Check out some similar questions!

heyyalliee · Oct 21, 2010, 11:58 AM

Given: A 3.0kg solid sphere (radius = 0.20m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.75m high and 5.5m long.

Find: When the sphere reaches the bottom of the ramp, what is its rotational kinetic energy? ( and same questions but Translational Energy)

So I know Krot= (1/2)Iw^2
then where can I go from here? How do I know what I is? And w?

Ktrans= (1/2)Mv^2
to find Ktrans= (1/2)(3.0kg) how can I find v^2?

Thanks in advance for help!

galactus · Oct 22, 2010, 03:24 PM

For a uniform solid sphere:

$I=\frac{2mr^{2}}{5}$ ... [1]

So, $v=\left(\frac{10gh}{7}\right)^{\frac{1}{2}}$ ... [2]

At the bottom of the ramp, the initial potential energy is the same as the final kinetic energy.

$\frac{1}{2}v^{2}(m+\frac{I}{r^{2}})=mgh$

$v=\sqrt{\frac{2gh}{1+\frac{I}{mr^{2}}}}$

Subbing [1] into v gives [2

The linear acceleration is given by $a=\frac{5}{7}g\cdot sin{\theta}$