What is the final velocity, in m/s, of a hoop that rolls without slipping down a 6.50 m high hill, starting from rest?
m/s
(b)
What would be the final velocity, in m/s, if a disk of the same mass and radius as the hoop rolled down the hill?
m/s
NCS
Favorite Answer
"without slipping" means that it is rolling and therefore has rotational KE:
KE = ½mv² + ½Iω²
(a) for a hoop, I = mr²
and "without slipping" means the ω = v/r, so
KE = ½mv² + ½(mr²)(v/r)² = mv²
PE becomes KE:
mgh = mv²
v = √(gh) = √(9.8m/s² * 6.50m) = 7.98 m/s
(b) for a disk, I = ½mr² and so
KE = ½mv² + ½(½mv²) = (3/4)mv²
mgh = (3/4)mv²
v = √(4gh/3) = 9.22 m/s
Hope this helps!