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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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