# Circular Motion

1. May 1, 2007

### danago

A 40kg gymnast is swinging on a horizontal bar. Her center of mass is 1.2m from the bar, and right at the top of the circle she is travelling in, her body has a speed of 1m/s.

What force must she hold onto the bar with right at the bottom of the swing if she is to continue swinging?

Well first thing i did was calculate the mechanical energy in the system right at the top of the swing.

$$\begin{array}{c} E_M = E_k + E_p \\ = \frac{{mv^2 }}{2} + mgh \\ = 980J \\ \end{array}$$

Since energy is conserved, i used this to calculate the tangental velocity at the bottom of the swing.

$$\begin{array}{l} 980 = 20v^2 \\ v = 7ms^{ - 1} \\ \end{array}$$

Using this, i can calculate the centripetal force (net force).

$$\begin{array}{c} F_c = \frac{{mv^2 }}{r} \\ = 1633.\overline {33} \\ \end{array}$$

At the bottom of the swing, the force she holds on with and the weight force act in opposite directions, and i can say that:

$$\begin{array}{c} \sum F = F - mg \\ F = \sum F + mg \\ = 2033.\overline {33} \\ \end{array}$$

The answer the book gives is different however. Im not really sure what i have done wrong. Any help?

Thanks,
Dan.

2. May 1, 2007

### neutrino

The total energy is off by 20J.

3. May 1, 2007

### danago

If i use g=9.8ms-2, i get a total energy of 960.8J, but the book is using g=10. Is that the 20J difference youre talking about? Or have i missed something else?

4. May 2, 2007

### danago

anyone have any input?