Cylinder Rolling Down an Inclined Plane

Click For Summary

Homework Help Overview

The discussion revolves around a solid cylinder rolling down an inclined plane, with participants exploring the calculation of its rotational and translational kinetic energy. The problem involves concepts from dynamics and rotational motion, specifically focusing on energy conservation and the moment of inertia of the cylinder.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to apply the conservation of energy principle, setting up equations involving gravitational potential energy and kinetic energies. There are discussions about the correct moment of inertia and its implications on the calculations.

Discussion Status

Several participants have shared their calculations and results, but discrepancies in the values of kinetic energy and moment of inertia have led to confusion. Some guidance has been offered regarding the correct moment of inertia for the cylinder, and there is ongoing questioning about the values used in calculations, particularly the gravitational constant.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or receive. There are also varying interpretations of the problem setup and the values used in calculations, leading to different results among participants.

eagles12
Messages
76
Reaction score
0

Homework Statement



A 3.0kg solid cylinder (radius=.15m, length=.7m) is released from rest at a top of a ramp and allowed to roll without slipping. The ramp is .9m high and 5m long. Find the rotational and translational kinetic energy.

Homework Equations



krot=1/2Iw^2
Ktrans=1/2mv^2

The Attempt at a Solution



I already determined that the final kinetic energy is 26J

v=rw
v=(.15)w
Ktrans=1/2(3)(v^2)
 
Physics news on Phys.org
i set up
mgH=1/2mv^2+1/2Iw^2
and got
26=1/2(3)v^2+1/2(.1225)w^2
26=.03375w^2+.06125W^2
26=.095w^2
w=16.54
with this i got
Krot=16.76
but it is saying this is not correct
 
eagles12 said:
i set up
mgH=1/2mv^2+1/2Iw^2
and got
26=1/2(3)v^2+1/2(.1225)w^2
26=.03375w^2+.06125W^2
26=.095w^2
w=16.54
with this i got
Krot=16.76
but it is saying this is not correct

Wrong moment of Inertia!
 
would it be 1/2 MR^2
 
eagles12 said:
would it be 1/2 MR^2

If i tell that, then i would solve your homework.
Think about it, how the cylinder will roll down (about which axis)?
 
i got the rotational one and plugged in the w and r into v=rw
then squared v and multiplied by the mass
and divided by two, but i didn't get the right answer
 
eagles12 said:
would it be 1/2 MR^2

The moment of inertia of the cylinder with respect to the axis through its centre is 1/2 MR^2. It is not 0.1225. How did you get that value?

(The in a plane motion of a rigid body is equivalent with a translation of its CM and rotation around the axis through the CM.)

ehild
 
And can you tell us what value of g you used?
I don't get 26 neither by 10 nor 9.8.
 
i used 9.8 and it has already told me that Kf=26 and Krot=8.8 are correct.
using this, i used v=rw to get v=3.399 and plugged that into 1/2mv^2
doing that i got 17 but it is saying this is incorrect.
 
  • #10
eagles12 said:
i used 9.8 and it has already told me that Kf=26 and Krot=8.8 are correct.
using this, i used v=rw to get v=3.399 and plugged that into 1/2mv^2
doing that i got 17 but it is saying this is incorrect.
Seriously...why 17J is not correct answer? If total energy is 26.486J and rotational energy is 8.8J then the kinetic energy will be 26.486J-8.8J=17.687J~18J. I was always taught to use the value 9.81m/s^2 and not to use rounded values like 26J in calculations. :wink:
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 18 ·
Replies
18
Views
6K
Replies
39
Views
4K
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
4K