Cylinder Rolling Down an Inclined Plane

  • Thread starter eagles12
  • Start date
  • #1
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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)
 

Answers and Replies

  • #2
76
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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
 
  • #3
3,812
92
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!!
 
  • #4
76
0
would it be 1/2 MR^2
 
  • #5
3,812
92
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)?
 
  • #6
76
0
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
 
  • #7
ehild
Homework Helper
15,543
1,909
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
 
  • #8
3,812
92
And can you tell us what value of g you used?
I don't get 26 neither by 10 nor 9.8.
 
  • #9
76
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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
380
7
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:
 

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