M2 Energy Question of a rotating rod Watch

kennz
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#21
Report Thread starter 3 years ago
#21
(Original post by the bear)
i am getting v = 4 √ ( ag ) for B

either their answer is wrong or i am :dontknow:
I was thinking that if you go from the rod point of view.

The mass has maximum velocity when the rod is vertical i.e. B falls a height of 2a and A rises a height of a.

KE=GPE
so
2mga-mga (since one rises and one falls) = mv^2+0.5mv^2 when you consider the centre of mass . ...

cancel masses.

ga=1.5v^2

a=sqrt(2/3 ga) But this seems wrong...
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the bear
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#22
Report 3 years ago
#22
(Original post by kennz)
I was thinking that if you go from the rod point of view.

The mass has maximum velocity when the rod is vertical i.e. B falls a height of 2a and A rises a height of a.

KE=GPE
so
2mga-mga (since one rises and one falls) = mv^2+0.5mv^2 when you consider the centre of mass . ...

cancel masses.

ga=1.5v^2

a=sqrt(2/3 ga) But this seems wrong...
nearly there i think...

if the velocity of A is v then the velocity of B must be 2v as it is twice the distance from the pivot compared to A.

2mga-mga (since one rises and one falls) = 0.5m{2v}^2+0.5mv^2
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kennz
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#23
Report Thread starter 3 years ago
#23
(Original post by the bear)
nearly there i think...

if the velocity of A is v then the velocity of B must be 2v as it is twice the distance from the pivot compared to A.

2mga-mga (since one rises and one falls) = 0.5m{2v}^2+0.5mv^2
they are travelling in opposite directions but this doesnt matter as the velocities are squared. Just cant see to get that 2/3 haha
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