Rotational motion conservation of energy; time

AI Thread Summary
A hollow cylinder is rolling at 3.3 m/s when it reaches a 15-degree incline, and the problem involves determining how far it will ascend and the time spent on the incline. The first part was solved, finding the height reached to be 4.3 meters using conservation of energy. The second part posed challenges regarding the calculation of time, with confusion over the radius and mass of the hoop, which are not necessary for the solution. The average speed while decelerating was clarified to be the average of initial and final speeds, leading to the correct time calculation. Ultimately, the correct time on the ramp was confirmed to be 5.21 seconds after addressing misunderstandings in the calculations.
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Homework Statement


A hollow cylinder (hoop) is rolling on a horizontal surface at speed v=3.3m/s when it reaches a 15 degree incline. How far up the incline will it go? How long will it be on the incline before it arrives back at the bottom.


Homework Equations


SOH CAH TOA
MGH=KEtrans +KErot
w=v/r
x=theta*r
\omega^2=wo^2 +2\alpha\vartheta

The Attempt at a Solution


I have finished the first part of the problem and solved for 4.3 m using conservation of energy and a little trig. The second part is giving me trouble because I'm not sure if I calculated the radius of the hoop correctly (.525m) which could be throwing off my substitutions and calculations. I used the last equation listed with 2\pi for theta. I know the time should be 5.2 seconds but I'm doing something wrong. I know i have to relate speed and time where vf=0 and take that time and multiply is by 2 to get the total time on ramp. Please help?
 
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What's the mass of the hoop? (this is only relevant to check your work)

(and with your work, I arrived at 5.21 seconds on the ramp)You can simplify this problem immensely for the second part. Just treat it like an object decelerating from 3.3 m/s to 0 m/s over 4.3 meters. Once you've solved for that acceleration, you have a nifty formula relating velocity, acceleration, and time.
 
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The mass of the hoop is not given, i was under the impression mass does not matter because it all cancels out? Which equation can I use to determine the mass of the hoop? I'll do the work but I just don't know which one to use. I don't see an equation relating rotational motion and mass..
 
0338jw said:
The mass of the hoop is not given, i was under the impression mass does not matter because it all cancels out? Which equation can I use to determine the mass of the hoop? I'll do the work but I just don't know which one to use. I don't see an equation relating rotational motion and mass..

Yeah, ignore that entirely. I misread what you wrote. You don't need to solve for the mass.
 
0338jw said:
The second part is giving me trouble because I'm not sure if I calculated the radius of the hoop correctly (.525m) which could be throwing off my substitutions and calculations. I used the last equation listed with 2\pi for theta. I know the time should be 5.2 seconds but I'm doing something wrong. I know i have to relate speed and time where vf=0 and take that time and multiply is by 2 to get the total time on ramp. Please help?
To find the time on the ramp you do not need to know (nor do you have sufficient information to determine) the radius of the hoop. (Nor do you need the mass of the hoop.)

Hint: What's the average translational speed of the hoop as it goes up the ramp?
 
hmm. I'm sort of confused by what you mean by average translational speed of the hoop, do you mean to plug my numbers back into PE=KEtrans +KErot? would the average translational speed be Vf-Vo/2?
 
0338jw said:
I'm sort of confused by what you mean by average translational speed of the hoop, do you mean to plug my numbers back into PE=KEtrans +KErot?
No. All you need is simple kinematics.
would the average translational speed be Vf-Vo/2?
Almost: Vave = (Vf + Vo)/2
 
I plugged my numbers back into x=1/2at^2 +Vo*T +X0 substituting (Vf-Vo)/T But I still get the wrong answer. What am i missing?
 
The only formula you need is distance = ave speed X time. You don't need the acceleration.
 
  • #10
when I use that eq I get (0-3.3m/s)/2=-1.65m/s.
4.3m=-1.65*T I get -2.6 for T which isn;t right. Sorry if I'm not getting something this simple but I am just not seeing something.
 
  • #11
0338jw said:
when I use that eq I get (0-3.3m/s)/2=-1.65m/s.
4.3m=-1.65*T I get -2.6 for T which isn;t right.
(1) The average speed is 1.65 m/s, not -1.65 m/s. (Why are you subtracting?)
(2) Going up the incline is just half the time.
 
  • #12
yes! thank you! I forgot to multiply it by two! It was negative because I thought it was Vf-V0 leaving -3.3m/s. Thanks a bunch! :biggrin:
 
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