Determine the translational speed of the cylinder

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SUMMARY

The translational speed of a solid cylinder rolling without slipping down an inclined plane can be determined using energy conservation principles. The equation derived is v = √[(4/3)gh], where m is the mass, r is the radius, and h is the height from which the cylinder starts. The rotational inertia used in the calculations is I = 1/2mr². This method confirms that the approach to solving the problem is correct, as validated by participants in the discussion.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgh)
  • Knowledge of kinetic energy equations (K = 1/2mv²)
  • Familiarity with rotational inertia (I = 1/2mr²)
  • Concept of rolling motion and the relationship between translational and angular speed (v = rω)
NEXT STEPS
  • Study the derivation of energy conservation in rolling motion
  • Explore the implications of different shapes and their rotational inertia on rolling dynamics
  • Learn about the effects of friction in rolling without slipping scenarios
  • Investigate advanced problems involving multiple objects rolling down inclines
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for clear examples of energy conservation in rolling motion scenarios.

k-rod AP 2010
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Homework Statement


A solid cylinder w/ mass m, radius r, and rotational inertia 1/25mr2 rolls without slipping down an inclined plane. THe cylinder starts form rest at height h. The inclined plane makes an angle θ w/ the horizontal.Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.


Homework Equations


v=velocity
ω=angular speed
I=moment of inertia
U=gravitational potential energy
K=Kinetic Energy

The Attempt at a Solution


U=Ktranslational + Krotational

mgh=(1/2mv2) + (1/2Iω2)

mgh=(1/2mv2) + (1/2 (1/2mr2)(v2/r2))

gh=(v2/2) + (v2/4)

gh=(2v2/4) + (v2/4)

gh= (4/3)v2

v2=(4/3)gh

v= √[(4/3)gh] would this be the correct procedure to solve this?
 
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hi k-rod AP 2010! :smile:
k-rod AP 2010 said:
A solid cylinder w/ mass m, radius r, and rotational inertia 1/25mr2 rolls without slipping down an inclined plane. THe cylinder starts form rest at height h. The inclined plane makes an angle θ w/ the horizontal.Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.

v2=(4/3)gh

v= √[(4/3)gh] would this be the correct procedure to solve this?

(i assume you meant rotational inertia 1/2 mr2 ? :wink:)

yes, that's the right method and result! :smile:

(btw, another way is to say the "rolling mass" is I/r2, = m/2, so the total "effective mass" is 3m/2, so the "effective gravity" is 2g/3, so 1/2 mv2 = 2mgh/3 … but i don't think the examiners would like that! :redface:)
 

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