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MAVERICK [17]
3 years ago
15

A sphere of radius 1.59 cm and a spherical shell of radius 7.72 cm are rolling without slipping along the same floor. The two ob

jects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the sphere's angular speed to the spherical shell's angular speed be?
Physics
1 answer:
valina [46]3 years ago
3 0

Answer: 4.86

Explanation:

sphere moment of Inertia Iₑ = (2/5)mrₑ²

Let the sphere of radius 1.59 cm be x

Let the spherical shell of radius 7.72 cm be y, so that

Iₑ(x) = 2/5 * m * 1.59²

Iₑ(x) = 2/5 * m * 2.5281

Iₑ(x) = 1.011m

Iₑ(y) = 2/5 * m * 7.72²

Iₑ(y) = 2/5 * m * 59.5984

Iₑ(y) = 23.84m

Also, the angular speed of the sphere's would be ωₑ(x) and ωₑ(y)

total k.e = rotational k.e + linear k.e

for sphere = ½Iₑωₑ² + ½mωₑ²rₑ²

For sphere x

{ωₑ²[ 1.011 + 1.59²]} =

ωₑ²(1.011 + 2.5281) =

ωₑ²(3.5391)

For sphere y

{ωₑ²[ 23.84 + 7.72²]} =

ωₑ²(23.84 + 59.5984) =

ωₑ²(83.4384)

If the ratio of x/y = 1, then

ωₑ(x)²(3.5391) / ωₑ(y)²(83.4384) = 1

ωₑ(x)²(3.5391) = ωₑ(y)²(83.4384)

[ωₔ(x)/ωₑ(y)]² = [83.4384] / [3.5391] ~= 23.5762

[ωₔ(x)/ωₑ(y)] = √(23.5762)

[ωₔ(x)/ωₑ(y)] = 4.86

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I think the key here is to be exquisitely careful at all times, and
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We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
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Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
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Time per unit angle =  (1/2ω)  sec/rad
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At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
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What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
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Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
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The planets are in line again as soon as the faster one has 'lapped'
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      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

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You're welcome.  The generous bounty of 5 points is very gracious,
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