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

`T = (gM)/((2b^2)/(R^2) + 1)`

Step-by-step 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 = (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 = ((gM-T)MR^2)/(2b^2M)`

`T = (gMR^2)/(2b^2) - (TR^2)/(2b^2)`

`T\left(1 + (R^2)/(2b^2)\right) =  (gMR^2)/(2b^2)`

`T(2b^2 + R^2)/(2b^2) = (gMR^2)/(2b^2)`

`T = (gMR^2)/(2b^2 + R^2)`

`T = (gM)/((2b^2)/(R^2) + 1)`

Answer 2

The tension in the cord is
`T=(MR^(2)g )/(2b^(2)+R^(2) )`

Step-by-step explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

`bT=((MR^(2) )/(2) )((a)/(b) )\\T=(MR^(2)a )/(2b^(2) )` (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

`Mg=(M+(MR^(2) )/(2b^(2) ) )a\\a=(2b^(2)g )/(2b^(2)+R^(2) )`

Replacing in eq. 1

`T=(MR^(2) )/(2b^(2) ) ((2b^(2)g )/(2b^(2) +R^(2) ) )\\T=(MR^(2)g )/(2b^(2)+R^(2) )`