Answer:
Speed of the center of the disc is 4.96 m/s
Explanation:
As we know that moment of inertia of the hollow disc is given as
[tex]I = \frac{m(R_1^2 + R_2^2)}{2}[/tex]
now we know when one wire is break then disc will roll down so here loss in gravitational potential energy will convert into kinetic energy of the system
So we will have
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]
so we will have
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R_2})^2[/tex]
so we have
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}m(\frac{0.60^2 + 0.30^2}{2})\frac{v^2}{0.60^2}[/tex]
[tex](10)(2) = 0.5v^2 + 0.3125 v^2[/tex]
[tex]20 = 0.8125 v^2[/tex]
[tex]v = 4.96 m/s[/tex]
