Rotational Motion of a cylinder

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SUMMARY

The discussion focuses on the rotational motion of a hollow cylinder rolling up a 15-degree incline at a speed of 4.3 m/s. The user successfully calculated the distance the cylinder travels up the incline as 7.3 meters using conservation of energy principles. However, they encountered difficulties in determining the time taken to ascend and descend the incline, noting that the rotational aspect complicates the problem compared to a non-rotating object. The solution involves applying kinematic equations and understanding the effects of acceleration on the cylinder's motion.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with kinematic equations
  • Knowledge of rotational dynamics, specifically for hollow cylinders
  • Basic trigonometry (SOH CAH TOA)
NEXT STEPS
  • Study the effects of rotational inertia on motion for different shapes
  • Learn how to apply kinematic equations to rotating objects
  • Explore the concept of potential energy and its conversion to kinetic energy in rotational systems
  • Investigate the relationship between angular velocity and linear velocity in rolling motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators seeking to clarify concepts related to energy conservation and kinematics in rolling objects.

Carpe Mori
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Homework Statement


A hollow cylinder (hoops) is rolling on a horizontal surface at speed v = 4.3 m/s when it reaches a 15 degree incline (a) how far up the incline will it go? (b) how long will it be on the incline before it arrives back at the bottom?


Homework Equations



PE = KE + RE
SOH CAH TOA

w = v/r

I = mr^2


The Attempt at a Solution



A) I figured out letter A with conservation of Energy (yawn) distance = 7.3 m

B)I am not sure exactly how to solve this problem. If it was a non rotating cube i could do it very easilly =D. I am sure the fact that it is rotating though affects it somehow. For a cube I would use the Kinetics equations (y = -.5gt^2 + vt + y0) to solve for time. Could someone help me out thanks =D
 
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the time taken to go up the incline=the time taken to come down.
calculate time taken to go up sing simple kinamtic relations and then double it to get the desired result.

use velocity final=0
velocity[initial]=4.3 m/s
distance=7.3 m

now the important part is accln- i leave it to you to figure it out.
use =s=ut+1/2at^2
 

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