Question

A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

Answer 1

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

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'$

as we know that ball start from rest

$v_i = 0$

height of the ball initially is given as

$h = Lsin\theta$

$h = 1200sin53 = 960 cm$

also we know that

$I = \frac{2}{5}mR^2$

also for pure rolling

$v = r\omega$

also we know that

$480 = m*9.8$

$m = 49 kg$

now plug in all data in above equation

$480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0$

$4608 = 3.92\omega^2 + 1.568\omega^2$

$\omega^2 = 839.65$

$\omega = 29 rad/s$

So speed at the bottom of the inclined plane will be 29 rad/s