# Mechanics QuestionWatch

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Thread starter 2 weeks ago
#1
Question
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Thread starter 2 weeks ago
#2
Question:
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Thread starter 2 weeks ago
#3
ll
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2 weeks ago
#4
(Original post by AlishaWhite)
ll
What have you tried?
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Thread starter 2 weeks ago
#5
(Original post by RDKGames)
What have you tried?
normally you would equate potential energy to kinetic but it wants the end of the rod and it is a rotation. So it wouldnt be mgh = 0.5mv²
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2 weeks ago
#6
(Original post by AlishaWhite)
normally you would equate potential energy to kinetic but it wants the end of the rod and it is a rotation. So it wouldnt be mgh = 0.5mv²
Yep, so you need to consider rotational K.E.

How much GPE does the rod lose in the process of motion?

This must be equal to the rotational KE when the rod is vertical, and this is given by , where is the moment of intertia of the rod about the end which is fixed to the ceiling, and is the angular speed.

Rearrange this equality for . Then clearly, velocity of the other end of the rod is
Last edited by RDKGames; 2 weeks ago
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Thread starter 2 weeks ago
#7
(Original post by RDKGames)
Yep, so you need to consider rotational K.E.

How much GPE does the rod lose in the process of motion?

This must be equal to the rotational KE when the rod is vertical, and this is given by , where is the moment of intertia of the rod about the end which is fixed to the ceiling, and is the angular speed.

Rearrange this equality for . Then clearly, velocity of the other end of the rod is
In E = mgh

would h = L or would it be 0.5L
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2 weeks ago
#8
(Original post by AlishaWhite)
In E = mgh

would h = L or would it be 0.5L
It's a uniform rod, so the CoM is halfway along the rod.
Last edited by RDKGames; 2 weeks ago
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Thread starter 2 weeks ago
#9
(Original post by RDKGames)
It's a uniform rod, so the CoM is halfway along the rod.
Ok thanks.

Not rlly sure on this on either:
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2 weeks ago
#10
(Original post by AlishaWhite)
Ok thanks.

Not rlly sure on this on either:
1/total resistance for the LHS would be (1/R1+R2)+1/R3 and so you equate that to Rtot of the RHS which is (1/Rp+2R3), make Rp the subject
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Thread starter 2 weeks ago
#11
(Original post by Varss)
1/total resistance for the LHS would be (1/R1+R2)+1/R3 and so you equate that to Rtot of the RHS which is (1/Rp+2R3), make Rp the subject
I got the LHS but for RHS did u mean 1/(Rp + 2R3) or (1/Rp) + 2R3
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2 weeks ago
#12
(Original post by AlishaWhite)
I got the LHS but for RHS did u mean 1/(Rp + 2R3) or (1/Rp) + 2R3
sorry that's my fault lol I meant 1/(Rp)+2/(R3)
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