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AP FRQ, cue stick strikes a billiard ball, torque, COM, etc

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data
    this question is from 1980.
    1980M3. A billiard ball has mass M, radius R, and moment of inertia about the center of mass I=2MR2/5.
    The ball is struck by a cue stick along a horizontal line through the ball's center of mass so that the ball initially slides with velocity v0 as shown above. (There's an accompanying diagram. It just shows a cue stick hitting a ball with radius R, not much to see there.) As the ball moves across the rough billiard table (coefficient of sliding friction μk), its motion gradually changes from pure translation through rolling with slipping to rolling without slipping.
    a. Develop an expression for the linear velocity v of the center of the ball as a function of time while it is rolling with slipping.
    b. Develop an expression for the angular velocity ω of the ball as a function of time while it is rolling with slipping.
    c. Determine the time at which the ball begins to roll without slipping.
    d. When the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table. During the subsequent motion the angular momentum about point P remains constant despite the frictional force. Explain why this is so.

    2. Relevant equations
    vcm=ωr while rolling.
    I don't know, all those torque and angular momentum equations?
    Τ=Iα, etc.

    3. The attempt at a solution
    No idea where to start conceptually on this one. Why isn't it just rolling from the atart? Why would friction help it to start rolling? Could comeone just help me to get started, and then maybe I could figure it out from there?
  2. jcsd
  3. May 7, 2010 #2
    that is truely intriguing. see it in two parts. first the ball is hit by the cue and after a short period ( the period it remains in contact with the cue) it acquires an angular speed and a translational speed both having initial value zero. set up eqn of the form v= u + ft for translational and rotational motion. u will get two eqns with assumed final values.
    now come to the 2nd part. now there is no external force except friction. the key to the problem is friction in this case actually increasing the rotational motion but decreasing linear speed. now set up eqns, like in the previous case, involving linear and rotational motion. here the initial values of ang. and linear speed will be the final values of those of the first part. now do a bit of maths to eliminate or evaluate the parametars according to yr need.
    for q a) when it is rolling without slipping v = wr (it is not true for any motion which is not pure rolling).
    for q d) when it is in pure rolling the friction, being a kinetic sort, can not act as a dissipative force, as the ball has no translational motion. we neglect the rolling friction here. so the ball will role without any external torque although friction is present.
    just think over it and u will get the key.
    Last edited: May 8, 2010
  4. May 8, 2010 #3
    I have decided to be a bit elaborate this time.
    Ans. a) think of a ball sliding over a horizontal surface. The lower most point alongwith the all other points of the ball will experience a forward motion. As the only touching point is the lower most point, it will experience a frictional force opposing the motion, backward in this case. It will reduce the forward speed (decln.). The equation is simple, of the form v0 = v –Mu*g*t. this describes v as a function of t.
    Ans. B) the ball starts with a sliding motion and no rolling. as the speed of the lower most point (where friction actually telling its effect) reduces, the upper point will try to maintain its speed of vo due to inertia. So the instantaneous vertical axis of the ball through the touching point will experience a forward tilt and the rolling motion will start. This is how ‘friction starts rolling’. Now, as rolling starts the lowermost point will now have a backward motion (just imagine a ball rolling, the lower most point is the only point which have a ‘backward’ motion), such that kinetic friction in case of the rolling motion will have ‘forward’ value, as friction always works against motion. That value of friction multiplied by the torque arm, the radius here, will produce a torque in counter-clockwise direction for a ball moving to your right. That torque divided by the MI will give the angular accln. For rolling motion. Just set up the eqn as in a) (but this time the rolling counterpart and with accln and not decln.) and get w as a function of time. the ball will not start rolling from the begining as the ball has been struck at the middle horizontally. there will be no initial torque. the friction torque will come into play afterwords.
    Ans c) after u set up the previous two eqn.s put v = w.R, and t = T, the time taken to start pure rolling. V = w.R is the condition for pure rolling.
    For d) as I have said earlier sliding friction does not do work against rolling motion. So the angular momentum will remain same.
  5. May 11, 2010 #4
    WOW!! You totally made everything conceptually clear and answered all my questions! Torque and angular stuff has always been difficult for me, so thanks so much for your help! It was an interesting problem ;)
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