Solid disk rolling up a hill problem

  • #1
I'm somewhat clueless on how to solve the following problem, since a solid disk is rolling up a hill rather than downhill.

A solid disk (mass = 5 kg, R = 0.6m) is rolling across a table with a translational speed of 6 m/s. The disk then rolls up a hill of height 2 m, where the ground again levels out. Find the translational and rotational speeds now.

I tried using KEi + PEi = KEf + PEf, including both KEt and KEr in KEf, and I calculated that KEi = 135J (45J are KEr, and 90J are KEt). I'm not sure if I'm plugging in my values correctly, since I got a translational speed of 3.847, and my answer was wrong.
 

Answers and Replies

  • #2
dav2008
Gold Member
609
1
You included gravitational potential energy and translational kinetic energy, but what are you forgetting?
 
  • #3
Rotational kinetic energy? I thought I included that too...KEr.
 
  • #4
dav2008
Gold Member
609
1
Oh well then you should be getting the right answer. You probably have a mistake in your arithmetic.

I also got 135J for the total initial kinetic energy but I got a different answer for the final speed.
 
Last edited:
  • #5
Okay...can you check my work?

I used KEi + PEi = KEr + KEt + PEf, which I used 135 + 0 = 45 + (0.5)(5)(v^2) + (5)(9.8)(2), and solved from there. Am I using these values incorrectly?
 
  • #6
dav2008
Gold Member
609
1
You have the wrong final rotational kinetic energy. The disk is still rolling when it's at the top of the hill but the angular velocity is not the same as it was before. (If the translational speed is now v, what's the angular speed?)
 
  • #7
How would I go about correcting my KEr, and what would the fact that it's still rolling at the top of the hill change in my equation?
 
  • #8
dav2008
Gold Member
609
1
You would write your final translational kinetic energy in the same manner you wrote your final translational kinetic energy.

If the object has a translational speed of v then what is it's rotational kinetic energy? You already said that the translational kinetic energy would be [itex]\frac{1}{2}(5)(v^2)[/itex] so now write the final rotational kinetic energy in terms of v.
 
Last edited:
  • #9
Do you mean I should write my final rotational kinetic energy in the same way? Sorry, I am getting lost, since your previous 2 posts seem to be about 2 different things wrong with the problem. Would you be able to clarify what you are referring to?
 
  • #10
dav2008
Gold Member
609
1
You said 135 + 0 = 45 + (0.5)(5)(v^2) + (5)(9.8)(2).

Here you are assuming that the final kinetic rotational energy is the same as the initial kinetic rotational energy (45J).

I'm saying that that's not the case because the object is now traveling at at a speed v.

You wrote conservation of energy:

KEroti+KEtransi+PEi=KErotf+KEtransf+PEf

You then calculated the initial energies to be 135J:

135J=KErotf+KEtransf+PEf

Now you just need to write the kinetic energies in terms of an unknown speed "v" and then solve for v
 
Last edited:
  • #11
Oh, okay. So I would ahve 2 unknowns in this problem? How would I solve each one?
 
  • #12
dav2008
Gold Member
609
1
Well you would have angular velocity and translational velocity. What is the relationship between the two?
 
  • #13
v=rw? How would I incorporate that into the problem? I just simpliefied my equation and got 37 = 1.25w^2 + 2v^2
 
  • #14
dav2008
Gold Member
609
1
Well I don't know if you simplified your equation right or not but you know what w is in terms of v from the equation you just posted...
 
  • #15
LOL I just figured that out...so w = 4.148, and then I plug that into v=rw to get v, which is 2.4888? Eek...these answers are wrong! Did I do something wrong?
 
Last edited:
  • #16
dav2008
Gold Member
609
1
Well since youre solving for v it would just be easier to substitute v/.6 for w in your final rotational energy.

I think you're just coming across arithmetic errors at this point.
 
Last edited:
  • #17
I got the same answer...can you check if my work is wrong? I'm using:

37 = 1.25(v/0.6)^2 + 2.5v^2, which becomes 37 = 3.472v^2 + 2.5v^2, then 37 = 5.972v^2, 6.195 = v^2, then v = 2.489.
 
  • #18
dav2008
Gold Member
609
1
I think you divided by .62 before you plugged in v/.6 for w because I am getting [itex]37 = 1.25(v)^2 + 2.5v^2[/itex]
 
Last edited:
  • #19
I had that 1.25 before I even divided by 0.6. I'm really confused now. Would you be able to show me your work for this problem?
 
  • #20
dav2008
Gold Member
609
1
[tex]135=\frac{1}{2}(9.8)(2)+\frac{1}{2}(5)v^2+.5(.9)(\omega)^2[/tex]
[tex]37=\frac{1}{2}(5)v^2+.5(.9)(\frac{v}{.6})^2[/tex]
[tex]37=2.5v^2+1.25v^2[/tex]
 
  • #21
Where did you get the .9?
 
  • #22
dav2008
Gold Member
609
1
That is the moment of inertia that I calculated for the disc.

[tex]I=.5mr^2[/tex] for a solid disc, axis through center
[tex]=.5(5)(.6)^2[/tex]
[tex]=.9[/tex] kg-m2
 
Last edited:
  • #23
Is inertia involved in this problem? I thought I was working with kinetic and potential energy only?
 
  • #24
dav2008
Gold Member
609
1
Show me how you solved for the initial rotational kinetic energy. You had to have used the moment of inertia.

The rotational kinetic energy is equal to [tex]\frac{1}{2}I\omega^2[/tex] where I is the moment of inertia and [itex]\omega[/itex] is the angular velocity. You obviously used this formula when you calculated the initial rotational kinetic energy
 
Last edited:
  • #25
Oh okay, I figured out that inertia part! I realized I used mass instead of inertia in KEr on accident! So Is my velocity if 3.14 m/s correct? And that gives me an angular velocity of 5.23?
 

Related Threads on Solid disk rolling up a hill problem

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
16K
Replies
14
Views
3K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
3
Views
4K
Top