Hi everyone,(adsbygoogle = window.adsbygoogle || []).push({});

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:

[tex]\overline{I} = 15.246 kg m^2[/tex]

[tex]I_o = 103.576 kg m^2[/tex] (this is moment of inertia about the fixed point of rotation)

[tex]m = 73 kg[/tex]

[tex]\omega_0 = 0 rad/s[/tex]

[tex]\omega = 3.9 rad/s[/tex]

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:

[tex]Fx = mr\omega^2 = 1221 N[/tex]

[tex] \sum{M_G} = F_y(1.1) = 15.246\alpha[/tex]

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:

[tex]mg(1.1) = I_o\alpha[/tex]

This yields [tex]\alpha = 7.605 rad/s^2[/tex].

METHOD 2: Using kinematic equations:

[tex]\omega^2 = \omega_0^2 + 2\alpha(\theta)[/tex]

This gives me [tex]\alpha = 4.84 rad/s^2[/tex]. (The [tex]\theta[/tex] in this case is 90 degrees, or [tex]\pi/2[/tex])

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?

Thanks in advance!

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# Homework Help: Angular Acceleration

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