Finding Velocity and Angular Speed of a Rolling Cylinder

AI Thread Summary
To find the velocity and angular speed of a solid cylinder rolling down a 2.5-meter incline, the mass is not required for the calculations since it cancels out in the energy equations. The potential energy at the top, E = mgh, converts entirely into kinetic energy at the bottom, which includes both translational and rotational components. The moment of inertia for the cylinder is I = mr^2, and the final kinetic energy is the sum of translational and rotational energy. By applying these principles, the final velocity and angular speed can be determined without needing the mass. Understanding the relationship between potential energy and kinetic energy is key to solving this problem.
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Homework Statement



A solid cylinder of radius 20cm is released from a 2.5 high incline. If it rolls down without losing any energy to friction, find the cylinder's velocity at the bottom of the incline and the angular speed at the bottom of the incline.

Homework Equations





The Attempt at a Solution



I need to be given a mass to solve this! If I had a mass, I'd plug it into I = mr^2, find the inertia. With that, I'd plug it into E = mgh, and after i'd find W (angular velocity). Help!
 
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Are you sure the masses don't cancel out? :)
 


gamer1319 said:

Homework Statement



A solid cylinder of radius 20cm is released from a 2.5 high incline. If it rolls down without losing any energy to friction, find the cylinder's velocity at the bottom of the incline and the angular speed at the bottom of the incline.

Homework Equations





The Attempt at a Solution



I need to be given a mass to solve this! If I had a mass, I'd plug it into I = mr^2, find the inertia. With that, I'd plug it into E = mgh, and after i'd find W (angular velocity). Help!

The mass is m. You don't need to know the mass with this problem any more than you need to know the mass of an object to find its final speed after falling 2.5 meters, if it was released at rest.

Don't forget, the final Kinetic Energy, KE, includes a rotational part as well as a translational part, i.e. KE = (1/2)·I·ω2 + (1/2)·m·v2 .
 

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