Angular velocity (rads/sec)?
Please solve, Thank You!
A uniform circular disk with mass (20kg and radius = 5cm) rolls (without slipping) down a hill with a vertical drop of 2 meters, what is the disk’s angular velocity (rads/sec) at the bottom of the hill. Find the moment of inertia
- EdwardLv 71 decade agoFavorite Answer
The ph_yo has the right start, however it is a angular velocity that is in question
So the potential energy at the top equal to the kinetic energy (Ke) on the bottom.
Pe=Ke (but Ke is rotational energy!)
mgh = .5 I w^2
m – mass of the cylinder
g - acceleration due to gravity
h – height at the top
I - moment of inertia of the cylinder
w – angular velocity
I= .5m r^2
r – radius of the cylinder
since mgh = .5 I w^2
we have w=sqrt(2mgh/I)
w=2/ .05 sqrt (9.81 x 2)=
w= 177 rad /sec
now moment of inertia
I=.5 (2) (.05)^2
I=0.025 kg m^2
- Anonymous1 decade ago
potential = kinetic, assuming the disk start from rest
mgh = 1/2 m v^2
v^2 = 2gh ==> v = sqrt( 2*9.81*2) = 6.26
w = v/r = 6.26/.05 = 125.3 rads/sec.