# Tangental Force

1. Apr 11, 2010

### Johnny0290

1. The problem statement, all variables and given/known data

In this problem we want to learn a little bit about what is sometimes called dynamical loading. Our simple system consists of a uniform stick of length L and mass M hinged at one end. We would like to calculate the forces on the (frictionless) hinge when the stick is released from rest at an angle theta_0 with respect to the vertical. You may find it useful to combine work and energy equations with torque (N II) equations.

3.2 Show that the tangential (tangent to the direction of motion, perpendicular to the stick) force exerted on the stick by the hinge is F_t = Mg/4*sin(theta).

2. Relevant equations

F=ma
I=1/3mr^2
torque=r x F=I*alpha

3. The attempt at a solution

F*d=I*alpha
(L/2)mgsin(theta)=(1/3)mL^2 * alpha

((3/2)mgsin(theta))/L = m*alpha

alpha = L*a

(3/2)mgsin(theta)=ma

Am I approaching this problem in the completely wrong way or am I missing something that factors into the tangental force? Any help is appreciated. Thanks!

2. Apr 12, 2010

### tiny-tim

Hi Johnny0290 !

(have a theta: θ and an omega: ω and an alpha: α and a tau: τ and try using the X2 and X2 tags just above the Reply box )

Yes, that's fine until …
… but all that does is give you is the tangential acceleration of the end of the stick.

You need to find the acceleration of the centre of mass of the stick, so that you can apply good ol' Newton's second law to find the "missing" reaction force