Genius's! Help Needed

Author	Topic: Genius's! Help Needed
Sticky Infiltrator Member # 4256	written Thursday, October 28 2004 10:04 Profile #0 Ok this is going to sound pathetic- I'm tutoring this guy for a freshman physics course and he just brought me this problem for help.- I told him I would work on it but I'm getting stumped and I hate to go to faculty for something that I should totally know how to do. SO...... Since I believe that the average active Spiderwebber is rather intelligent I decided to post it here. A uniform pool ball of radius R and mass M is motionless on a flat surface. You hit it horizontally with a cue, with a large force F for a short time, (delta)t, with the point of impact being 6R/5 above the table. Sketch the subsequent motion of the ball with a graph on the y-axis and t on the x-axis (not hard). Give the (t,v) coordinates for any discontinuities or other critical points in the graph. You may use the variables F, (delta)t, M, R, g, u(mu)static, and u(mu)kinetic (n.b. v is the velocity of the center of mass). Ok..... Posts: 564 | Registered: Wednesday, April 14 2004 07:00
*i The Establishment Member # 6	written Thursday, October 28 2004 12:03 Profile #1 Just looking at this, I can see the following things: F = dp/dt Assume the force is constant over the impact of the ball with the cue. This allows you to do the following: F = (delta)p/(delta)t = m(delta)v/(delta)t v = F(delta)t/m This is the velocity after the external force of the cue is applied. From this point, we have rolling motion. I suggest using energy methods. Keep in mind the object has both translational and rotational kinetic energy. The conservation of mechanical energy: Ui + Ki + W = Uf + Kf There is no change of any kind of potential after the impact, so those terms can be cancelled out. This leaves that the work done is equal to the change in kinetic energy. Set the final kinetic energy as zero. Observe the work is done by friction. Recall: W = integral(F dot dl) Assume constant friction force and the cue does not slip (i.e. contact is only at a single point) allowing you to use the static coefficient of friction. From your force balance, you should get that N = mg and that the friction force f = mu(s)*mg. Since the force is constant with respect to distance, you can pull it out of the integral and just integrate to the distance 'd' it will travel. W = mu(s)*mg*d K = 1/2*m*v^2 + 1/2*I*omega^2 Solve for omega similarly with the rotational torque-angular momentum relation: T = dL/dt = I*omega/(delta)t T = r x F = rFsin(theta) theta is the angle with respect to the lever arm which originates at the center of mass. I is the moment of inertia for a sphere which I do not recall at this moment. Use the part of where the ball was hit with respect to the table to solve for this. Solve for omega: omega = rFsin(theta)*(delta)t/I Go back to the conservation of energy, solve for the distance 'd' traveled by the ball. As for the velocity, you know the distance d traveled and at the force is a constant. The equation you may want to use is: x(t) = at^2 + v0*t + x0 a = -mu(s)*g v0 is the velocity immediately after impact. Solve for the time at which the distance 'd' is reached. This is the time the ball moves allowing you to draw the endpoint of the graph. The velocity likewise is: v(t) = -mu(s)*g*t + v0 This should give you the answer you want. Note that you probably do not need the energy stuff. As you could theoretically solve for it using forces alone. You would just have to consider the torque friction exerts on the cue ball. This is a quick draft, so there may be mistakes. -------------------- Your flower power is no match for my glower power!! Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Sticky Infiltrator Member # 4256	written Thursday, October 28 2004 13:40 Profile #2 quote: From this point, we have rolling motion. Problem- There is rolling motion but at the same time it is actually skidding due to rotational inertia. Until enough rotational momentum has accumulated due to the kinetic friction the rotational velocity(omega) will increase while translational velocity(v) will decrease At impulse F(delta)t = (delta)(Mv) therefore V(initial)= F(delta)t/m the graph is obvious- it goes up sharply for the accelleration/contact time- gradually decreases as Energy is transfered to rotation from translation. Once V = w(omega)R it will stay flat/no acelleration (assuming no air resistance or irregularities) and so it can't have a "d" without a time limit Oh rotational I of a sphere is 2/5 MR^2 The torque due to the cue * contact time t is equal to the change in the product of Inertia * rotational velocity. T(delta)t = (delta)(Iw) The torque is RF/5 (its one fifth R above the center) by r x F = T where r = R/5 so Iw = RF(delta)t/5 = 2wMR^2/5 so initial angular velocity is w = (F(delta)t)/(2MR) So I think that I have the initial angular velocity and the initial translational velocity but how they change- thats whats getting me. Only that they stop changing when V = wR- Sorry I didn't say thanks for your long explanation up there. It got me this far...... and I think that I see something now, so tomorrow.. [ Thursday, October 28, 2004 18:15: Message edited by: m's chosen ] Posts: 564 | Registered: Wednesday, April 14 2004 07:00
