Rolling cylinder on an incline

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SUMMARY

A cylinder with mass M and radius R rolls down an incline from a height h. The linear velocity at the bottom is determined to be sqrt((4/3)gh). The linear momentum is equivalent to the momentum of the center of mass (CM) of the cylinder, while the angular momentum requires further clarification as it is not simply mvsin(theta). The discussion emphasizes the importance of using conservation laws to derive these quantities without needing the angle of inclination.

PREREQUISITES
  • Understanding of rotational dynamics
  • Familiarity with conservation of energy principles
  • Knowledge of linear and angular momentum definitions
  • Basic physics of rolling motion
NEXT STEPS
  • Study the conservation of energy in rolling objects
  • Learn about the relationship between linear and angular momentum
  • Explore the equations of motion for rolling cylinders
  • Investigate the effects of incline angles on rolling dynamics
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Physics students, mechanical engineers, and anyone interested in the dynamics of rolling objects on inclines.

andrewp7
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A cylinder of a mass M and a radius R starts at the top of a hill at a height h, and rolls to the bottom. At the bottom of the hill, what is its linear velocity, linear momentum, and angular momentum?


I believe the the velocity is sqrt((4/3)gh) and the the angular momentum is mvsin(theta) but I am not sure if those are right and I still d not know the linear momentum
 
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You have to give all quantities in terms of M, R, and h. Theta is not given. Look after the definition of angular momentum, it is not mvsin(theta). The linear momentum is the same as the momentum of the CM of the cylinder.

ehild
 
wouldn't I need some type of angle because it is on an incline?
 
Do you? You got the speed without the angle, don't you? How did you got it? Using what law?

ehild
 

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