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Conservation of energy in case of rotating falling rod

by Perpendicular
Tags: case, conservation, energy, falling, rotating
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Perpendicular
#1
Nov1-12, 06:25 PM
P: 49
Suppose we have a rod standing vertically and then slightly disturbed so it begins to fall. After it falls through some height or angle assuming a clockwise rotational fall I can see that the left end is sliding on the surface ( for simplicity I'm ignoring friction ) horizontally. Here I assume that I know the change in height although knowing length of rod and angular displacement would work as well.

In that case, would I be right to say that the change in PE equals

1/2 I w^2 + 1/2 M v^2

For I , v and w at the Center of mass ? While that is normally the case for combined translation and rotation, I'm having doubts using that expression here as the leftmost end doesn't seem to rotate about the CM, it's sliding instead. Should I try to take the CM's source of torque ( the normal force on the leftmost point of contact ) and work from there to obtain some results ?
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tiny-tim
#2
Nov1-12, 06:40 PM
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Hi Perpendicular!

(try using the X2 button just above the Reply box )
Quote Quote by Perpendicular View Post
In that case, would I be right to say that the change in PE equals

1/2 I w^2 + 1/2 M v^2

For I , v and w at the Center of mass ?
Yes, that is the correct formula for KE of a general body.
Keep calm and carry on!
Perpendicular
#3
Nov2-12, 04:25 AM
P: 49
well it is but I'm confused regarding what to do with the normal force N...supposing some angular movement theta I am thinking of expressing the torque due to N as some trigonometric function and then integrating over dtheta to get work done due to torque by normal force.

Then I think the eqn will look like (mgh) + (Work done due to the torque by normal force) = (Rotational energy from CM frame) + (KE w.r.t CM )

Am I right ? Or am I wrong in taking into account the work done by the normal force ?

tiny-tim
#4
Nov2-12, 05:05 AM
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Conservation of energy in case of rotating falling rod

Hi Perpendicular!
Quote Quote by Perpendicular View Post
well it is but I'm confused regarding what to do with the normal force N...supposing some angular movement theta I am thinking of expressing the torque due to N as some trigonometric function and then integrating over dtheta to get work done due to torque by normal force.

Am I right ? Or am I wrong in taking into account the work done by the normal force ?
the normal force is vertical, but the displacement of its point of application is horizontal, so the work done is zero

(only the friction force would do work, so only the friction force affects the energy equation)
Then I think the eqn will look like (mgh) + (Work done due to the torque by normal force) = (Rotational energy from CM frame) + (KE w.r.t CM )
(mgh) = (Rotational energy from CM frame) + (KE w.r.t CM )
Perpendicular
#5
Nov2-12, 05:14 AM
P: 49
The body rotates clockwise and the normal force applies a clockwise torque which must do some work over the angular displacement, right ? And from the CM frame, where else would you even get torque ?

after all work = torque.angular change as well.
tiny-tim
#6
Nov2-12, 05:40 AM
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Quote Quote by Perpendicular View Post
The body rotates clockwise and the normal force applies a clockwise torque which must do some work over the angular displacement, right ?
right

but that's only using the component of the normal force that makes the torque

you're ignoring the rest of the normal force (and it's very difficult to work out what that is!)
please calculate work done from forces, not from torques!!
Perpendicular
#7
Nov2-12, 06:07 AM
P: 49
So basically one component of N does work while the other cancels that work out ?
tiny-tim
#8
Nov2-12, 06:27 AM
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basically, N does no work because it is perpendicular to the displacement!

but if you resolve it into perpendicular components at an angle to N (not 0 or 90), then yes, the work done by each component cancel out!


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