Calculating Final Position of Billiard Ball with Rolling Friction

[mickg77]

My problem involves rolling a billiard ball with an initial x velocity of 2.4ms and an initial y velocity of 0.7ms. The rolling friction is 0.1 and I have to work out the final resting position of the ball. There is no sliding friction involved. I have to break this into 0.1seconds and return the results, i think that I have to work out the initial decelleration then the speed and new position, and then the next decelleration and so on?

the mass of the ball is 0.17kg

I'm pretty new to this, so sorry if its a bit basic.

Michael

 

[Sythalist]

You must assume that the ball begins in the state of pure rolling. Pure rolling occurs when the tangential speed of the circumference of the ball equals the linear speed: v = rw. Where r is the radius of the ball, while v and w are the linear and angular velocities.

The force of friction acts at the point of contact between the ball and the surface and thus applies a torque on the circumference of it. The torque induces angular acceleration similar to the way a force induces linear acceleration:

1.) T = I@, where T is torque, I is the moment of inertia of the object and @ is angular acceleration.
This is related to the Newtonian form: F = ma

From the definition of torque: 2.) T = fr, where f is the force of friction and r is the radius of the ball.

Force of friction is determined from the coefficient of rolling friction and the normal force on the ball: 3.) f = CFn = Cmg , where C is the coefficient of friction.

The moment of inertia is taken from literature values: 4.) I = 2mr^2/5

As described at the beginning, pure rolling exhibits a relationship between the angular acceleration and the linear acceleration: 5.) a = r@ <--> @ = a/r

Substituting equations 2-5 into 1 gives:
Cmgr = (2mr^2/5)(a/r)

Simplifying gives:
6.) a = 5Cg/2

Acceleration remains constant, so you only have to determine the new speeds at each time interval to determine the distance traveled using the equations:
v = u + at
v^2 = u^2 + 2ad

Sorry ran out of time I'll try to finish up later.