P-dog's blog: boring but important: 201709

20170930

Presentation: work and energy

Tarp-lined ramp leading down to a pond? Check. All-terrain vehicle? Check. Pulling a cord attached to a raft and rider? Whee! (Video link: "Human Slingshot Slip and Slide - Vooray.")

In this presentation we will introduce a new connection between forces and motion, in terms of how forces can do work on or against an object in order to speed up or slow down its motion. This is a new approach to connecting forces and motion, compared to the previous discussion in this course of using Newton's laws to relate how forces on an object result in a net force that may or many not change its motion.

First, defining the "amount of motion" of an object in terms of its translational kinetic energy, and then expressing how a force may do work on or against the motion of an object.

Translational kinetic energy KE_tr is the energy of motion. (We'll deal with rotational kinetic energy KE_rot later on.)

Translational kinetic energy KE_tr depends on the mass m and the square of the speed v of the object, and the resulting units of kg·m²/s² are also expressed as joules. A stationary object has no translational kinetic energy, and the faster an object moves, the more translational kinetic energy it has.

Instead of finding out how much translational kinetic energy KE_tr an object has, often we are more concerned with its initial-to-final change ∆KE_tr, which is the final amount of translational kinetic energy minus the initial amount of translational kinetic energy. Notice how the common factors of (1/2) and mass m (which is presumed to be constant) are pulled out, leaving a "difference of squares" for the final and initial speeds in the parenthesis.

In order to change the translational kinetic energy of an object (speeding it up or slowing it down), a force (such as that exerted by the horses) must do a certain amount of work W either on, or against the motion of the object (here the loaded sledge the horses are pulling).

Work is accomplished by exerting a force on an object, but in such a way that the object moves through a displacement s (note the unusual use of "s" for a generic ∆x, ∆y, or other direction displacement!), and the "tail-to-tail" angle θ between the force and displacement vectors (when their "bases" are drawn touching together) is anything besides 90°. The units of work are given in N·m, or yet again, joules, so keep in mind that work can be done on or against any mechanical energy form or forms.

Second, let's now explicitly make the connection between the work done by a force acting on or against the motion of an object, and the resulting changes in translational kinetic energy of the object.

This is an incomplete form of the total energy conservation equation we will utilize later, but this "work-energy theorem" shows how the transfer of energy (as work is done by a force acting on or against the motion of the object) causes a corresponding change in the translational kinetic energy of the object.

If the force does work on the object (by being exerted along the direction of its motion), then the work will have a positive sign, and the translational kinetic energy of the object will increase, making the sign of the ∆KE_tr term positive. Note for this case how the left- and right-hand sides of this equation must have the same positive sign.

If the force does work against the object (by being exerted opposite to the direction of its motion), then the work will have a negative sign, and the translational kinetic energy of the object will decrease, making the sign of the ∆KE_tr term negative. Note for this case how the left- and right-hand sides of this equation must have the same negative sign.

Squirrel. Catapult! Squirrel catapult! (Note the person behind the sliding glass door, cutting the release cord with a scissors to launch the squirrel.) (Video link: "squirrelcatapult.gif.")

To keep things simple, let's consider a strictly horizontal version of this contraption, which would make the work done by any vertical forces (such as the weight force of Earth on the squirrel) zero, as the angle between these vertical forces and the horizontal direction of motion is 90°.

The bungee cord (and basket) exerts a force on the squirrel directed to the right, along the direction of motion, so the bungee cord does work on the squirrel. As a result, the squirrel picks up speed (starting from rest), and since translational kinetic energy depends on the square of the speed, since speed increases, then the squirrel's translational kinetic energy increases.)

In the work-energy theorem equation:

W = ∆KE_tr,

the work will have a positive sign (as work was done on the squirrel by the bungee cords), causing the squirrel's translational kinetic energy to increase (and also have a positive sign), so the +/– signs of both the left-hand side and the right-hand side of the equation are consistent with each other:

(+) = (+).

(If you were to calculate the numerical values for the work done (in J) and the resulting numerical value for the increase in translational kinetic energy (in J), then they would have to be equal to each other (along with having the same sign).

For the catapulted squirrel, the bungee cord force does work __________ the squirrel, which __________ the squirrel's translational kinetic energy.)
(A) on; increases.
(B) against; decreases.
(C) (Unsure/lost/guessing/help!)

Note as this car stops, the brakes glow from the heat generated! (Video link: "McLaren SLR review - Top Gear - BBC.")

For the braking car, the brakes do work __________ the car, which __________ the car's translational kinetic energy.
(A) on; increases.
(B) against; decreases.
(C) (Unsure/lost/guessing/help!)

Don't blink, or you'll miss Mrs. P-dog in action! (Video link: "110530-1230869-excerpt.")

