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Rotational Momentum

  • Thread starter Zokudu
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[SOLVED] Rotational Momentum

This is my first attempt at asking a question on here so please be gentle. I've been working on this problem for almost an hour and a half

Homework Statement



In a circus performance, a large 3.0 kg hoop with a radius of 1.6 m rolls without slipping. If the hoop is given an angular speed of 5.6 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 18° with the horizontal, how far (measured along the incline) does the hoop roll?



2. Relavent Equations

1/2Iα^2=Rotational Momentum
1/2mv^2=kintetic Momentum
mgh=potential Momentum
V(tangent)=rω
I=1/2mr^2



3. Attempt a Solution

First i tried getting the rotational momentum and when i put the numbers in it comes out to about 60.221 kgm/s and then i set it equal to the potential momentum equation and got the height is 2.048m. However this answer came out as wrong.

So then i tried getting the tangential velocity and got 8.960 m/s and put that into the kentetic momentum equation and got 120.422 kgm/s (double what i got for rotational momentum) and set that to the potential energy equation and got 4.096m. Which is also wrong.

I think im missing something to do with the ramp being 18 degrees above the horizontal but im not sure where to add that into my math.

Please any help would be appreciated thank you.
 
Last edited:

Answers and Replies

  • #2
Doc Al
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2. Relavent Equations

1/2Iα^2=Rotational Momentum
1/2mv^2=kintetic Momentum
mgh=potential Momentum
These are expressions for energy, not momentum: Rotational KE (which should be 1/2Iω^2), translational KE, and gravitational PE.
V(tangent)=rω
That's the condition for rolling without slipping.
I=1/2mr^2
That's the rotational inertia of a cylinder. What's the rotational inertia of a hoop?

Set up a conservation of energy equation. Hint: What's the total kinetic energy of the rolling hoop while it's on the horizontal?
 
  • #3
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Oops my fault wrong words i guess =/ and i ment ω not α i'm sorry. I'm not very good at physics :D

Well rotational inertia of a hoop=mr^2 correct?

well that makes more sense then so the total kinetic energy of the hoop is 120.422

so then i need to solve for the potential energy to get the height since i have the mass of the hoop. so:

120.422=mgh
120.422=(3)*(9.8)*(h)
and solve for an h of: 4.096m

which when i plug it into the online assignment thing comes out as wrong. =/ Am i missing somethign else? where would that 18 degrees come in? I'm sorry if im being dense.
 
  • #4
Doc Al
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44,882
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Well rotational inertia of a hoop=mr^2 correct?
Right.

well that makes more sense then so the total kinetic energy of the hoop is 120.422
Please show exactly how you got this. (Did you include both kinetic energy terms?)

so then i need to solve for the potential energy to get the height since i have the mass of the hoop. so:

120.422=mgh
120.422=(3)*(9.8)*(h)
and solve for an h of: 4.096m
h is the vertical distance the hoop travels; you need the distance along the ramp. (You'll need a bit of trig. Hint: The distance along the ramp is the hypotenuse of a right triangle.)
 
  • #5
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1/2Iω^2=Rotational KE
I=mr^2
I=3*(1.6)^2
I=7.68
1/2(7.68)(5.6)^2=120.422+120.422 for the KE
solve for an h of 8.193
divide by sin(18) and got 26.514

OOO i figured it out it does have both.

Thank you soo much ive been stressing over this problem forever
 
Last edited:

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