Peoples Physics Concepts

9.7. Rolling Energy Problems http://www.ck12.org

Ei=Ef

PE=KE+KErot start with conservation of energy

mgh=

1

2

mv^2 +

1

2

Iω^2 substitute the proper value for each energy term

mgh=

1

2

mv^2 +

1

2

mr^2 ω^2 substitute in the moment of inertia of the hoop

mgh=

1

2

m(ωr)^2 +

1

2

mr^2 ω^2 we can putvin terms of omega because the hoop is rolling without sliding

2 gh= (ωr)^2 +r^2 ω^2 now we will multiply by 2 and divide by m to simplify the equation

2 gh= 2 ω^2 r^2

ω=

√

gh

r

solving forω

ω=

√

9 .8 m/s^2 ∗3 m

.75 m

plug in the known values

ω= 7 .2 rad/s

Watch this Explanation

MEDIA

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URL: http://www.ck12.org/flx/render/embeddedobject/411

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