If a billiard ball is hit in just the right way by a cue stick...

Question

If a billiard ball is hit in just the right way by a cue stick, the ball will roll without slipping immediately after losing contact with the stick. Consider a billiard ball (radius r, mass M) at rest on a horizontal pool table. A cue stick exerts a constant horizontal force F on the ball for a time t at a point that is a height h above the table’s surface (see Fig. 10–78). Assume that the coefficient of static friction between the ball and table is μₛ . Determine the value for h so that the ball will roll without slipping immediately after losing contact with the stick.
<IMAGE>

Accepted Answer

Welcome back. Everyone in this problem. In a classroom, a teacher pushes a globe of radius R and mass M to demonstrate the rotation of the earth. The globe is set on a horizontal surface and a constant force F is applied at a point H above the ground for a duration. T The goal is for the globe to roll without slipping, mimicking the earth's rotation smoothly with the coefficient of static friction between the globe and the surface. As MU S identify the eight H for applying the force that achieves this non slipping, rolling motion for our answer choices. A says that it's three R divided by four F multiplied by the expression F minus. Mu GM B says that it's seven R divided by five F multiplied by the expression mu GM minus FC says it's seven R divided by five F multiplied by the expression F minus mu Gm and D says it's seven R divided by the expression five F minus mu GM. Now let's just make sure we understand what we're trying to figure out here. So we want to identify the height edge for applying the force that would achieve this non slipping rolling motion. OK? Now, if we're going to figure out what that height is, first, it would help for us to understand all of the forces that are acting on our globe here during this demonstration. So let's try to resolve our forces for translational and rotational motion and then we'll see where we can go from there. OK. So in other words, we are going to resolve our forces for translational motion for our forces in the Y direction. We know that it's going to be equal to zero. OK. For our forces in the X direction, we know that they are going to follow Newton's second law. OK. So we'll get to that in a bit. And that means we're also looking at our rotational motion. So that is the sum of the torque. Now, when we look at our globe here, we know that the weight of the globe is acting downwards as MG. A normal force is acting upwards, OK, perpendicular to our surface. And we can call that force FN. Like we said earlier, we know that a constant force F OK is applied at a point above the ground. OK. And we know that for our non slipping rolling motion, our frictional force then is going to be acting against that applied constant force F. So we can have that force as FFR now here for our forces in the Y direction, as we said before, we know that those forces will be equal to zero, our ball is rolling along our surface. So that means our normal force FN minus the mass, sorry, the weight of our globe, which is MG would be equal to zero. Therefore, our normal force FN equals MG. Next, let's try to resolve our forces in the X direction. Like I was started saying earlier, we can apply Newton's second law. So applying Newton's second law to the X axis recall, Newton's second law tells us that the net force is directly proportional to the acceleration by the formula that says the net force equals the mass multiplied by the acceleration. Here, our net force in the X direction is going to be equal to F minus our frictional force. Fr So F minus fr equals ma. OK. Now, what do we know about our frictional force? Well, if I make a note here, recall that our frictional force is equal to our coefficient of static friction multiplied by our normal force. OK. So this is pretty important because no, that tells us then that four forces in the X direction F minus mu FN is going to be equal to M. Again, we know our normal force is MG. So we can rewrite that as F minus mu multiplied by MG equals M. And now we can solve for our acceleration here because this tells us our acceleration equals F minus mu MG all divided by M which if we were to simplify that we get the acceleration to be F divided by M minus mu G because M cancels M. Ok. So we resolved our forces in the Y direction in X direction. Now, let's start to look at our torque for a bit. Now, what do we know about this summer? For a talk? Well, we know that our torque represents our rotational force applied over a distance and that torque, the sum of the torque is equal to the moment of inertia multiplied by the angular acceleration. Now, what do we know about our torque here? Well, we know that there is going to be a torque applied for our force F and we also know that our frictional force will generate its own torque. We are here. That means we're going to have F multiplied by H minus R for our applied force. F plus our frictional force. FFR multiplied by our radius of our globe. And that is going to be equal to I alpha. Now, again, we already know our frictional force is MU FN or MU MG. So we can use that idea here to write it as F multiplied by H minus R plus. OK. Uh Mu MG. Let's replace our fictional