Question

A yo-yo of mass M = 0.210 kg and outer radius R that is 4.50 times greater than the radius of its axle r is in equilibrium if a mass m is suspended from its outer edge as shown in the figure below. Determine the tension in the two strings, T1 and T2 and the mass m. T1 = N T2= N kg m = m

Answer 1

Solution:

The Total weight (m+0.210) kg is supported by inner string

Thus, T1 = (m+0.210)g N

And, T2*R = T1*r

Thus, T2 = (T1*r) / R = T1 / 4.50 N [ Outer Radius R = 4.50*inner radius(r)]

Now, T2 = mg

Therefore, T1 = 4.50*mg = (m+0.210)g

Or, 3.50mg = 0.210g

Then, m = 0.06 kg [Answer]

Therefore, T2 = 0.06kg*9.8 m/s^2 = 0.588 N[Answe]

And T1 = (0.06+0.210)*9.8 = 2.646 N[Answer]