What is the minimum speed needed for a pendulum mass to reach its highest point?

[OVB]

A mass is attached to a pendulum of length L that is at an angle theta (8) with respect to the normal. The mass has an initial velocity Vo. What is the minimum speed (symbolically) needed for the mass to go all the way down to its lowest point and all the way back up its highest point (as in, the rope of the pendulum is projected vertically upwards).

I set 1/2(m)Vo^2 + mgL(1-cos(8)) = 2mgL

and got Vo = square root of 2gL + 2gLcos(8), but my book says it is the square root of 3gL + 2gLcos(8). I don't see what's wrong with my set up, as I have checked the result of this set up several times and got the same answer.

 

[OVB]

no, I meant that it was originally at an angle with respect to what it would be at rest (i.e., bob below pivot) , and the velocity is supposed to be like that since it is supposed to be solved symbolically.

 

[OlderDan]

no, I meant that it was originally at an angle with respect to what it would be at rest (i.e., bob below pivot) , and the velocity is supposed to be like that since it is supposed to be solved symbolically.

OOPs. I inadvertantly deleted my first message. And OOPs again, I did misinterpret the potential energy term in your original equation. You are quite correct about the initial potential energy. However, I was correct that you have neglected the velocity that the bob must have at the higest point. The bob cannot possibly reach that point with zero velocity. In order to reach that point, the bob must travel on a circular path. The net force at the top of the circle must be just enough to keep it in circular motion. If the bob had zero kinetic energy at the top, it would fall straight down, but it could never even get to the top if it didn't have enough kinetic energy in the first place. You need a kinetic energy term on the right hand side of your equation, and you need to find the velocity from circular motion considerations.