 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
Natural Science Forum / Physics / General Physics / January 2008Bowling Ball Problem
Hello: On my freshman year physics exam was this problem: A bowling ball is sent down an alley with an initial velocity of v0. When it first starts to roll without slipping, it has a velocity of v1. The question: what fraction is v1 of v0? For this problem, assume a zero coefficient of friction. I didn't come anywhere close to solving this but I answered that when the ball stops slipping, its angular velocity equals its linear velocity and I wrote down the formulas. I got partial marks. I was fascinated when the professor took up the solution. The proportion of v1 to v0 depended on only one thing: a bowling ball is a sphere and has its own specific moment of inertia. In the equations, almost everything cancelled out and the final result was v1 = 5/7 v0. I've forgotten all the calculations in deriving this. Can anyone help? Thanks in advance, Denny > Hello: > [quoted text clipped - 18 lines] > Thanks in advance, > Denny The only way for the ball to stop slipping and start rolling is if the coefficient of friction is not zero. Otherwise there is no force applied to the ball once it has been released. Tom Davidson Richmond, VA > > Hello: > [quoted text clipped - 27 lines] > > - Show quoted text - Yes, this makes sense. So I guess I forget what the prof. said about the cof. Maybe it doesn't matter what the value of mu is and it cancels out in the equations as most everything did? > > Hello: > [quoted text clipped - 27 lines] > > - Show quoted text - Agreed: no friction, no conversion from sliding to rolling... However, if we take the "zero coefficient" as shorthand for "energy is conserved" then we can show: KE at start, while sliding = 0.5 M (V0)^2 KE at end, when rolling = 0.5 M (V1)^2 +0.5 L W^2 The second term is the rotational kinetic energy... L is the moment of inertia of a sphere, radius R, which is 0.4 M R^2 and W is the angular velocity. which is V1 / R if there is no slipping... Set the two energies equal, and see what happens... On Jan 28, 4:44 pm, "jmorr...@idirect.com" <jmorr...@idirect.com> wrote: > > > Hello: > [quoted text clipped - 44 lines] > > Set the two energies equal, and see what happens... In my mind a far more interesting problem in rolling/sliding friction is to answer the question "neglecting friction at the outset, (a) where on a billiard ball must the cue strike such that the subsequent motion will be pure sliding, and (b) where on a billiard ball must the cue strike such that the subsequent motion will be pure rolling?" The answer has practical application in determining how much "English" is placed on the ball, which will affect its path later as the effects of friction accumulate. Tom Davidson Richmond, VA | On Jan 28, 4:44 pm, "jmorr...@idirect.com" <jmorr...@idirect.com> | wrote: [quoted text clipped - 57 lines] | is placed on the ball, which will affect its path later as the effects | of friction accumulate. That's... err... "english", and about the only time you are allowed to use a first letter lower case "e" in Olde England using the English language. In England we have American muffins (which you can buy at Heathrow airport but nowhere else, they are just muffins) and put "side" or "top" on the snooker ball. We even have Danish pastries and Italian bread, but they are called "pastries" and "bread". > Hello: > [quoted text clipped - 4 lines] > v1. The question: what fraction is v1 of v0? For this problem, assume > a zero coefficient of friction. ? What force is going to cause it to start rolling if not friction? > I didn't come anywhere close to solving this but I answered that when > the ball stops slipping, its angular velocity equals its linear [quoted text clipped - 6 lines] > > I've forgotten all the calculations in deriving this. Can anyone help? I would approach it this way: Friction causes both a loss in linear KE and an increase in rotational KE. Suppose the forward velocity at some time is v. Then the change in linear KE is 0.5*m*(v0^2 - v^2). That much work has gone into rotational energy 0.5*I*w^2 You want to know when v and w are such that there is rolling without slipping, and this is called v=v1. Of course this occurs when w = v1/r. So we have 0.5*m*(v0^2 - v1^2) = 0.5*I*(v1/r)^2 = (1/5)*mr^2*(v1/r)^2 = (1/5)*m*v1^2 Multiply both sides by 2, divide by m: v0^2 - v1^2 = (2/5)v1^2 (7/5) v1^2 = v0^2 v1^2 = (5/7) v0^2 So I get that v1 = sqrt(5/7) v0, not 5/7 v0. - Randy >> Hello: >> [quoted text clipped - 7 lines] > ? What force is going to cause it to start rolling > if not friction? Now follow that observation