Conservation of energy and friction

[llandau]

If a body moves along a quarter of a circle and it is subject only to conservative forces, it is relatively easy to deduce the equation of motion. But if friction comes into action, it seems more difficult. Suppose that a body has initial velocity v_0 and that it moves along a quarter of a circle of radius R, subject also to friction (coefficient of friction is known). Is it still possible, even if conservation of energy can't be applied, to derive the final velocity at the top?

 

[Fewmet]

If a body moves along a quarter of a circle and it is subject only to conservative forces, it is relatively easy to deduce the equation of motion. But if friction comes into action, it seems more difficult. Suppose that a body has initial velocity v_0 and that it moves along a quarter of a circle of radius R, subject also to friction (coefficient of friction is known). Is it still possible, even if conservation of energy can't be applied, to derive the final velocity at the top?

The particle travels a distance of \piR/2. Friction does work to it along that path equal to \muN\piR/2, where \mu is the coefficient of friction and N is the normal force. (You need to work out the expression for N if the particle moves upward in a gravitational field...or do you want to make life simple and let the circle be horizontal?)

The work done by friction equals the reduction of kinetic energy.

 

[llandau]

Yes, I should have been more precise.. I meant that the circle is vertical so the normal force along the motion should be mv^2/R+mgcos(theta). Now, I wonder if it is possible to evaluate the work done by friction in this case elementary, so that the final velocity can be easily obtained... Thanks in advance

 

[aim1732]

For normal force you would need to write out the equations of Newton's second law in the tangential and normal directions.That should provide you an N dependent on angle traversed along the circle.Then you would need to perform an integration along the length of the path for work done by friction. That should be relatively easy considering friction and displacement always are 180 degrees to each other.

 

[llandau]

I actually tried that way:

- in the radial direction, the law of motion is: N-mgcos(theta)=mv^2/R
- in the tangential direction, the only force is friction, that is (mu)m(gcos(theta)+v^2/R).
Now, what I find confusing is this: I should integrate to find the work done by friction, but with respect to what? It seems here that friction depends on v and on theta and the relation between the two is not obvious (to me).

 

[aim1732]

in the tangential direction, the only force is friction, that is (mu)m(gcos(theta)+v^2/R).

Are you sure this is correct?

 

[llandau]

You are right, plus the component of gravity mg sin(theta). Being gravity conservative, the real problem is how to calculate the work done by friction. Do you have a solution?

 

[aim1732]

Acceleration in the tangential direction is dv/dt where v is speed of the particle by definition. Now that you have an equation of the Second Law in the tangential direction containing a velocity term(by direct substitution for N from first eqn,) and the differential dv integrate to get v(t).That should be the magnitude of speed as a function of time. Then put it back in the first equation to calculate N(t). That should help with calculating the work of friction.

Lev, the logic seems good enough to me but the math is scary looking. Have a go.I will try too.Hell maybe we can compare our solutions.