Cylinder Rolling Down an Inclined Plane

AI Thread Summary
A solid cylinder with a mass of 3.0 kg rolls down a 0.9 m high ramp, and the total final kinetic energy is calculated to be 26 J. The moment of inertia used in the calculations is incorrectly stated as 0.1225 instead of the correct value of 1/2 MR^2. Participants discuss the importance of using precise values for gravitational acceleration, with some suggesting that 9.81 m/s² should be used instead of rounded figures. The correct rotational kinetic energy is indicated to be 8.8 J, leading to confusion over the translational kinetic energy calculation. Accurate application of these principles is essential for solving the problem correctly.
eagles12
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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)
 
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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
 
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:
 
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