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Fairly simple rolling and slipping ball problem.

by cpfoxhunt
Tags: ball, fairly, rolling, simple, slipping
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cpfoxhunt
#1
Apr9-08, 01:34 PM
P: 16
1. The problem statement, all variables and given/known data

A ball radius r mass m starts rolling without slipping up a ramp inclinced at an angle phi to the horizontal, and reaches a maximum hight, h. Derive an expression for the angular velocity, omega, that the ball has at the base of the ramp (ignore rolling friction throughout question.).

b) The ball is initially launched without rotation towards the ramp on a horizontal surface with a coefficient of sliding friction, mu. The ball slides along the surface, begins to roll, and stops slipping before it reaches the ramp. Find an expression for the time, t, raken for it to stop sliding in terms of h, g and mu.

c) by considering the initial velocity v0 and the resistence felt by the ball before it begins to roll, derive an expression relating its initial and final energies in terms of the sliding distance s.

find x, v0, and t taking values h = 0.25m, r = 2cm, m = 50g and mu = 0.3


2. Relevant equations

When something undergoes pure rolling motion, v = rw

3. The attempt at a solution - w = omega u = mu

a) Energy at top = mgh. mgh = 1/2 * mv^2 + 1/2 Iw^2 . I know that as it is rolling, v = rw, and solve to get omega = (10gh/7(r^2))^(0.5) (using I = 2/5 ma^2 if a is the radius)

b) here's where it gets a little dodgy - I assume that there is a rotational equation even though it is sliding (i.e. that it is rotating a little bit but not pure rotation) - is this valid?

I(dw/dt) = umgr , integrate once w.r.t time to get w = t(5ug/2r) where the moment of inertia for a sphere has been used. I then equate this to my earlier figure from a), and get (8h/35gu^2)^(0.5) - is this right?

c) I say that initial energy = 0.5m(v0)^2 and this minus the final energy of slipping (equal to 1/2 * I * w^2 where w is the omega in a) ) = Fx where F is the resistive force during the slipping, and x is slipping distance.

I can then work out t pretty easily, and am not sure how to get X and v0.

Any help is very greatly appreciated,
Cheers
Cpfoxhunt

EDIT: i initially did this using I for a disk (where I is inertia), and have corrected it a bit hurriedly - some of my multiplying fractions might be a bit out?
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Shooting Star
#2
Apr10-08, 05:06 AM
HW Helper
P: 1,979
Quote Quote by cpfoxhunt View Post
c) I say that initial energy = 0.5m(v0)^2 and this minus the final energy of slipping (equal to 1/2 * I * w^2 where w is the omega in a) ) = Fx where F is the resistive force during the slipping, and x is slipping distance.

I can then work out t pretty easily, and am not sure how to get X and v0.
The sliding frictional force stops acting as soon as rolling without slipping is achieved, and v becomes equal to rw.

The work F*x done by the frictional force should be equal to the (initial KE-final KE). The final KE should be (1/2)Iw^2+(1/2)mv^2, where v=rw.

The rest of the work looks very correct.
cpfoxhunt
#3
Apr10-08, 10:06 AM
P: 16
Hi, thanks for the reply. I used that relation to work out part a), and now have a single equation Fx = unknown initial ke (v0 dependant) - known final ke. there are three unknowns which im asked to calculate here. How can I calculate them?
Cheers
Cpfoxhunt

Shooting Star
#4
Apr10-08, 11:43 AM
HW Helper
P: 1,979
Fairly simple rolling and slipping ball problem.

Quote Quote by cpfoxhunt View Post
and now have a single equation Fx = unknown initial ke (v0 dependant) - known final ke.
[(½)mv0^2 - {(½)mv^2 + (½)Iw^2} = µmg*x. This is the one you have used, I think.]

It starts to roll due to sliding friction. The sliding friction stops once v=rw is reached.

For the time when it is sliding and rolling, you have to use (for the rotational motion),

F*r = I(dw/dt) until w becomes v/r from zero, and,

F = m(d^2x/dt^2), until the velocity becomes v from v0, and x goes from 0 to x.
woodie37
#5
Oct23-09, 07:45 PM
P: 14
um...is v not constantly equal to rw? because the ratio of linear velocity and angular is constant?


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