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# 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

### 2 Answers

- 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)

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

- Anonymous1 decade ago
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.