How Do You Calculate the Kinetic Energy of a Rotating CD?

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eagles12
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Homework Statement



A 12g CD with a radius of 6cm rotates with an angular speed of 38 rad/s. What is kinetic energy?

Homework Equations



k=1/2mv^2
v=rw

The Attempt at a Solution



v=(6)(38)=228
K=1/2(12)(228^2)
K=311904
but its saying this is the incorrect answer
 
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It says that the CD rotates, so you are much better off modeling under angular speeds rather than constantly changing translational speeds (e.g. particles x distance from center of rotating disk.) When deriving these equations a term shows up which is called the moment of inertia. Have you gone over this?
 
eagles12 said:

Homework Statement



A 12g CD with a radius of 6cm rotates with an angular speed of 38 rad/s. What is kinetic energy?

Homework Equations



k=1/2mv^2
v=rw

The Attempt at a Solution



v=(6)(38)=228
K=1/2(12)(228^2)
K=311904
but its saying this is the incorrect answer

You are dealing with rotational motion, so you want rotational kinetic energy. This will involve the mass moment of inertia of the CD and angular velocity. Be careful about units, they have to be consistent; Work with meters and kilograms.
 
eagles12 said:
using this i got k=.0156 but this seems too small

Too small for what? And what are the units. Everything in physics has units.
 
eagles12 said:
the units are joules

Well then, the result looks reasonable :smile:
 
i had to use two significant digits but it said my answer was incorrect
 
eagles12 said:
i had to use two significant digits but it said my answer was incorrect

What was your answer? Did it require units as well?
 
eagles12 said:
my answer was .02 the units were already given

The leading zero is not a significant digit.
 
Rounded to the nearest hundredth, that should be correct. But as gneill stated, that is only one significant digit.

maureenw002 said:
you are much better off modeling under angular speeds rather than constantly changing translational speedshttp://www.infoocean.info/avatar1.jpg

Bot/Troll?
 
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Sefrez said:
Rounded to the nearest hundredth, that should be correct. But as gneill stated, that is only one significant digit.


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