M2 Energy Question of a rotating rod Watch

kennz
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Hi there, I'm having trouble with a question in the oxbox textbook for edexcel.

14. A particle of mass m is attached to the end A of a light rod OA of length a. O is freely hinged to a fixed point and the rod is held in a horizontal position before being released. If the particle had been intially projected downwards with speed u, find the value of u for which the rod would just tracel round a complete circle.

I think it would travel around a circle radius a and the angular velocity is constant. The change in GPE would be 0 as there is no change in height if it travels a complete circle. But this is all i can gather so far.

Thanks,
PS the answer is sqrt(2ga)
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EricPiphany
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(Original post by kennz)
Hi there, I'm having trouble with a question in the oxbox textbook for edexcel.

14. A particle of mass m is attached to the end A of a light rod OA of length a. O is freely hinged to a fixed point and the rod is held in a horizontal position before being released. If the particle had been intially projected downwards with speed u, find the value of u for which the rod would just tracel round a complete circle.

I think it would travel around a circle radius a and the angular velocity is constant. The change in GPE would be 0 as there is no change in height if it travels a complete circle. But this is all i can gather so far.

Thanks,
PS the answer is sqrt(2ga)
The question isn't clear, but the rod rotates in a vertical plane.
EDIT: The question does say projected downwards though, so it is quite clear.
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the bear
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when it reaches the horizontal again it must have sufficient KE to continue to the highest point.

the KE at the horizontal will equal the KE at the start which is 0.5 m u2
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kennz
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(Original post by the bear)
when it reaches the horizontal again it must have sufficient KE to continue to the highest point.

the KE at the horizontal will equal the KE at the start which is 0.5 m a2
So you would equate 0.5a^2 and 0.5u^2? I dont really understand what you mean, sorry.
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the bear
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(Original post by kennz)
So you would equate 0.5a^2 and 0.5u^2? I dont really understand what you mean, sorry.
oops... i mean 0.5 m u2

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kennz
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(Original post by the bear)
oops... i mean 0.5 m u2

Oh right haha. Yes I know this but i need to get an answer of sqrt(2ga) and I just cant get it. The rod would travel a distance of the circumference of the circle which is 2*pi*a...
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the bear
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(Original post by kennz)
Oh right haha. Yes I know this but i need to get an answer of sqrt(2ga) and I just cant get it. The rod would travel a distance of the circumference of the circle which is 2*pi*a...
so the KE as it is just about to move above the horizontal is 0.5 m u2

this must exactly match the increase in GPE as it reaches the top...

this increase is mgh... see if you can continue...
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EricPiphany
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(Original post by kennz)
Hi there, I'm having trouble with a question in the oxbox textbook for edexcel.

14. A particle of mass m is attached to the end A of a light rod OA of length a. O is freely hinged to a fixed point and the rod is held in a horizontal position before being released. If the particle had been intially projected downwards with speed u, find the value of u for which the rod would just tracel round a complete circle.

I think it would travel around a circle radius a and the angular velocity is constant. The change in GPE would be 0 as there is no change in height if it travels a complete circle. But this is all i can gather so far.

Thanks,
PS the answer is sqrt(2ga)
Hint: For the rod to travel in a full circle, the speed at the highest point has to be greater than zero. (Why?)
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kennz
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(Original post by the bear)
so the KE as it is just about to move above the horizontal is 0.5 m u2

this must exactly match the increase in GPE as it reaches the top...

this increase is mgh... see if you can continue...
got it thank you!
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kennz
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There's another question I'm having trouble with too.

A light rod AB of length 3a has particles of mass m attached at A and B. The rod rotates in a vertical plane about O where OA=a. The rod is held in a horizontal position and then released. Find the maxium speed of B in the subsequent motion.

Answer is sqrt(2ga/3)

Thanks again

I've had a go and got this:

change in GPE for particle at A = -mga and change in GPE for particle at B =2mga

Change in KE for particle at A = o.5mv^2 and change in KE for particle at B = mv^2

equate KE and GPE and you get the answer but Im not sure this is correct.
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kennz
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(Original post by the bear)
so the KE as it is just about to move above the horizontal is 0.5 m u2

this must exactly match the increase in GPE as it reaches the top...

this increase is mgh... see if you can continue...
There's another question I'm having trouble with too.

A light rod AB of length 3a has particles of mass m attached at A and B. The rod rotates in a vertical plane about O where OA=a. The rod is held in a horizontal position and then released. Find the maxium speed of B in the subsequent motion.

Answer is sqrt(2ga/3)

Thanks again

I've had a go and got this:

change in GPE for particle at A = -mga and change in GPE for particle at B =2mga

Change in KE for particle at A = o.5mv^2 and change in KE for particle at B = mv^2

equate KE and GPE and you get the answer but Im not sure this is correct.
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the bear
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is O between A and B or to one side ?
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kennz
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(Original post by the bear)
is O between A and B or to one side ?
between A and B sorry for the late reply
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the bear
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i would say replace the rod with a point mass at the midpoint, so 0.5a from 0.

this mass gains KE as it falls. the maximum KE must be at the bottom... it has fallen 0.5a downwards...

so you can find the v for this midpoint and then scale it up for the end point B
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kennz
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(Original post by the bear)
i would say replace the rod with a point mass at the midpoint, so 0.5a from 0.

this mass gains KE as it falls. the maximum KE must be at the bottom... it has fallen 0.5a downwards...

so you can find the v for this midpoint and then scale it up for the end point B
so you equate mga with mv^2 ?
since there
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the bear
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(Original post by kennz)
so you equate mga with mv^2 ?
since there
the COM only falls 0.5a

so 2m*g*0.5a = 0.5*2m*v2

which looks equivalent to your version...
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kennz
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(Original post by the bear)
the COM only falls 0.5a

so 2m*g*0.5a = 0.5*2m*v2

which looks equivalent to your version...
Ohhh ok sorry we havent done much stuff with reconceptualising it with a point mass. I find it quite weird. The answer is sqrt(2ga/3) so idk where the 3 comes from
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the bear
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(Original post by kennz)
Ohhh ok sorry we havent done much stuff with reconceptualising it with a point mass. I find it quite weird. The answer is sqrt(2ga/3) so idk where the 3 comes from
at the moment [no pun intended ] we have found the v for the COM.

as you move away from the pivot the v increases linearly... since B is 2a from the pivot compared to the COM which is 0.5a from the pivot you can work out the v for B
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kennz
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(Original post by the bear)
at the moment [no pun intended ] we have found the v for the COM.

as you move away from the pivot the v increases linearly... since B is 2a from the pivot compared to the COM which is 0.5a from the pivot you can work out the v for B
So for B

0.5&m*v^2 = 2a*m*g ?
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the bear
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(Original post by kennz)
So for B

0.5&m*v^2 = 2a*m*g ?
i am getting v = 4 √ ( ag ) for B

either their answer is wrong or i am :dontknow:
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