Answer:
The tension in the cord is [tex]T=\frac{MR^{2}g }{2b^{2}+R^{2} }[/tex]
Explanation:
Given:
M = mass
b = radius
R = spool of radius
The equation is:
[tex]bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} }[/tex] (eq. 1)
The sum of forces in y:
∑Fy = Mg - T = Ma
[tex]Mg=(M+\frac{MR^{2} }{2b^{2} } )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2} }[/tex]
Replacing in eq. 1
[tex]T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2} }[/tex]
Answer:
[tex]T = \frac{gM}{\frac{2b^2}{R^2} + 1}[/tex]
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:
[tex]Tb = I\alpha[/tex]
[tex]Tb = MR^2\alpha/2[/tex]
We can substitute angular acceleration as ratio of linear acceleration and spool radius
[tex]\alpha = a/b[/tex]
Therefore[tex]Tb = MR^2a/2b[/tex]
[tex] T = \frac{aMR^2}{2b^2}[/tex]
The system has 2 forces, gravity pulling it down and tension force upward. The net force would generate the linear acceleration a
[tex] Mg - T = ma[/tex]
[tex]a = (gM - T)/M[/tex]
now we can substitute for a into the tension equation
[tex]T = \frac{(gM-T)MR^2}{2b^2M}[/tex]
[tex]T = \frac{gMR^2}{2b^2} - \frac{TR^2}{2b^2}[/tex]
[tex]T\left(1 + \frac{R^2}{2b^2}\right) = \frac{gMR^2}{2b^2}[/tex]
[tex]T\frac{2b^2 + R^2}{2b^2} = \frac{gMR^2}{2b^2}[/tex]
[tex]T = \frac{gMR^2}{2b^2 + R^2}[/tex]
[tex]T = \frac{gM}{\frac{2b^2}{R^2} + 1}[/tex]
