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- Thread starter Darius Kaufmann
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BvU

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easier variation: how about a disk ? Well, kinetic energy from rotation is ##{1\over 2} I\omega^2## with ##I= {1\over 2} M R^2## for a disk.

Kinetic energy of a point mass M moving along at speed ##v## is ##{1\over 2} M v^2##

Equate the energies $${1\over 2} \left ( {1\over 2} M R^2\right) \omega^2 = {1\over 2} M v^2\ \Rightarrow \ v^2 = {1\over 2} \omega^2 R^2\ \Rightarrow \ v = {1\over 2} \sqrt 2\ \omega R$$

Can you now do the 3D sphere by yourself ?

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BvU

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Oops...makes me wonder where the question comes from ?what all the symbols mean in the above equations

M is mass, kilogram

R is radius, meter

##\omega## is angular speed, radians/second

v is speed, meter/second

##I## moment of inertia, kilogram meter

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

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For example, points on the equator of a spinning proton are presumed to be moving at c, according to current theories

This is nonsense. Why do you think that?

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sophiecentaur

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I'm wondering what it is, exactly that you want to know. We need some basic things about motion. The averageSummary:Is there a way to calculate the average linear speed of all points in the volume of a sphere rotating on a single axis?

Is there a way to calculate the average linear speed of all points in the volume of a sphere rotating on a single axis?

There are two quantities that indicate the Motion and 'stopability' of a body. In the past they were confused with one another but eventually they were acknowledged to be two different things. One is the Kinetic Energy and the other is the Momentum. KE involves the square of the velocity and that is always positive. Momentum involves Velocity Vectors which have direction as well as speed.

A stationary, spinning sphere clearly has Energy, despite a mean velocity of zero. @BvU discusses the KE idea, above. But the linear momentum is zero so what about the rotation?

Linear Momentum is just mv but angular momentum depends on the distribution of the mass as well as the total mass. Its Moment of Inertia is the result of adding all the individual particle masses times the (distance from the centre of rotation) squared. So the outer parts have a bigger contribution than the inner parts (flywheels usually have the mass in an outer ring for this reason).

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BvU

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Wholeheartedly agree.This is nonsense. Why do you think that?

Must admit I only read the first half of #1 and (re)acted on that.

Reading the second half feels like hearing a chainsaw go through a violin...

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