Simple pendulum gravitational potential energy at 30°

[marc_2094]

Homework Statement

A simple pendulum is swinging. Its mass is m and its length is L. What is its gravitational potential energy if it is oriented at 30° with respect to the vertical? (Let gravitational potential energy be zero if it is in the vertical orientation).

Homework Equations

U = mgh

The Attempt at a Solution

Given mass = m and length = L, the only thing left is to solve for h. Since it is oriented 30° wrt vertical, then h = Lcos30°. So U = mgLcos30°

But it's not the answer.

 

[gneill]

Homework Statement

A simple pendulum is swinging. Its mass is m and its length is L. What is its gravitational potential energy if it is oriented at 30° with respect to the vertical? (Let gravitational potential energy be zero if it is in the vertical orientation).

Homework Equations

U = mgh

The Attempt at a Solution

Given mass = m and length = L, the only thing left is to solve for h. Since it is oriented 30° wrt vertical, then h = Lcos30°. So U = mgLcos30°

But it's not the answer.

Hi marc_2094, Welcome to Physics Forums.

Did you draw a diagram showing the pendulum in both orientations? Pay close attention to the location of the line segment that represents the difference in height.

 

[marc_2094]

This is my diagram. Is it correct?

 

[gneill]

This is my diagram. Is it correct?

I don't see the original height (zero reference point for the gravitational potential) indicated, or the line segment representing the change in height.

Mark the original position of the pendulum bob and its new position (at 30°). Is the change in height equal to the length of the triangle leg as you've shown?