Conservation of energy in rotating bodies

  • #1
Screenshot 2021-03-01 at 2.29.09 AM.png


The conservation of energy equation is basically GPE is converted to KE of block and KE of cylinder.

To get the correct answer, the KE of the cylinder is 1/2mv^2, where m is its mass and v is the velocity of its COM (which is the centre of cylinder).
However, I viewed the cylinder as rotating about the point of contact with the step, and thus thought that its KE was 1/2Iw^2, where I = 3/2mr^2 (by parallel axis theorem), and rw = v, where v is the velocity of the centre of mass.

Why is my calculation of KE for the cylinder wrong?

Thank you for your help.

This problem can be found in Jaan Kalda's Mechanics problems, or in SS Krotov's Aptitude Test Problems.
 

Answers and Replies

  • #2
Dale
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Summary:: Why is the kinetic energy I ascribe to the cylinder in the following question wrong?

Why is my calculation of KE for the cylinder wrong?
Because the cylinder is not rotating.
 
  • #3
A.T.
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I viewed the cylinder as rotating about the point of contact with the step
Why about this contact and not the other one? Why rotating at all?
 
  • #4
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as they above say your work is wrong, but u have some ok intuition. cylinder not rotate about its mass centre, however the mass centre point do rotate in circular arc about the contact point on right (well, so long as it is still contact!). this you can use to advantage! say, mass centre of cylinder has rotate by ##\theta## about right point of contact. let ##v = |\underline{v}|## be speed of the mass centre. can first use the energy conserve$$2gr(1-\cos{\theta}) = v^2$$now, remember that mass centre move as if it were particle which is act upon by resultant force, so you now pretend mass centre is a particle, and resolve resultant of force toward right contact point (i.e. you calculate centripetal component):$$-N + mg\cos{\theta} + N_2 \sin{2\theta} = \frac{mv^2}{r}$$where ##N## is force of contact from right point, and ##N_2## force of contact from left point. now you must do examine cases when ##N = 0## and ##N_2 = 0## respective, i.e. set each to ##0## in turn and use two equation to eliminate ##v##, and find ##\theta_{\text{max}}## for each. then you can tell, which lose contact first.

see - much more easy than do use lot of messy coordinate and difficult equation, yes ;)
 

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