Rolling yo-yo and weight

In summary, the problem involves a 16lb yo-yo on a horizontal table with dimensions r = 4in and R = 12in, with a radius of gyration about the center of mass of 6in. A cord is wrapped around the inner radius, passes over a pulley, and is attached to an 8lb block. The acceleration of the block and tension in the cord are to be found, with answers of 4.830ft/s^2 and 6.792lb respectively. The key equations used are F=ma, torque=I*alpha, Iaxis=Icm+mh^2, and I=(radius-gyration)R^2. The missing equation needed to solve
  • #1
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Homework Statement


a 16lb yo-yo is resting on a horizontal table and is free to roll. The dimensions are r = 4in and R = 12in and the radius of gyration about the CM is 6in. A cord is wrapped around the inner radius, the passes over a perfect pulley, and is attached to a 8lb block.

Find acceleration of 8lb block and the tension in the cord

(answers are 4.830ft/s^2 and 6.792lb)

Homework Equations



F = ma
torque = I*alpha = F*moment-arm
Iaxis = Icm + mh^2
I = (radius-gyration)R^2




The Attempt at a Solution



there are two accelerations a1 (weight) and a2 (yo-yo).

(16/32slugs)*a2 = T

I at table contact point = 90slug*in^2

net torqe at table contact point = 90slug*in^2 * alpha = T(12in - 4in)

T = 11.25*alpha and 0.5slugs * a2 = T

for weight, (8/32slugs)a1 = 8lb - T

so far, I have 4 unknowns and 3 equations so I can't solve for anything. What equation am I missing? i don't know how to find a relationship between alpha and a2 (i thought a2 = 12in*alpha but that doesn't work out right).

thanks in advance
 
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  • #2
Consider pure rotation about axis passes through contact point. the center of mass is 12 in from contact point.
 
  • #3
i found out what i was missing: an equation describing the relationship between the acceleration of the weight and the accleration of the yo-yo.

a-weight = a-yo-yo - R*alpha.
 

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