When you get a bit ahead in your studies, cut classes some...

When you get a bit ahead in your studies, cut classes some afternoon and visit your local pool or billiards parlor to bone up on momentum conservation. Note that, no matter how complicated the collision of balls, the momentum along the line of action of the cue ball before impact is the same as the combined momentum of all the balls along this direction after impact, and that the components of momenta perpendicular to this line of action cancel to zero after impact, the same value as before impact in this direction. You'll see both the vector nature of momentum and its conservation more clearly when rotational skidding ("English") is not imparted to the cue ball. When English is imparted by striking the cue ball off center, rotational momentum, which is also conserved, somewhat complicates analysis. But regardless of how the cue ball is struck, in the absence of external forces, both linear and rotational momentum are always conserved. Both pool and billiards offer a first-rate exhibition of momentum conservation in action.

Key Concepts

Momentum Conservation

Momentum conservation is a fundamental principle in physics stating that the total momentum in a closed system remains constant over time if no external forces act on it. This principle applies to both linear and rotational momentum and is essential for analyzing collisions and interactions in mechanics.

Vector Nature of Momentum

Momentum is a vector quantity, meaning it has both magnitude and direction. This characteristic requires that when analyzing momentum, each directional component is treated separately, especially in collision problems where forces and momentum components along different axes must balance independently.

Linear Momentum

Linear momentum is the product of an object's mass and its velocity, representing the momentum associated with its translational motion. In any collision, the sum of the linear momentum vectors along a particular direction before the impact must equal the sum after the impact, assuming no external forces act in that direction.

Rotational Momentum

Rotational momentum, or angular momentum, pertains to the momentum associated with an object's rotation about an axis. In the absence of external torques, the total rotational momentum of a system remains constant. This concept is critical when analyzing situations where objects not only translate but also rotate, complicating the overall momentum conservation analysis.

Transcript

00:01 So before doing the parts of the problem, let me remind you that for a complete rolling motion, we can write the velocity, which is the velocity of the center of mass, to be equal to the radius of the ball times the angular velocity.

00:23 And this is only valid when the motion is completely rolling motion.

00:33 But in this problem, since the ball is initially slipping as well while it rolls, therefore this condition or this relation between v0 and omega -0 for the initial motion of the ball is not valid.

00:51 Therefore, we won't be applying this relation at all.

00:57 So whenever we have to use the initial velocity, we will be using b -0.

01:03 And whenever we have to use the initial angular velocity, we will be using omega -0 without using this relation.

01:11 Now, you must see that from the free body diagram, so basically this diagram, the voh, of the ball is after it has moved away from the initial point where o is the initial point.

01:31 Now there are three forces acting on the ball as you can see.

01:36 The fictional force in which i have drawn to be opposite to the direction of motion of the ball, direction of motion of the center of mass of the ball to be specific.

01:49 And there is a normal force from the ground at the point of contrast.

01:53 Of the ball and the ground and there is weight acting from the center of the ball.

02:03 Now, fn and mg as you can see are in opposite directions and each has the same liver arm about an axis passing through the point o perpendicular to the plane of this paper.

02:21 Now because of that, there will be no torque, no net torque.

02:29 This is because now if you calculate the torque about point o, so let me just do that.

02:36 So net torque about point o will be equal to i alpha or we can write that.

02:48 Let me use the other relation.

02:49 So we know that net torque can be written as change in angular momentum over time.

02:56 Now let's find the talks about this point o, which is the initial point of the ball.

03:01 So since this frictional force passes through that point, so we have no torque due to this frictional force.

03:11 Now if you take clockwise talks to be positive, then we have fn giving a counterclockwise torque, therefore a negative sign, and the distance is r.

03:25 And we have mg giving a clockwise torque and the distances are as well.

03:32 The distance r is the perpendicular distance of either force from o.

03:40 And that is equal to dl over dt, which is the right side that we wrote.

03:46 This means that fn negative fn.

03:53 Let me just write the other way.

03:55 So this gives r times n, g minus fn to be equal to d l over d t now if you write newton's second law for equilibrium you have so there is no acceleration of the ball in the vertical direction therefore we can say that the ball is an equilibrium in the vertical direction it's moving but it's moving to the right so it's moving horizontally, but vertically there is no acceleration.

04:38 So we can write the net force in y direction to be equal to 0.

04:43 So this gives fn to be equal to mg, which are the two vertical forces.

04:51 Now, if you apply this condition over here, we'll get that this term goes away, goes to 0.

04:59 This means we can write d l over d t to be equal to zero.

05:05 And therefore, l is a constant.

05:16 Now, we are done with the first part.

05:20 So for the second part, we follow the hint, this hint, and oops, we follow the hint given in the problem and express the total angular momentum as a sum.

05:43 Of two terms, so which is angular momentum with respect to the center of mass plus angular momentum, the rotational angular momentum about the blue.

06:02 So for angle, the angular momentum for the center of mass will be mr times the velocity of the center of mass.

06:15 So we have mass times the radius times the velocity of center of mass and the angular momentum of this will be due to the rotational motion of omega so that will be equal to i times omega now the angular momentum is a constant so we equate the angular momentum at the initial position with the center of mass equal to v -0 and omega is omega -not or we can also call it to be omega c and final angular momentum is zero velocity of center of mass is zero and omega is zero as well so if we do that so let's write since angular momentum is constant so initial angular momentum is final angular momentum now we use this relation over here so we have m r vcm plus i omega initially equal to m r e cm plus i omega for the final angular momentum now let's plug the values for the initial motions so we have mr v0 plus i omega not and this is equal to so finally we have this term velocity of center of mass zero and also omega is this means that the right hand side will be zero.

08:07 So from here we can express omega -0 or omega -c to be equal to mrp -0 over i and i forgot to write that this is the moment of inertia with the center of mass...

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