Calculate the velocity of a cylinder as it leaves the top of a slope (using Energy)

In summary, the conversation discusses a problem involving gain in gravitational potential energy and loss in translational and rotational kinetic energy. The equations and calculations are shown, with a focus on identifying an error and correcting it. The importance of using a consistent reference equation is emphasized.
  • #1
jisbon
476
30
Homework Statement
Cylinder with mass 1.5kg and 0.1m radius roll without slipping up a slope of length 3m and height 1m. Cylinder has an initial translational velocity of 4m/s. Calculate velocity of the cylinder as it leaves the top of the slope.
Relevant Equations
Gain in GPE = Loss in translational KE + loss in rotational KE
I seem to be able to do this problem (at least from what I think, but my answer is still wrong according to the answer key, please do help check.)

Since:
Gain in GPE = Loss in translational KE + loss in rotational KE
##\left(m\cdot g\cdot h\right)=\left(\frac{1}{2}\left(m\right)\left(v_{f}^2\right)-\frac{1}{2}\left(m\right)\left(v_{i}^2\right)\right)+ \left(\frac{1}{2}\left(I\right)\left(\omega_{f}^2\right)-\frac{1}{2}\left(I\right)\left(\omega_{i}^2\right)\right)##
Whereby
##I = 0.5 (mr^2) = 0.5 (1.5*0.1^2)##
##\omega = v/r = 4/0.1##

##\left(1.5\cdot 9.8\cdot 1\right)=\left(\frac{1}{2}\left(1.5\right)\left(v_{f}^2\right)-\frac{1}{2}\left(1.5\right)\left(4^2\right)\right)+\left(\left(\frac{1}{2}\left(0.5\cdot \:1.5\cdot \:0.1^2\right)\left(\frac{v_{f}}{0.1}\right)^2\right)\:-\:\left(\frac{1}{2}\left(0.5\cdot 1.5\cdot 0.1^2\right)\left(\frac{4}{0.1}\right)^2\right)\right)##

Are the equations correct/did I miss something out? Thanks
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Since it does not slip, what happens to rotational speed in relation to translational speed? Ok, i think you took care of that. I'm on my phone so it's hard to read all of the symbols.
 
  • #3
I think I see it now. You really have what appears to be just an error of swapping the signs. Try writing it like this: (Total energy at beginning) = (Total energy at top) to see how that works, and if you spot the error.
 
  • Like
Likes jisbon
  • #4
scottdave said:
I think I see it now. You really have what appears to be just an error of swapping the signs. Try writing it like this: (Total energy at beginning) = (Total energy at top) to see how that works, and if you spot the error.
Solved the error. Thanks for your help. Will use your method in the future
 
  • #5
jisbon said:
Solved the error. Thanks for your help. Will use your method in the future
Either method is fine. The problem was that your reference equation was in terms of loss of energy, but you turned that into final - initial, which would be a gain in energy.
It is generally safer to record reference equations in terms of gain.
 

Related to Calculate the velocity of a cylinder as it leaves the top of a slope (using Energy)

1. How do you calculate the velocity of a cylinder as it leaves the top of a slope?

The velocity of the cylinder can be calculated using the formula v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the slope.

2. What is the formula for calculating the potential energy of the cylinder at the top of the slope?

The potential energy of the cylinder can be calculated using the formula PE = mgh, where PE is the potential energy, m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the slope.

3. Does the mass of the cylinder affect its velocity as it leaves the top of the slope?

Yes, the mass of the cylinder does affect its velocity. According to the formula v = √(2gh), the velocity is directly proportional to the square root of the mass. This means that a heavier cylinder will have a greater velocity compared to a lighter cylinder when leaving the top of the slope.

4. How does the height of the slope affect the velocity of the cylinder?

The height of the slope directly affects the velocity of the cylinder. As the height increases, the velocity also increases. This is because the potential energy of the cylinder increases with height, and this potential energy is converted into kinetic energy as the cylinder moves down the slope.

5. Is there a minimum height required for the slope in order for the cylinder to reach a certain velocity?

Yes, there is a minimum height required for the slope in order for the cylinder to reach a certain velocity. This can be calculated by rearranging the formula v = √(2gh) to solve for h. The minimum height for a given velocity can then be determined by plugging in the values for v, g, and solving for h.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
500
  • Introductory Physics Homework Help
Replies
30
Views
461
  • Introductory Physics Homework Help
Replies
7
Views
697
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
843
  • Introductory Physics Homework Help
Replies
17
Views
426
  • Introductory Physics Homework Help
Replies
11
Views
338
  • Introductory Physics Homework Help
Replies
28
Views
399
  • Introductory Physics Homework Help
Replies
15
Views
313
  • Introductory Physics Homework Help
Replies
8
Views
852
Back
Top