A cylindrical roll of coins with a radius of 0.02 meter rolls up a 15° ramp with an initial angular speed of 45 rad/sec. What will its vertical elevation be when it comes to rest before it starts rolling back down the ramp?
This is a conservation of energy problem. The tricky bit is they omit to tell you the mass of the coins, but just write M for the mass, and it will cancel out in the end. They don't tell you
The roll of coins is a cylinder of at least approximately constant density. This lets you look up the momentum of inertial.
And when things of constant radius roll along a surface, the linear velocity is proportional to the angular velocity.
There are three types of energy in this problem:
* Linear kinetic energy
* Rotational kinetic energy
* Gravitational potential energy
And the sum of these is constant. That means it's the same at the beginning and end of the problem, so you don't have to solve the equations of motion.
The roll of coins is a cylinder of at least approximately constant density. This lets you look up the momentum of inertial.
And when things of constant radius roll along a surface, the linear velocity is proportional to the angular velocity.
There are three types of energy in this problem:
* Linear kinetic energy
* Rotational kinetic energy
* Gravitational potential energy
And the sum of these is constant. That means it's the same at the beginning and end of the problem, so you don't have to solve the equations of motion.
rpenner is correct, this problem can be solved through the conservation of energy. Since they don't say anything about friction then the only thing that is going to slow down the roll of quarters is the pull of gravity as it moves up the incline. But initially it has a kinetic energy of 0.5*mass*velocity^2. They give you the angular velocity which can be converted into its linear velocity (or more precisely the tangential velocity on any point of the circle, imagine it as the tangential velocity at the top of the roll of quarters always so you can picture an arrow pointing off tangentially on the top of the circle pointing always in the direction of travel) by the product of the angular velocity times the radius of the roll of quarters. So that would be v = omega*radius = 0.02*45 = 0.9 meters/sec.
So you can picture the point on the top of the roll of quarters moving at a speed of 0.9 m/s in the direction that the roll of quarters is traveling in.
So its initial kinetic energy is 0.5*mass*0.9^2. When it comes to a stop all the initial kinetic energy of the roll of quarters has been converted into a potential energy due to its vertical height from the level surface that the incline lies upon. So you can just set the initial kinetic energy equal to the final potential energy and solve for the height. So you get: 0.5*mass*0.9^2 = mass*acceleration due to gravity*vertical height,
0.5*mass*0.9^2 = mass*9.8*height, the mass ratios away from each side of the equation leaving
0.5*0.9^2 = 9.8*height, solving for the height yields:
(0.5*0.9^2)/9.8 = height = 0.04133 meters or about 1.63 inches
It travels for a much longer distance up the incline, but the incline is only 15 degrees so vertically its final height is not all that much.
So you can picture the point on the top of the roll of quarters moving at a speed of 0.9 m/s in the direction that the roll of quarters is traveling in.
So its initial kinetic energy is 0.5*mass*0.9^2. When it comes to a stop all the initial kinetic energy of the roll of quarters has been converted into a potential energy due to its vertical height from the level surface that the incline lies upon. So you can just set the initial kinetic energy equal to the final potential energy and solve for the height. So you get: 0.5*mass*0.9^2 = mass*acceleration due to gravity*vertical height,
0.5*mass*0.9^2 = mass*9.8*height, the mass ratios away from each side of the equation leaving
0.5*0.9^2 = 9.8*height, solving for the height yields:
(0.5*0.9^2)/9.8 = height = 0.04133 meters or about 1.63 inches
It travels for a much longer distance up the incline, but the incline is only 15 degrees so vertically its final height is not all that much.