As we know that sphere roll without slipping so there is no loss of energy in this case
so here we can say that total energy is conserved
Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy
[tex]\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'[/tex]
as we know that ball start from rest
[tex]v_i = 0[/tex]
height of the ball initially is given as
[tex]h = Lsin\theta[/tex]
[tex]h = 1200sin53 = 960 cm[/tex]
also we know that
[tex]I = \frac{2}{5}mR^2[/tex]
also for pure rolling
[tex]v = r\omega[/tex]
also we know that
[tex]480 = m*9.8 [/tex]
[tex]m = 49 kg[/tex]
now plug in all data in above equation
[tex]480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0[/tex]
[tex]4608 = 3.92\omega^2 + 1.568\omega^2[/tex]
[tex]\omega^2 = 839.65[/tex]
[tex]\omega = 29 rad/s[/tex]
So speed at the bottom of the inclined plane will be 29 rad/s
