HS Thomas

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posted 13 years ago

What happens if you roll these objects up hill ? So the objects are bowled up hill by some mechanism that releases the object from the the same line.

Which of these gets up the hill the highest ?

I pinched these from another thread.

Include three disks small, medium, large.

regards

[ November 17, 2003: Message edited by: HS Thomas ]

Which of these gets up the hill the highest ?

I pinched these from another thread.

Originally posted by Jim Yingst:

Bert's answer looks good to me - "spinning inertia" is generally called "moment of inertia", at least in the US.

Let's expand this a bit. Assume the following objects are all released from the top of an inclined ramp at the same time. Assume there's no air resistance and each object rolls straight down the incline with no slippage. What order do the objects arrive at the bottom?

Solid steel cylinder, 10 cm diameter, 10 cm long Solid steel cylinder, 10 cm diameter, 1 cm long Solid plastic cylinder, 10 cm diameter, 10 cm long Steel pipe segment, 10 cm outer diameter, 9 cm inner diameter, 10 cm long Plastic pipe segment, 20 cm outer diameter, 18 cm inner diameter, 1 m long Plastic pipe segment, 20 cm outer diameter, 19 cm inner diameter, 1 m long Solid steel sphere, 10 cm diameter Solid plastic sphere, 10 cm diamteter Solid steel sphere, 1 cm diameter

Let's say that plastic has a density 1/10 that of steel. (Dunno, I just made that up because I'm too lazy to look it up.) Assume that the conditions are such that each object musttravelthe same total distance down the ramp - larger objects don't get an advantage because their leading edge arrives earlier. One way to do this is to make sure that the leading edges are all aligned at start, and then look at arrival times of leading edges. Or, align all the centers of mass initially, and then meaure arrival times of certers of mass. Either way - just don't mix the two methods up.

Include three disks small, medium, large.

regards

[ November 17, 2003: Message edited by: HS Thomas ]

Jim Yingst

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posted 13 years ago

If "some mechanism" were to start each object out with the same speed it arrived at the bottom with (in the first problem), then the journey up the hill would be a mirror image of the behavior in the first problem. (Assuming again that gravity and the slope are the only two things affecting each rolling object after it's started.) If the starting mechanism does something different, well, results will be different. What did you have in mind?

"I'm not back." - Bill Harding, *Twister*

HS Thomas

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posted 13 years ago

I was afraid you'd say that. The mind boggles.

Here's another uphill problem just to break the original problem down a bit.

A uniform, solid disk and a uniform, solid sphere each having the same mass, M and radius R are rolling on a horizontal surface with the same translational speed, v. The two objects roll up a hill. Which one will roll higher up the hill and by exactly what factor will its maximum height exceed the height attained by the other?

regards

Here's another uphill problem just to break the original problem down a bit.

A uniform, solid disk and a uniform, solid sphere each having the same mass, M and radius R are rolling on a horizontal surface with the same translational speed, v. The two objects roll up a hill. Which one will roll higher up the hill and by exactly what factor will its maximum height exceed the height attained by the other?

regards

Jim Yingst

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posted 13 years ago

For a rolling body, total kinetic energy is

KE = mv^2/2 + Iω^2/2

where

m = mass

v = translational velocity

I = moment of intertia

ω = angular velocity = v/R for rolling body of radius R

So

KE = v^2/2 * (m + I/R^2)

Since I'm too lazy to try to show integrals in HTML I'll just google up a table of moments of intertia for various bodies... ah, here is one. The relevant parts are:

solid cylinder/disk (along axis): I = mR^2/2

solid sphere: I = 2mR^2/5

Pluggin those into KE above, we get

cylinder: KE = 3/4 mv^2

sphere: 7/10 mv^2

To get the maximum height rolled up an incline, set potential energy at the top equal to kinetic energy at the bottom:

KE = mgh

where h = vertical height above start, and g = 9.81 m/s^2 acceleration from gravity. Then:

cylinder: 3/4 mv^2 = mgh ==> h = 3/4 v^2/g

sphere: 7/10 mv^2 = mgh ==> h = 7/10 v^2/g

So, the cylinder will roll (3/4) / (7/10) = 15/14 times farther up the incline as the sphere will. Because the started with more energy (due to higher moment of inertia) it takes longer to shed that energy.

Note that we never even had to assume that m or R was the same for the cylinder and sphere. Those two variables canceled out before we needed to plug anytying in. As long as the sphere and cylinder start with the same initial translational velocity v, the above results will hold.

[ November 17, 2003: Message edited by: Jim Yingst ]

KE = mv^2/2 + Iω^2/2

where

m = mass

v = translational velocity

I = moment of intertia

ω = angular velocity = v/R for rolling body of radius R

So

KE = v^2/2 * (m + I/R^2)

Since I'm too lazy to try to show integrals in HTML I'll just google up a table of moments of intertia for various bodies... ah, here is one. The relevant parts are:

solid cylinder/disk (along axis): I = mR^2/2

solid sphere: I = 2mR^2/5

Pluggin those into KE above, we get

cylinder: KE = 3/4 mv^2

sphere: 7/10 mv^2

To get the maximum height rolled up an incline, set potential energy at the top equal to kinetic energy at the bottom:

KE = mgh

where h = vertical height above start, and g = 9.81 m/s^2 acceleration from gravity. Then:

cylinder: 3/4 mv^2 = mgh ==> h = 3/4 v^2/g

sphere: 7/10 mv^2 = mgh ==> h = 7/10 v^2/g

So, the cylinder will roll (3/4) / (7/10) = 15/14 times farther up the incline as the sphere will. Because the started with more energy (due to higher moment of inertia) it takes longer to shed that energy.

Note that we never even had to assume that m or R was the same for the cylinder and sphere. Those two variables canceled out before we needed to plug anytying in. As long as the sphere and cylinder start with the same initial translational velocity v, the above results will hold.

[ November 17, 2003: Message edited by: Jim Yingst ]

"I'm not back." - Bill Harding, *Twister*

HS Thomas

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Posts: 3404

posted 13 years ago

The answer I have is that one object goes 7.1 % higher. That's while only considering a disk and a sphere. I shall look at the other solution and see if I can spot any holes in that compared to Jim's solution. I'd be darned if Jim's wrong.

regards

[ November 18, 2003: Message edited by: HS Thomas ]

regards

[ November 18, 2003: Message edited by: HS Thomas ]

Jim Yingst

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It is sorta covered in the JavaRanch Style Guide. |