A uniform ball, of mass M = 6.00 kg and radius R, rolls smoothly from rest down a ramp at angle `theta = 30.0^(@)` .
(a) The ball descends a vertical height h = 1.20 m to reach the bottom of the ramp. What is its speed at the bottom?

Explanation

The mechanical energy E of the ball-Earth system is con served as the ball rolls down the ramp. The reason is that the only force doing work on the ball is the gravitational force, a conservative force. The normal force on the ball from the ramp does zero work because it is perpendicular to the ball.s path. The frictional force on the ball from the ramp does not transfer any energy to thermal energy because the ball does not slide (it rolls smoothly).
Thus, we conserve mechanical energy (`E_(f)= E_(i)`):
`K_(f)+ U_(f)= K_(i) + U_(i)` (11-11)
where subscripts f and i refer to the final values (at the bottom) and initial values (at rest), respectively. The gravitational potential energy is initially `U_(i) = Mgh` (where M is the ball.s mass) and finally `U_(f) = 0`. The kinetic energy is initially `K_(i) = 0`. For the final kinetic energy K, we need an additional idea: Because the ball rolls, the kinetic energy involves both translation and rotation, so we include them both by using the right side of eq 11-5.
Calculations: Substituting into Eq. 11-11 gives us
`((1)/(2)I_("com")omega^(2)+ (1)/(2)Mv_("com")^(2))+ 0= 0+ Mgh` (11-20)
where `I_("com")` is the ball.s rotational inertia about an axis com through its center of mass, `v_("com")` is the requested speed at the bottom, and `omega` is the angular speed there.
Because the ball rolls smoothly, we can use Eq. 11-2 to substitute `v_("com")R` for `omega` to reduce the unknowns in com Eq. 11-12. Doing so, substituting `2//5MR^(2)` for `I_("com")` ( from Table 10-2f), and then solving for `v_("com")` com give us
`v_("com")= sqrt((10)/(7)gh)= sqrt((10)/(7)(9.8m//s^(2))(1.20m))`
`= 4.10 m//s`
Note that the answer does not depend on Mor R.