to its natural conclusion. >> I didn't come anywhere close to solving this but I answered that when >> the ball stops slipping, its angular velocity equals its linear [quoted text clipped - 13 lines] > > That much work has gone into rotational energy 0.5*I*w^2 That assertion does not pass the sniff test. As you point out, we cannot dispense with friction. The rate at which work reduces linear KE is given by f * linear v. The rate at which work increases rotational KE is given by f * rotational v. f is, by your observation, non-zero. We are given that linear v != rotational v. It follows that kinetic energy is not conserved! So we're forced to a different line of attack. Let's try conservation of momentum. As the bowling ball rolls/slides down the lane, the momentary force contributes to a linear decelleration of the ball and to a proportional rotational acceleration of the ball. We should be able to use that constant of proportionality to determine the linear velocity at which the rotation rate will have sped up to match the ball's reduced linear velocity. Intuitively, that feels right, but let's back it up with an equation. We're trying to get an equation of the form: delta linear velocity = k * delta rolling velocity delta linear velocity/dt = delta (momentum/mass)/dt = delta momentum /dt / m = force/m delta rolling velocity/dt = r * delta angular velocity/dt = r * delta angular-momentum/moment-of-inertia/dt = r * torque/moment-of-inertia/dt = r^2 * force/moment-of-inertia = r^2 * force/(2/5mr^2) = force/(2/5 m) = 5*force/2m = 2.5*force/m So for every 1 m/s that the ball slows down linearly, it's rolling rate will speed up by 2.5 m/s Call its linear velocity v and its rolling velocity l v0 = initial linear velocity l0 = initial rolling velocity v1 = final linear velocity l1 = final rolling velocity v1 = v0 - x l1 = l0 + 2.5x But v1 = l1 by the problem statement. v1 = v0 - x v1 = l0 + 2.5x But l0 = 0 by the problem statement. v1 = v0 - x v1 = 2.5x Solve for x v0 - x = 2.5x v0 = 3.5x x = v0/3.5 Back-substitute giving v1 v1 = 2.5x v1 = 2.5*v0/3.5 v1 = 5/7*v0 QED > Hello: > [quoted text clipped - 18 lines] > Thanks in advance, > Denny The ball starts by slipping so that all of its kinetic energy is translational. However, when it starts rolling, this same kinetic energy is shared between translational and rotational, with the lock between them given by the relationship that w = v/r when it is rolling. The ratio of the sharing between translational and rotational energy is then determined (convince yourself) completely by the ratio of the moment of inertia to the mass (times r^2). The thing that completely determines this ratio is the geometry factor that appears as a numerical coefficient in the moment of inertia. For a sphere, this geometry factor is 2/5. For other geometries, the factor is different. This means that, just by looking at the geometry factor in the moment of inertia, you can tell which rollable object would win a race in the bowling alley (say, a hollow ball, a solid ball, a hollow cylinder, a solid cylinder, a hollow wheel, a solid wheel, all of the same mass M), and by how much. In fact, it's worthwhile to demonstrate to yourself as an exercise that, for an object whose geometry factor in the moment of inertia is, say, c, then in general v1 will be related to v0 by v1 = [1/(1+c)] v0. For c = 2/5, the thing in brackets is 5/7. PD > Hello: > [quoted text clipped - 18 lines] > Thanks in advance, > Denny The vertical forces on the ball, due to gravity and the normal force from the ground, are equal and opposite, and since they act along the same line (ideally), the torques on the ball due to these forces cancel each other out. The remaining force is a horizontal force due to friction. It acts on the ball at the point where it touches the ground. So pick a point on the ground along the trajectory of the ball, and the torque on the ball about that point will be zero. Thus the angular momentum of the ball about that point is conserved. Equating the initial angular momentum with the final angular momentum: M = mass of ball r = radius omega = angular velocity M v0 r = M v1 r + (2/5) M r^2 omega M v0 r = M v1 r + (2/5) M r^2 (v1/r) M v0 r = (7/5) M v1 r v1 = (5/7) v0 In general, finding quantities that are conserved is a very useful way of solving physics problems.  SignatureJim E. Black (domain in headers) How to filter out stupid arguments in 40tude Dialog: !markread,ignore From "Name" +"<email address>" [X] Watch/Ignore works on subthreads |
Reply to this Message |
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2009 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.