Part II : Rotational Dynamics of Rolling Objects In this part of the experiment,

Question

Part II : Rotational Dynamics of Rolling Objects

In this part of the experiment, you will calculate an inclined ring, solid sphere. or solid disk the bottom of plane first Use method you wish to calculate this but you may try to use the conservation of rotational energy (pages 300- 301 of the PHYS-110 text). when you feel that you have the problem solved, see your instructor to get the results verified an experimental set-up. Some helpful things to measure and look up in the text may be mass of ring here, and disk; height moments of inertia each object Your instructor will be happy to accept bets on which reaches the ground first since there is only one right answer This is one race where the outcome can always be predicted.

Explanation / Answer

When an object is rolling down a ramp, its energy is made up of three components:

mgh=12mv2+12I2mgh=12mv2+12I2

The first term is the potential energy; this is the energy is takes to lift the object up the ramp. This is equal to mgh,mgh, with mm being the mass, ggthe acceleration due to gravity, and hh the height of the ramp.

The second term is the translational kinetic energy; this is the energy it takes for the object to move down the ramp. This is equal to 12mv212mv2, with mm being the mass and vvbeing the translational velocity.

The third term is the rotational kinetic energy; this is the energy it takes for the object to roll. This is equal to 12I212I2, with II being the moment of inertia (the object’s resistance to being rotated) and being the angular velocity.

When an object is rolling such that a point on its edge has a velocity vv, the angular velocity is given by vRvR. Because none of these shapes are slipping down the slope (that complicates your life), this is true.

For reasons, the moment of inertia of an object about an axis is r2dmr2dm. The intution behind this is that, as a tiny piece of mass is further away from the axis of rotation, it must move faster. The mathematics behind this is a little more complicated but you can probably find an elegant derivation online somewhere.

So, let’s find these moments of intertia:

For the tube, this is best done is cylindrical coordinates. m=2rlm=2rl, dm=Rdzddm=Rdzdand r=Rr=R, so I=20l0R2dzdI=020lR2dzd.

This evaluates to mR2mR2 once you solve the integrals and substitute in the mass.

For the cylinder, this is also best done in cylindrical coordinates. m=r2lm=r2l,dm=rdrddzdm=rdrddz and r=rr=r, so I=l020R0r3drddzI=0l020Rr3drddz.

This evaluates to 12mR212mR2once you solve the integrals and substitute in the mass.

For the ball, switch to spherical coordinates, because it is a sphere. m=43r3m=43r3, dm=r2sindrdddm=r2sindrdd and r=rsin,r=rsin,

so I=200R0(rsin)2r2sindrddI=0200R(rsin)2r2sindrdd

This evaluates to 25mR225mR2 once you solve the integrals and substitute in the mass.

And for a sphere, because I’m nice and also ridiculously pedantic, m=4r2m=4r2, dm=R2sindddm=R2sindd and r=Rsinr=Rsin, so I=200(Rsin)2R2sindd.I=020(Rsin)2R2sindd.

This evaluates to 23mR223mR2 once you solve the integrals and substitute in the mass.

Now it’s the final stretch. Rearrange the energy equation above to find the velocity:

mgh=12mv2+I2R2v2mgh=12mv2+I2R2v2

v2=mgh12m+I2R2v2=mgh12m+I2R2

You’ll notice that, in this expression, as the moment of intertia gets larger, the velocity gets smaller. This means that, for the four shapes I’ve mentioned, they arrive at the bottom in the order

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