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Height of an different objects along a curved incline

  • Thread starter sara167
  • Start date
8
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1. Homework Statement
the five objects of various masses, each denoted m, all have the same radius. They are all rolling at the same speed as they approach a curved incline.
Solid sphere - m = 1.0 kg
Hollow Sphere - m = 0.2 kg
Solid Cylinder - m = 0.2 kg
Solid Disk - m = 0.5 kg
Hoop - m = 0.2 kg

Rank the objects based on the maximum height they reach along the curved incline.

2. Homework Equations
Hoop - I=mr^2
Solid Disk and Cylinder - I=.5mr^2
Hollow Sphere - I=2/3mr^2
Solid Sphere - I=2/5mr^2

Where I is the moment of inertia, m is the mass, and r is the radius.

3. The Attempt at a Solution
I am unsure of where to go from here. I know the equations for inertia, but when I used the equation leaving out r^2 since they all have the same radius they were in the wrong order according to the website. What am I supposed to do?

I also know that the linear kinetic energy and the rotational kinetic energy are converted into gravitational potential energy. The equation I believe is 1/2mv^2 +1/2Iw^2 = mgh.
M is the mass, v is linear velocity, I is moment of inertia, w is angular velocity, g is gravity, and h is the height.
If I rearrange the equation to find height,
h=(1/2mv^2+1/2Iw^2)/mg
the masses cancel out leaving,
h=(1/2v^2 +1/2(r^2)(w^2))/ g

Am I on the right track? Am I forgetting something?
 
Last edited:

TSny

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You're on the right track. Can you relate w to v for a rolling object?
 
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w=v/r so the equation would look like
h=(1/2v^2 +1/2v^2)/g
h=2(1/2v^2)/g

I am unsure of how this equation will help relate the different masses to different height. I was also told that they don't all reach the same height because the incline is curved.
 

TSny

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h=(1/2mv^2+1/2Iw^2)/mg

the masses cancel out leaving,

h=(1/2v^2 +1/2(r^2)(w^2))/ g
In going from the first to the second equation above, did you handle I properly? Different shaped objects will have different expressions for I.

If the mass m cancels out, then what does that tell you?
 
8
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So from the first equation, if I put in the different equations for I i can find the different heights?
for example: for a hoop the equation would be
h=(1/2mv^2 + 1/2(mr^2)(v/r)^2/mg
This equation the mass would cancel out saying the height is?

for a solid disk or cylinder:
h=(1/2mv^2 + 1/2(.5mr^2)(v/r)^2/mg
Im getting the mass would cancel out again even with a different I because all that changed was the coefficient.

Im getting stuck some how with the same answers, but masteringphysics says its wrong with the heights all being the same.
 

TSny

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Yes, the mass always cancels out. That just tells you that the mass of the object is irrelevant in getting the height. (Same is true for just throwing an object straight up - the height depends on the initial speed but not the mass.)

But, you should find that you get different heights for different shapes when rolling up a hill.

What else cancels out besides the mass? You essentially have the answers for the heights. Just simplify the expressions for each type of object. You'll easily be able to compare the heights then.
 
8
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I think I get it.
So for a solid disk or cylinder:
h=2v^2/2g
This shows the same height as the hoop correct?
The height for the hollow sphere would be 4/3?
The height for the solid sphere would be 4/5?
 

TSny

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Hmm. For disk: h = [(1/2) m v^2 + (1/2)(.5 m v^2)]/mg = [(1/2) v^2 + (1/4) v^2 ]/g= ?

What does this simplify to?
 
8
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1/8(v^2+v^2)/g = 1/16v^2/g?
 

TSny

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Let's see. (1/2) v^2 + (1/4) v^2 = (1/2 + 1/4) v^2 = ?
 
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3/4 v^2
 

TSny

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Bingo! Good. So, h = (3/4) v^2/g for the disk or cylinder. See if you can now work out the others.
 
8
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Hoop= v^2/g
hollow sphere= 5/6v^2/g
solid sphere= 7/10v^2/g

If I switch them to a common denominator then the order from greatest to smallest height would be:
Hoop
Hollow Sphere
Disk and Cylinder
Solid Sphere

Correct?
 

TSny

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Looks good. Nice work.
 
8
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Thank you so much for all your help
 

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