Answer:
linear acceleration
[tex]a = \frac{2g}{2 + \frac{R}{b}}[/tex]
angular acceleration
[tex]\alpha = \frac{2g}{R(2 + \frac{R}{b})}[/tex]
Explanation:
As we know that the force due to tension force is upwards while weight of the disc is downwards
so we will have
[tex]2mg - T = 2ma[/tex]
also we have
[tex]Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha[/tex]
now we have
[tex]Tb = mR^2(\frac{a}{R})[/tex]
[tex]T = \frac{mRa}{b}[/tex]
now we have
[tex]2mg = (2ma + \frac{mRa}{b})[/tex]
[tex]a(2 + \frac{R}{b}) = 2g[/tex]
so we have
linear acceleration
[tex]a = \frac{2g}{2 + \frac{R}{b}}[/tex]
angular acceleration
[tex]\alpha = \frac{2g}{R(2 + \frac{R}{b})}[/tex]