Rotational Dynamics Question

  • Thread starter magnummd
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  • #1
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Homework Statement



Problem: A massless string is wrapped around a solid cylinder as shown in the diagram at the right. A block of mass m = 2.0 kg hangs from the string. When released, the block falls a distance of 82 cm in 2.0 s. Starting with a free-body diagram, calculate the mass of the cylinder.

Homework Equations



d = Vot + 0.5at^2
Tension = ma - mg
Torque = (I)(alpha) = F(r)

The Attempt at a Solution



Conceptually, I figure I need to find the force being exerted on the wheel, and the acceleration. That will leave nothing but the mass to be determined. But the fact that it's rotational, throws a wrench into it, especially since no radius is provided.

using d = Vot + 0.5at^2 we get a = 0.41 m/s^2 for the block.
Tension is therefore, 18.78 N in the opposite direction of the block's movement and is the force being exerted on the wheel (I think).

Torque = 18.78(r) = 0.5Mr^2(a/r)

If my approach is right, I need M, but r is in the way. What to do?
 
Last edited:

Answers and Replies

  • #2
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Torque = 18.78(r) = 0.5Mr^2(a/r)

If my approach is right, I need M, but r is in the way. What to do?
Wait a minute... the radii all cancel... therefore M = 37.56/0.41 = 91.6 kg

Did I figure it out, or did I just dig myself into a deeper hole? Help please!
 
  • #3
Pyrrhus
Homework Helper
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Could you give more information about the geometry?
 
  • #4
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I don't really know where I could attach a photo of it, but it essentially looks like a string hanging off a pulley with a mass on the end.

Any suggestions on how I can show it to you?
 
Last edited:
  • #5
Pyrrhus
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When you make a new reply below "submit reply" look for a button called "Manage Attachments".
 
  • #8
I'm new to this forum but just a general hint i post my pictures on photobucket.com and then just click the image button where i want it in the post and paste in the link ...
 
  • #10
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Anything guys (or ladies)?

The attachment's been approved. So does anyone have any ideas?
 
  • #11
Doc Al
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Your solution looks good to me.
 
  • #12
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Alright then! Thanks a lot!
 

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