force with MU MG multiplied by R being equal to I alpha. Now, we can solve for alpha because this tells us alpha equals F multiplied by H minus R plus Mu Mgr all divided by I or moment of inertia. No, we resolved our forces we have expressions for our normal force, our acceleration and our angular acceleration. But again, the question begs or it begs the question, what does that have to do with anything? Why, why would we need expressions for our acceleration here? Well, if we think about it, the object begins motionless and both its acceleration and angular acceleration remain unchanged over time. Thus, equations of constant acceleration can be applied to determine the velocity and angular velocity as functions of time, which may help us to figure out the height so far, we know that we have one equation that's expressed in terms of our height age that we want to know. So let's find our velocity and angular velocity. OK. Now, for our velocity, we know that our velocity is going to be equal to our initial velocity plus the acceleration multiplied by time. We know our initial velocity is zero. OK? And we have an expression for acceleration. So if we substitute it, we get our velocity V OK to be equal to acceleration F divided by M minus mu G. OK. All multiplied by our time. T. Next, we know that our angular velocity equals our initial angular velocity plus our angular acceleration multiplied by T again, our initial angular velocity would be equal to zero. So that means our angular velocity then is going to be equal to alpha, which we have an expression for alpha in terms of H. So that's F multiplied by H minus R plus mu mgr all divided by I, OK. And all of that is being multiplied by T. Now, at the precise moment when the globe ceases to be pushed by the teacher and makes no constant at that release point, it continues to rule without any slipping. Therefore, at this instant or relationship, OK. As I add, T release our relationship that says our angular velocity equals our velocity divided by R was true. Now, since it holds true, that means we can substitute it here because we have an expression for V into our formula for our angular velocity to the sulfur or height edge. Because what this means then what this means, no, let's put this in red. We're getting to that point because what this means is that Omega which we have here f multiplied by H minus R plus mu mgr all divided by I, OK, multiplied by T is going to be equal to our velocity V CM which is F divided by M minus mu G all multiplied by T all divided by R or multiplied by the reciprocal of R. In this case, notice that we can cancel T from both sides and we can express I in terms of M and R because the globe's moment of inertia, the globe's moment of inertia, I is going to be equal to 2/5 of Mr squared. So with that idea, then no, we can go ahead and try to solve for OK. So let's try to do that. So no, that means if we expand F multiplied by H minus R, we'll get F minus Fr plus mu mgr all divided by I, which is 2/5 of Mr R squared being equal to one divided by R multiplied by F divided by M minus mu G. Again, we're solving for H so we can multiply both sides by 2/5 of M squared. So we're gonna get FH minus F I plus Mu Mgr being equal to one divided by R multiplied by F divided by M minus mu G multiplied by 2/5 of Mr squared. Now, if you notice here, we can cancel R from R squared, R goes into R squared R times and no to isolate for H, we can leave F alone on our left hand side, which means that we're going to add Fr and subtract new Mgr from both sides. So that means FH OK, is going to be equal to F divided by M minus MGK multiplied by 2/5. Let me move this over a bit. It's gonna be multiplied by 2/5 of Mr, OK? Plus Fr minus MGMR. Sorry. Mmgr, let me just keep it looking the same way. Now notice that we can uh factor out R here, OK? Because all of these terms have uh all of these terms have R OK? So now when we do that, OK, we get R multiplied by two thieves of FF divided by M. Oh You know what, before we do that, let's just take a quick stop here. Let's just make sure we, we, we, we get our values right. So let's go ahead and expand just to make it easier while we work. So we can multiply through by M in our first term. OK. So that's gonna be two R divided by five, multiplied by F divided by M multiplied by M. That's just F mu G multiplied by M. So that's mu GM, OK? Minus Mu Mgr plus Fr, OK. And if we were to get H by itself, we could, at this point, we could divide both sides by F. So we can have all of this being divided by F, OK. Which now if we expand, if we go ahead and expand here, eh then we're going to have H being equal to two R divided by five FK multiplied by F minus Mu Gm. OK. Minus the mule. Let's say here we have our term, let me mgr, OK? Divided by F plus fr divided by F which would just be R and no, we can rewrite our entire expression then as age to be equal to seven fifths, OK? Seven fifths of R divided by F multiplied by F minus MGM. OK. Which we can write it as several R divided by five F multiplied by F minus MGM. So this could be our final expression for our height. Therefore, when we go back to our answer choices. Ok. That means that c is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

Watch next

Torque & Acceleration (Rotational Dynamics) practice set