Conservation of Energy with kinematics

  • Thread starter joej24
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  • #1
joej24
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Homework Statement



(a) If R = 12 cm, M = 570 g, and m = 50 g in Figure 10-18 (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles.

M is the mass of the mounted uniform disk.
A block with mass m hangs from a massless cord that is wrapped around the rim of the disk. The cord does not slip, and there is no friction at the axle.


(b) Repeat (a) with R = 5.0 cm.

Homework Equations



F = ma
alpha * R = a

torque = I * alpha

U+K = U+K

The Attempt at a Solution




I understood how to solve this problem (#9) with kinematics ( solving for acceleration and then using v^2 = vnot^2 + 2ad) but am having trouble with using conservation of energy.


This was my attempt

U + K = U + K
//It is initially at rest
0 + 0 = -mgh + Krotational + Ktangential

mgh = 05*I*omega + 0.5mv

I = 0.5mr = .004104 kg * m

a = 1.4626 so alpha = 12.189 rad/s

I don't think my initial setup is correct though because I do not get the correct answer



If someone could show me how I can post the image up here that'd be great too
Thanks!
 

Answers and Replies

  • #2
ehild
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Check the equations for the kinetic energies and moment of inertia.


ehild
 
  • #3
collinsmark
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A few things to get you started:
mgh = 05*I*omega + 0.5mv
I think you mean
mgh = ½2 + ½mv2

I'm guessing you just made a typo.
a = 1.4626 so alpha = 12.189 rad/s
There's no point in solving for the linear acceleration or the angular acceleration, when using conservation of energy for this problem.

All you need to do is solve for the velocity. You'll notice that there is an ω in your conservation of energy equation. Given the way the setup is described, with a cord wrapped around the rim of the disk, there is a very simple relationship between the linear velocity v and the angular velocity ω. Make the appropriate substitution of ω in terms of v. Then simply solve for v.
bump
Per the forum rules we can't help you unless you show that you've put an effort into solving the problem. You mentioned, "I don't think my initial setup is correct though because I do not get the correct answer." So, .... show us your work, and how you got to an answer. :smile:
 

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