What Determines the Angular Momentum of a Rolling Boulder?

[cbarker1]

Dear Every one,
Here is the question. How to get started with this question?
A spherical boulder of mass 90.2 kg and radius 20 cm rolls without slipping down a hill 16 m high from rest.
(a)
What is its angular momentum about its center when it is half way down the hill? (Enter the magnitude in kg [FONT=&quot]· m²/s.)
kg [FONT=&quot]· m²/s


(b)
What is its angular momentum about its center when it is at the bottom? (Enter the magnitude in kg [FONT=&quot]· m²/s.)
kg [FONT=&quot]· m²/s


Thanks,
Cbarker1

 

[skeeter]

A spherical boulder of mass 90.2 kg and radius 20 cm rolls without slipping down a hill 16 m high from rest.
(a)
What is its angular momentum about its center when it is half way down the hill? (Enter the magnitude in kg · m²/s.)
kg · m²/s


(b)
What is its angular momentum about its center when it is at the bottom? (Enter the magnitude in kg · m²/s.)
kg · m²/s

conservation of energy (assuming the boulder starts from rest) ...

initial gravitational potential energy = final gravitational potential energy + translational kinetic energy + rotational kinetic energy

$mgh_0 = mgh_f + \dfrac{1}{2}mv^2 + \dfrac{1}{2}I \omega^2$

note $v = r\omega$ ...

$mgh_0 = mgh_f + \dfrac{1}{2}m(r\omega)^2 + \dfrac{1}{2}I \omega^2$

$2mg(h_0-h_f) = m(r\omega)^2 + I \omega^2$

$2mg(h_0-h_f) = \omega^2(mr^2 + I)$

$\omega = \sqrt{\dfrac{2mg(h_0-h_f)}{mr^2 + I}}$

finally, note $L = I\omega$