How Is Acceleration Calculated for a Freely Falling Yoyo?

[vu10758]

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.

 

[OlderDan]

Isn't there something else you know about the construction of that yo-yo?

 

[vu10758]

I know that the yoyo is a cylinder that spins counter clockwise going down and clockwise going up. It is attached to a fixed place a tthe top. I don't know what I am missing.

 

[OlderDan]

I know that the yoyo is a cylinder that spins counter clockwise going down and clockwise going up. It is attached to a fixed place a tthe top. I don't know what I am missing.

Isn't there an inner radius for the axle and an outer radius for the cylinder?

 

[vu10758]

Yes. So I have to include both in the problem. The given R is the outer radius, and the inner radius is not explicitly mentioned.

 

[OlderDan]

Yes. So I have to include both in the problem. The given R is the outer radius, and the inner radius is not explicitly mentioned.

I don't see how you can do the problem without it. Maybe the problem assumes the string is wound around the outside of the cylinder. If that is the case, then the problem can be done with α = a/R. That is not very realistic for a yo-yo, but it does simplify the calculation.

That is what they assumed to get their answer. I wouldn't call this a yo-yo. It's just a cylinder with a string wrapped around it, but the problem can be done and their answer is OK. I will go back to the first post and look at what you did. You were off to a good start.

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 <== this F is T

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

F = (1/2)M*R*alpha = (1/2)M*a = T
Ma = Mg - (1/2)M*R*alpha = Mg - (1/2)M*a
Ma + (1/2)M*a = Mg
(3/2)M*a = Mg
a = (2/3)g
T = (1/2)M*a = (1/2)M*(2/3)g = (1/3)M*g

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a =(1/2)*alpha *R == The rest is replaced ==

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.

Sorry if I led you astray. I just didn't get this being a yo-yo problem.