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M2 Energy Question of a rotating rod

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Reply 20
Avatar for kennz
kennz
OP
8
Original post by the bear
i am getting v = 4 √ ( ag ) for B :colondollar:

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...
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
Reply 22
Avatar for kennz
kennz
OP
8
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
(edited 8 years ago)

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