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## Homework Statement

A pendulum of mass m and length L is released

from rest at an angle Ф to the vertical. There is no

friction and g = 9.81 m/s2.

Determine the normal force in the cord when Ф =

0°.

http://img87.imageshack.us/my.php?image=pend.jpg

Note that phi is giving in degrees.

**2. The attempt at a solution**

Decomposing in the radial direction, and taking T as the tension in the cord:

[tex]\sum F_{r} = m*a_{r} = - T + mgcos\Phi = m(-L\omega[/tex]²) (1)

Decomposing in transverse direction:

[tex]\sum F_{\theta} = m*a_{\theta} = mgsin\Phi = m(L\alpha)[/tex]

[tex]\alpha = d\omega/d\Phi * d\Phi/dt = d\omega/d\Phi * \omega [/tex]

So:

[tex]mgsin\Phi = m(L(d\omega/d\Phi)\omega)[/tex]

[tex]sin\Phi d\Phi = (L/g) \omega d\omega[/tex]

When the angle is phi, the angular velocity is 0. When the angle is 0, the angular velocity is omega.

[tex](L/g) \int^{\omega}_{0}\omega d\omega = \int^{0}_{\Phi} sin\Phi d\Phi[/tex]

[tex]\omega^{2} = -2(g/L)(1 - cos \Phi)[/tex]

So I end up with a negative term for my angular velocity square. So I can't calculate my angular velocity. Even if I use this omega square and stick it into (1), the resulting equation for T is wrong.

This was easily solvable with the principle of work and energy, however this method specifically is required.

Can someone give me a headsup of where I've gone wrong? Thanks!

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