Solving Forces Acting on a Swinging Gymnast

[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 traveling 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.

<br /> \begin{array}{c}<br /> E_M = E_k + E_p \\ <br /> = \frac{{mv^2 }}{2} + mgh \\ <br /> = 980J \\ <br /> \end{array}<br />

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

<br /> \begin{array}{l}<br /> 980 = 20v^2 \\ <br /> v = 7ms^{ - 1} \\ <br /> \end{array}<br />

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

<br /> \begin{array}{c}<br /> F_c = \frac{{mv^2 }}{r} \\ <br /> = 1633.\overline {33} \\ <br /> \end{array}<br />

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:

<br /> \begin{array}{c}<br /> \sum F = F - mg \\ <br /> F = \sum F + mg \\ <br /> = 2033.\overline {33} \\ <br /> \end{array}<br />

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

Thanks,
Dan.

 

[neutrino]

The total energy is off by 20J.

 

[danago]

The total energy is off by 20J.

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?

 

[danago]

anyone have any input?