Translational and Rotational speeds

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SUMMARY

The discussion focuses on calculating the rotational and translational speeds of a cylinder rolling down a 3.04-meter ramp inclined at 30 degrees. The key equations used include the conservation of energy principle, expressed as mgh = K_rotational + K_translational, where K_rotational is defined as 1/2*I*omega^2 and K_translational as 1/2*m*v^2. The moment of inertia for the cylinder is I_cylinder = 1/2*m*r^2. The user successfully derives the translational speed but struggles to isolate omega due to the radius dependency in the equations.

PREREQUISITES
  • Understanding of rotational dynamics and energy conservation
  • Familiarity with moment of inertia concepts, specifically I_cylinder = 1/2*m*r^2
  • Knowledge of angular velocity and its relationship to linear velocity (v = r*omega)
  • Basic algebraic manipulation skills to solve equations
NEXT STEPS
  • Study the derivation of rotational motion equations in physics
  • Learn about the relationship between linear and angular motion in rolling objects
  • Explore energy conservation in different mechanical systems
  • Investigate the effects of varying radius on rotational dynamics
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Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for examples of energy conservation in rolling objects.

hawk320
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Homework Statement


If a cylinder rolls down a 3.04 meter ramp with a 30 degree incline without slipping, how fast will the rotational and translational speeds be when the cylinder reaches the bottom of the ramp?


Homework Equations


mgh = K_rotational + K_translational
K_rotational = 1/2*I*omega^2
I_cylinder = 1/2*m*r^2
K_translational = 1/2*m*v^2
v= r*omega



The Attempt at a Solution


I can find out the translational speed by using the equation mgh = \frac{1}{2} * (\frac{1}{2} m r^2) \omega^2 + \frac{1}{2} m v^2
Which you can reduce down to mgh = 1.5 v^2 Since we know m,g and h we can solve for v.
But I can't seem to find a way to solve for omega without knowing the radius, when i reduce that equation down to solve for omega I always left with the radius somewhere. It may just be that you cannot solve for omega, but if you can find a way the help will be really appreciated.
 
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