# Homework Help: Angular Acceleration

1. Apr 9, 2004

### AngelofMusic

Hi everyone,

I'm just a bit confused about calculating angular acceleration and knowing when it is constant or not. Here's the problem:

A gymnast is swinging on a horizontal bar, and we're asked to calculate the forces in his hands when he is at 90 degrees with the bar. He was originally completely vertical. [ http://img23.photobucket.com/albums/v68/AngelOfMusic/FBDGymnast.jpg [Broken] ] (I basically represented the gymnast by a bar, but it doesn't matter in this case, since the moment of inertia is given.)

Information I've already found & confirmed:
$$\overline{I} = 15.246 kg m^2$$
$$I_o = 103.576 kg m^2$$ (this is moment of inertia about the fixed point of rotation)
$$m = 73 kg$$
$$\omega_0 = 0 rad/s$$
$$\omega = 3.9 rad/s$$

In the diagram, the distance between G and the location of the forces in his hands is 1.1 m. Now, when calculating the forces:

$$Fx = mr\omega^2 = 1221 N$$
$$\sum{M_G} = F_y(1.1) = 15.246\alpha$$

The part where I get stuck is in calculating the angular acceleration. I tried with two methods and got two different answers.

METHOD 1: Taking the moment about the fixed centre of rotation:
$$mg(1.1) = I_o\alpha$$
This yields $$\alpha = 7.605 rad/s^2$$.

METHOD 2: Using kinematic equations:
$$\omega^2 = \omega_0^2 + 2\alpha(\theta)$$
This gives me $$\alpha = 4.84 rad/s^2$$. (The $$\theta$$ in this case is 90 degrees, or $$\pi/2$$)

The first method gives me the correct answer for the final force according to the book, but I have no idea why the two methods would produce different answers. I think the angular acceleration *should* be uniform because the only external force applied is the gravitational force, which is constant. So why couldn't I use the kinematics version of the equation to find the angular acceleration?

Last edited by a moderator: May 1, 2017
2. Apr 9, 2004

### Staff: Mentor

The angular acceleration is not constant. While the gravitational force is constant, the torque it exerts depends on the angle. The torque only equals mgx when the angle of the gymnast is 90° (horizontal). In general the torque is:
$$\tau = \vec{r} \times \vec{F} = rFsin\theta$$
where F is the force (mg, acting down) and θ is the angle between r (the vector distance from the pivot to the point of application of the force) and the force.

3. Apr 9, 2004

### AngelofMusic

Ah, okay! Thanks a lot! That makes sense now.