Recall that in collisions, momentum is always conserved. Energy is conserved is the collision is elastic, but not conserved if the collision is inelastic. All the examples I showed last week had initial and final velocities oriented along one axis. But clearly this is not required, and the initial and final velocities can be at angles to each other. (As long as there are only 2 objects involved in the collision, it can be shown that all the velocities must lie in a plane.)
In glancing collisions, the initial and final velocities can be at angles to each other. In this case, the x and y components of momentum are separately conserved. The additional equation (conservation of y momentum) compensates for the additional angle variables.
The following example should be added to your list of homework problems.
A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30 degrees with respect to the original line of motion. (a) Find the velocity (magnitude and direction) of the second ball after the collision. (b) Was this an inesastic collision or an elastic collision?
Call the initial velocity vi = 5.00 m/s, and the final speeds v1 = 4.33 m/s and v2. Both balls have the same mass, m. Call the unknown angle q.
(Drawing) For part a, apply conservation of momentum. Call the incoming direction of the first ball the x direction. Then the inital momentum is mvi in the x direction, the initial y component is zero. After the collision, the x component of momentum is mv1cos30 + mv2cosq. Cancel the common m factor and set final and inital equal to get v2cosq = 5.00 m/s - 4.33cos30 = 1.25 m/s.
After the collision, the y component of momentum is mv1sin30 - mv2sinq. Cancel the common mass factor and set final equal to inital to get v2sinq = 4.33sin30 = 2.16 m/s.
This means that v2 = Sqrt(1.252 + 2.162) = 2.50 m/s, and q = Asin(2.16/2.50) = 60 degrees opposite to the first ball.
For part (b), let's just calculate the initial and final kinetic energies, since we know the speeds. Notice that the mass, although unknown, will be a common factor to all the kinetic energy terms, so I will divide it out. KEi/m = (5.00 m/s)2/2 = 12.5 J/kg. KEf/m = ((4.33 m/s)2 + (2.50 m/s)2)/2 = 12.5 J/kg. Since the initial and final kinetic energies are equal, the collision is elastic.
In the equations we will be dealing with in this and the next chapter, angles are measured in radians instead of degrees. By using radians, the equations are substantially simplified. Radians are defined such that 2p radians = 360 degrees.
An easy way to visualize what this means is to look at a circle of radius 1 (Drawing). An angle of 360 degrees makes one complete turn around the circle, covering the entire circumfrence of 2pr = 2p. The length of the perimeter on the r=1 circle equals the angle in radians. If we use an angle of less than 360 degrees, then we sweep out less than the full circumfrence. For instance, an angle of 180 degrees will sweep around half the perimeter of the circle, or 2p/2 = p. And for an angle qd, the fraction of the perimeter it will sweep around is qd/360, or a length of 2pqd/360 = pqd/180.
This gives us the transformation from degrees to radians:
qr = pqd/180
where qr is the angle in radians, and qd is the angle in degrees.
When discussing linear motion, we began by defining displacement. We will now define an analogous displacement for circular motion. Imagine a point P at a radius r from the center of a circle, and make a reference line that passes through P. If the circle rotates by an angle qr, (the reference line doesn't move), the point P moves an arc length s = (qr/2p) 2pr = qr/r. That is the fraction of the full circumfrence intercepted by the angle.
If P moves from angle q1 to q2 in a time Dt, then it moves with an average angular speed wavg = (q2 - q1) / (t2 - t1) = Dq / Dt. As was the case with linear velocity, taking the limit as Dt goes to zero gives the instantaneous angular speed. The units of angular speed are radians per second, rad/s. NOTE: Radians are not a dimension in the same sense as meters or seconds, and are often omitted from an answer. However, keeping radians in place is often helpful to see that things cancel appropriately.
The earth makes 1 revolution in 24 hours. Therefore w = (1 rev/24 hr)(2p rad/rev)(1 hr/3600 s) = 7.27x10-5 rad/s.
Recall that in chapter 2, after defining velocity we defined acceleration. We'll do the same thing now for the case of circular motion. We define angular acceleration as the change in angular speed divided by time, aavg = (w2 - w1)/(t2 - t1) = Dw/Dt. The greek letter a is used for angular acceleration. The units of angular acceleration are rad/s2. Instantaneous angular acceleration is obtained by taking the limit as Dt goes to zero.
Note that when a rigid object is rotating, every point on the object has the same angular speed and the same angular acceleration. The linear speed and accelerations change, depending on the radius from the axis of rotation, but the angular quantities are the same everywhere.
Note the similarities between the equations for linear velocity and acceleration, and angular speed and acceleration.
| Rotational Motion About a Fixed Axis with a Constant | Linear Motion with "a" Constant |
| w = w0 + at | v = v0 + at |
| q = w0t + (1/2)at2 | x = v0t + (1/2)at2 |
| w2 = w02 + 2aq | v2 = v02 + 2ax |
By measuring angles in radians, and using angular speed and acceleration, the equations for rotational motion are very similar to the familiar equations for linear motion. Therefore, techniques similar to those you have used for solving linear motion problems can be used to solve rotational motion problems. Let's work an example to demonstrate.
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At this point the person doing the laundry opens the lid, and a safety switch stops the spinning. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 20.0 s interval? Assume constant angular acceleration.
NOTE: In solving this problem we will determine the angle through which the tub rotates, and this angle will be larger than 2p. An angle greater than 2p simply means that the object rotated by more than one revolution. For instance, if the object makes 3 revolutions then it rotates through an angle of 3x2p = 6p.
First, convert the angular speed to rad/s: w = (5.0 rev/s)(2p rad/rev) = 10p rad/s = 31.4 rad/s. Next, we need the angular accelerations: ai = w / t = (31.4 rad/s)/(8.0 s) = 3.9 rad/s2, and af = -w / t = (-31.4 rad/s)/(12.0 s) = -2.6 rad/s2.
Now we calculate the angular displacements: q1 = w0t + (1/2)a1t2 = (1/2)(3.9 rad/s2)(8.0 s)2 = 125 rad, and q2 = (31.4 rad/s)(12.0 s) + (1/2)(-2.6 rad/s2)(12.0 s)2 = 190 rad. The total angular displacement is qtot = q1 + q2 = 125 + 190 rad = 315 rad. Or, in terms of revolutions, qtot = (315 rad)(1 rev/2p rad) = 50 rev.
Consider a rotating object with a point P a radius r from the axis of rotation. When the object makes an angular displacement Dq, then from the definition the radian, we know that the point P moves along the arc of a circle or radius r, and the arc length is Ds = rDq. We know that Dq/Dt = w, and Ds / Dt = v, in the limit that Dt goes to zero.
Upon dividing the previous equation on both sides by Dt, we get Ds/Dt = r Dq/Dt, or v = rw. This linear speed, v, is the instantaneous speed of point P. The direction of v is tangent to the circular path, and therefore this speed is often referred to as the tangential speed, or vt = rw.
While every point on the object has the same angular speed, w, only points at the same radius from the axis have the same tangential speed, vt.
Here is my first real chance to point out that radians are not a "real" unit, in the sense of meters or seconds. Let's dimensionally analyse the expression vt = rw. [vt] = L/T. [rw] = L angle/T. The only way these two can be the same is if angle is not a unit. This is true if (and only if) the angle is measured in radians, since then Dq = Ds/r, which has dimensions of L/L = no unit. If the angle is measured in degrees, these equations do not work. Remember to convert angles from degrees to radians!