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A yoyo is allowed to drop freely with a string held fixed in place at the top. Assume that the yoyo is a uniform disk of mass M and radius R.

1) Use Newton's Second Law to find the linear acceleration of the yoyo and the tension in the rope.

My answer key says that the answers are 2/3 g for acceleration and (1/3)Mg for tension

I have the net force = ma, and net torque = I*alpha.

so

Net force : mg - T = ma

Net torque = I*alpha

since this is a cylinder

I = (1/2)MR^2

I also know that torque is F*R

I have (1/2)M*R^2*alpha = F*R

F = (1/2)M*R*alpha

Ma = (1/2)M*R*alpha

a =(1/2)*alpha *R

alpha = a/R

so a= (1/2)a

This is not true.

I know how to find tension. I plugged in the correct value for a into the mg-T=ma. But I can't find the acceleration.

1) Use Newton's Second Law to find the linear acceleration of the yoyo and the tension in the rope.

My answer key says that the answers are 2/3 g for acceleration and (1/3)Mg for tension

I have the net force = ma, and net torque = I*alpha.

so

Net force : mg - T = ma

Net torque = I*alpha

since this is a cylinder

I = (1/2)MR^2

I also know that torque is F*R

I have (1/2)M*R^2*alpha = F*R

F = (1/2)M*R*alpha

Ma = (1/2)M*R*alpha

a =(1/2)*alpha *R

alpha = a/R

so a= (1/2)a

This is not true.

I know how to find tension. I plugged in the correct value for a into the mg-T=ma. But I can't find the acceleration.

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