Apollyon Warrior Member # 3188	written Friday, October 29 2004 06:40 Profile Homepage #3 Wow, I'm a senior in HS in Physics right now. I like it and all, but all this is what I'm going to do next year? *shudder* -------------------- And that, my Leige, is how we know the Earth to be banana-shaped. -Sir Bedevere Denn du bist was du isst und ihr wisst was es ist. -Rammstein Posts: 154 | Registered: Tuesday, July 8 2003 07:00
Sticky Infiltrator Member # 4256	written Friday, October 29 2004 07:48 Profile #4 Well this problem was asigned for an honours course and is one of the tougher problems that I have seen from the course. And I don't get asked about the easy questions that much so.... Of course it would be possible to complicate the problem further- put on a slope add air resistance and irregularities, make the sphere one of variable density, but anyways, I solved the problem last night thanks to *i's pointing in the correct direction, :) But if you are interested in physics definitly take a look, I'm obviously biased but it is one of the more enjoyable subjects- Are you in an advanced course or a conceptual physics course? Posts: 564 | Registered: Wednesday, April 14 2004 07:00
*i The Establishment Member # 6	written Friday, October 29 2004 08:14 Profile #5 Yeah, this is a tricky problem. I have a paper called "Rolling Cylinder on a Slanted Surface" that discusses this problem in depth. It is actually quite complicated. As far as I have seen, this type is one of the worst, if not the worst, problem that can reasoably asked in an introductory mechanics class. Often this type of problem is reserved for an intermediate 300-400 level mechanics course. Do not let yourself by scared out of considering physics by this problem. It seems hard, but it really falls down to some basic principles (forces, momentum, and energy). This just combines pretty much all of those. When we teach you to swim, we don't start with double roll dives into the deep end of the pool, we start with the basics and build from there. Physics is the same way. Also, what did you actually do to solve the problem? -------------------- Your flower power is no match for my glower power!! Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Sticky Infiltrator Member # 4256	written Friday, October 29 2004 09:35 Profile #6 Well I was pretty confident about my initial angular velocity(w) and my initial translational velocity(v) that I had gotten as shown above. v(initial) = F(delta)t/m and w(initial) = [F(delta)t]/(2MR) And after (delta)t contact was over there was only one force on the ball, friction. This force is entirely tangenital to the ball and thus creates torque without actually taking energy from the system of the ball(ie. as w increases v decreases, they can't both decrease) This torque(T) is given by T = u(kinetic)Mg x R ( The coefficient of kinetic friction times the normal force cross the perpendicular distance, which is R as the force is tangenital) This torque of course is equal to the inertia times alpha(a) T = Ia and I = (2MR^2)/5 so a = [5u(kinetic)Mg]/(2MR^2) M will then cancel out so as the velocity function for constant a is w = w(initial) + a(angular)t w at time t can be discovered by w = [F(delta)t] + [5gu(kinetic)t]/(2R) The translational velocity will be decreased by the same force and: F = Ma(translational) also F = u(kinetic)Mg so a = u(kinetic)g the standard form equation for v is: v = v(initial) + a(translational)t so v is determined by v = F(delta)t/m - u(kinetic)gt so..... v = wR when: F(delta)t/m - u(kinetic) = F(delta)tR/(2mR) 5/2 * gR/Ru(kinetic)t all the Rs conveniently cancel out so solving for this t: t = F(delta)t/(7mu(kinetic)g) and to get the v coordinate on the graph we plug this value for t into the v equation getting: v = 6F(delta)t/(7m) That was a silly problem looking over it- like *i says- its a very simple application of torques and forces that when broken down shouldn't be that hard- its just the breaking down part. Thanks again for the help- and I hope that you stay interested in PHYSICS! :D ( Heh- it might be interesting to solve it with energy now and see if I get the same answer- if I had that much energy) [ Friday, October 29, 2004 18:11: Message edited by: m's chosen ] Posts: 564 | Registered: Wednesday, April 14 2004 07:00