Cylinder Rolling Down an Inclined Plane

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SUMMARY

The discussion focuses on calculating the rotational and translational kinetic energy of a solid cylinder rolling down an inclined plane. The cylinder has a mass of 3.0 kg, a radius of 0.15 m, and a height of 0.9 m on the ramp. The correct moment of inertia for the cylinder is 1/2 MR^2, which is crucial for accurate calculations. The final kinetic energy is established at approximately 26 J, with the rotational kinetic energy calculated as 8.8 J, leading to a translational kinetic energy of around 17.7 J.

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  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with energy conservation principles in physics
  • Knowledge of kinematic equations relating linear and angular motion
  • Proficiency in using gravitational acceleration values (9.81 m/s²)
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  • Study the derivation and application of the moment of inertia for various shapes
  • Explore the relationship between linear and angular velocity in rolling motion
  • Investigate energy conservation in systems involving rotation and translation
  • Learn about the effects of different gravitational acceleration values on energy calculations
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators teaching concepts related to rotational motion and energy calculations.

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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