Speed of pendulum at certain angles

[Bronzer]

I'm having no problem finding the maximum speed of a pendulum but I'm not sure when I move its start point and measure point.

I would like to know the speed of a pendulum at 30 degrees that has been dropped at 140 Degrees. I'm just giving some numbers to explain that it is above the horizontal and doesn't pass through equilibrium.

Any thought's on this would be useful.

Thanks.

 

[phinds]

You need to give more background. Where are you in school? What math have you taken?

 

[Bronzer]

Out of school, AS level Math.

This isn't homework, just a practical problem I'm trying to get my head around.

All of my usual point of calls don't like pendulums that go above the horizontal (I guess GCSE assumes it's made of string) and I can only find equations for Vmax.

 

[sophiecentaur]

As with many such problems, you can either approach this problem by considering the Forces involved or by considering the Energy transfer. The Maths is pretty straightforward if you work out the change in gravitational potential energy and the corresponding gain in Kinetic Energy. That is an enormous clue!
That assumes the pendulum uses a solid rod (like most practical pendulums, aamof) If you want to assume string then you would have to know an awful lot more about the set up - the shape of the bob, for instance.

 

[Bronzer]

As with many such problems, you can either approach this problem by considering the Forces involved or by considering the Energy transfer. The Maths is pretty straightforward if you work out the change in gravitational potential energy and the corresponding gain in Kinetic Energy. That is an enormous clue!
That assumes the pendulum uses a solid rod (like most practical pendulums, aamof) If you want to assume string then you would have to know an awful lot more about the set up - the shape of the bob, for instance.

Right, so if I take energy of pendulum equation as

PE = mgL(1 – COS θ)

Mass should be irrelevant for this(?)

9.81*1*(1-cos(140))=13.2
9.81*1*(1-cos(30))=7.5

= 5.7j change

KE=1/2mv^2
2KE=mv^2
sqrt(2KE)=mv

Ignore M, so that V = sqrt(2*5.7)
=3.38m/s

Is that looking ok?

 

[sophiecentaur]

You lost L somewhere? But that's the sort of thing I meant.

 

[Bronzer]

Yeah L=1, sorry.

 

[sophiecentaur]

By choosing a magic number for L you have thrown away an important part of the relationship. The speed cannot be independent of L, can it, so you should include the length even if you later happen to give it a value of 1m, 1cm, 1km.