Question

A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickness of the string can be neglected. The yo-yo is released from rest. (a) What is the tension in the cord as the yo-yo descends and as it ascends

Answer 1

Answer:

$T = \frac{gM}{\frac{2b^2}{R^2} + 1}$

Explanation:

If we ignore the thickness of the string, then the tension T of the string on the yoyo would cause a torque with a magnitude of Tb. This torque would then generate an angular acceleration according to Newton's 2nd law:

$Tb = I\alpha$

$Tb = MR^2\alpha/2$

We can substitute angular acceleration as ratio of linear acceleration and spool radius

$\alpha = a/b$

Therefore$Tb = MR^2a/2b$

$T = \frac{aMR^2}{2b^2}$

The system has 2 forces, gravity pulling it down and tension force upward. The net force would generate the linear acceleration a

$Mg - T = ma$

$a = (gM - T)/M$

now we can substitute for a into the tension equation

$T = \frac{(gM-T)MR^2}{2b^2M}$

$T = \frac{gMR^2}{2b^2} - \frac{TR^2}{2b^2}$

$T\left(1 + \frac{R^2}{2b^2}\right) = \frac{gMR^2}{2b^2}$

$T\frac{2b^2 + R^2}{2b^2} = \frac{gMR^2}{2b^2}$

$T = \frac{gMR^2}{2b^2 + R^2}$

$T = \frac{gM}{\frac{2b^2}{R^2} + 1}$

Answer 2

Answer:

The tension in the cord is $T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

$bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} }$ (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

$Mg=(M+\frac{MR^{2} }{2b^{2} } )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2} }$

Replacing in eq. 1

$T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2} }$