley and the mass separately. What they have in common are the tension in the string and the acceleration. Draw a FBD for each and write out the equations (ƩF=ma etc.)
 

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