For Mrs. P-dog being catapulted upwards, the bungee cords do work __________ Mrs. P-dog, while the weight force does work __________ Mrs. P-dog.
(A) on; on.
(B) on; against.
(C) against; on.
(D) against; against.
(E) (Unsure/lost/guessing/help!)

For Mrs. P-dog's translational kinetic energy to be increased while being catapulted upwards, the amount of work from the bungee cords must be __________ the amount of work from the weight force.
(A) less than.
(B) the same as.
(C) greater than.
(D) (Not enough information is given.)
(E) (Unsure/lost/guessing/help!)

20170929

Physics presentation: impulse and momentum

Whuuuuuuut. (Video link: "bowling strike with a ping pong ball.")

In this presentation we will introduce another new connection between forces and motion, in terms of how the net force can exert an impulse on an object in order to change its momentum. This is yet another new approach to connecting forces and motion, compared to the previous discussion in this course of using Newton's laws to relate how forces on an object result in a net force that may or many not change its motion, and analyzing how forces can do work on or against an object in order to speed up or slow down its motion.

First, defining the momentum of an object, and then expressing how the net force can exert an impulse on this object.

The introduction slide showing a ping-pong ball knocking over all ten bowling pins should seem very strange to you, as the mass of the ping-pong ball is too small to effectively bowl a strike, even if it were traveling with a supersonic speed. In order to fully account for the "knocking-over" strength of a moving object, then, we must include mass as well as its speed (and direction) to define its momentum p.

Momentum p is a vector quantity (so don't forget to draw an arrow over it) whose magnitude depends both on the mass and speed of the object, with the combined units of both mass and speed (kg·m/s).

We also need to introduce the concept of impulse J, which is the product of the net force acting on an object and the duration of time that the net force acted on this object (whether for a brief instant, or for a prolonged period). (Video link: "Teaching Tee Ball Hitting.")

Impulse has the combined units of both force and time (N·s). Here we use the somewhat obscure (but totally legit) "J" symbol for impulse, remembering to draw an arrow over it (as it is a vector quantity). (It turns out that "I" is already reserved for rotational inertia in the next chapter.)

Second, let's now explicitly make the connection between the impulse acting on an object, and the resulting change in the momentum of the object.

This "impulse-momentum theorem" emphasizes how the impulse (exerted by the net force acting over a specific duration of time) causes a corresponding initial-to-final change in the momentum of the object. And vice versa, where the initial-to-final change in the momentum of an object is caused by the impulse on the object.

Let's apply these concepts to several objects that undergo changes in momentum, with an emphasis on the directions (+/– signs) of these quantities, and how they all must be consistent with each other, starting with a golf ball initially at rest, and then has a speed of 97 m/s after being hit by a golf club. (Video link: "The Moment of Impact. An Inside Look at Titleist Golf Ball R&D.")

This golf ball is initially at rest, so its initial momentum p₀ (mass times its initial velocity) is 0.

We'll define the horizontal direction to be positive to the right (and negative to the left). After it is hit by the golf club, its final momentum (mass times its final velocity) p_f points to the right (and will be a positive quantity).

The initial-to-final change in momentum ∆p of the golf ball is given by:

∆p = p_f – p₀,

and since get a positive quantity minus zero, then ∆p must be positive (thus pointing to the right).

Since the impulse "J" on the golf ball causes this initial-to-final change in momentum:

"J" = ∆p,

the impulse must also have the same direction as ∆p, and so it must also point to the right. (Also since the impulse "J" is the net force ΣF on the golf ball times the contact time ∆t, the net force of the golf club on the golf ball is also directed to the right.)

Now let's have you look at the directions involved in the impulse-momentum theorem for this catapult-launched F/A-18E-F Super Hornet, initially at rest, and then has a speed of 74 m/s after being it is catapulted. (Video link: "F/A-18E-F Super Hornet Catapult Launches.")

Super Hornet's initial momentum p₀ direction? (left (–), none (0), or right (+)?)
Super Hornet's final momentum p_f direction?
Direction of Super Hornet's initial-to-final change in momentum ∆p?
Direction of catapult's impulse "J" on the Super Hornet?

Finally, consider the directions involved in the impulse-momentum theorem for this Ford Ranger, hitting a crash barrier with a speed of 11.0 m/s, and then rebounding off the crash barrier with a speed of 2.2 m/s. (Video link: "Crash Test Ford Ranger 2012....")

Ford Ranger's initial momentum p₀ direction? (left (–), none (0), or right (+)?)
Ford Ranger's final momentum p_f direction?
Direction of Ford Ranger's initial-to-final change in momentum ∆p?
(Hint: watch your signs!)
Direction of crash barrier's impulse "J" on the Ford Ranger?