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KE of Rotation & Moment of Inertia

  • Thread starter valeriex0x
  • Start date
  • #1
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Homework Statement


A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string, as shown in the figure . When the system is released from rest, you determine that the stone reaches a speed of 3.50 m/s after having fallen 2.50 m.

What is the mass of the cylinder?


Homework Equations



KE=1/2(1/2MR^2)ω^2

v=rω

The Attempt at a Solution



v=rw
3.50=.375w
w=9.3

ke=1/2(1/2M(.375)^2)(9.3)^2

I am doing this right and
is 30406kg correct?
 

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Answers and Replies

  • #2
I like Serena
Homework Helper
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Hi again valeriex0x! :smile:

What did you do with the kinetic energy of the stone?

Btw, the mass you found looks a bit large doesn't it?
 
Last edited:
  • #3
43
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i got it
 

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