How Long Does It Take for a Child to Swing Back and Forth on a Tire Swing?

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Homework Statement

A horizontal tree branch is directly above another horizontal tree branch. The elevation of the higher branch is 9.45 m above the ground, and the elevation of the lower branch is 5.58 m above the ground. Some children decide to use the two branches to hold a tire swing. One end of the tire swing's rope is tied to the higher tree branch so that the bottom of the tire swing is 0.46 m above the ground. This swing is thus a restricted pendulum. Starting with the complete length of the rope at an initial angle of 14.6° with respect to the vertical, how long does it take a child of mass 28.5 kg to complete one swing back and forth?

Homework Equations

T=2∏√I/mgd

The Attempt at a Solution

I used the T=2∏√I/mgd equation, but I'm having trouble finding d because I don't know where the center of mass is. Also, What branch would I use?

 

[JHamm]

Did this problem come with a picture? It sounds to me that at the bottom of the swing the rope then comes in contact with the lower branch and this branch becomes the new pivot point. If that is the case then the period would be double the time it takes the child to reach the bottom of the swing + double the time it takes the child to reach their peak height when the pivot changes to the lower branch (the doubling comes from the symmetry of the pendulum).

 

[technician]

I agree with JHamm. Unless there is some essential information missing I think this is a long pendulum for 1/2 of the cycle and a short pendulum for the other 1/2.
I make the length of the long pendulum 8.99m and the length of the short pendulum 5.12m.
The time period for a pendulum is a straight forward expression.
T = 2∏√l/g