Solve Rotational Momentum Problem: 3.0kg Hoop, 5.6 rad/s, 18° Ramp

In summary, the problem is asking for the distance a 3.0 kg hoop rolls up a ramp when given an initial angular speed of 5.6 rad/s without slipping. The relevant equations include rotational kinetic energy, translational kinetic energy, potential energy, and the condition for rolling without slipping. The rotational inertia of a hoop is mr^2 and the total kinetic energy of the hoop is 120.422 kgm/s. To find the distance along the ramp, you need to use trigonometry to find the hypotenuse of a right triangle.
  • #1
Zokudu
4
0
[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 I am missing something to do with the ramp being 18 degrees above the horizontal but I am not sure where to add that into my math.

Please any help would be appreciated thank you.
 
Last edited:
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  • #2
Zokudu said:
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
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 I am being dense.
 
  • #4
Zokudu said:
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
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 I've been stressing over this problem forever
 
Last edited:

1. What is rotational momentum?

Rotational momentum, also called angular momentum, is a measure of an object's tendency to keep rotating around a fixed point. It is calculated by multiplying the object's moment of inertia by its angular velocity.

2. What is the equation for rotational momentum?

The equation for rotational momentum is L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

3. How do I solve a rotational momentum problem?

To solve a rotational momentum problem, you need to know the values for the moment of inertia, angular velocity, and any external torques acting on the object. Then, you can use the equation L = I * ω to calculate the angular momentum.

4. What is the moment of inertia?

The moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the object's mass distribution and the axis of rotation. For a hoop, the moment of inertia is equal to MR^2, where M is the mass and R is the radius of the hoop.

5. How can I apply rotational momentum to real-world problems?

Rotational momentum is important in many real-world situations, such as understanding the movement of objects in space or designing machines that use rotating parts. It is also used in sports, such as figure skating and gymnastics, to control and manipulate the body's rotation. By understanding rotational momentum, we can better predict and control the movements of rotating objects in our daily lives.

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