Angular velocity (rads/sec)?

Question

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

Answer 1

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)
or
w= sqrt(2mgh/.5mr^2)
w=2/r sqrt(gh)
w=2/ .05 sqrt (9.81 x 2)=
w= 177 rad /sec

now moment of inertia
I=.5m r^2
I=.5 (2) (.05)^2
I=0.025 kg m^2




Have fun

Answer 2

energy

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.