Forces and Motion Questions

Question 1. Newton realized the force that made apples fall to the ground was the same force that kept the Moon in orbit around the Earth. From this, he deduced that any object could therefore be placed in motion around our planet in a circular orbit. Using the fact the centripetal acceleration is for an object of mass m moving in a circle of radius r with constant velocity v, find an expression for the velocity of any object in a circular orbit around another massive object, otherwise known as the orbital speed.

Answer: The force due to centripetal acceleration is simply . However, we also know that the gravitational force, Fg, is also given by . As the Moon rotates around the Earth or any object orbits another, the forces the two exert on one another are equal and opposite. This is Newton’s Third Law, which keeps popping up everywhere. There’s no getting around that one.

All we have to do is make the two forces equal and then solve for v. Starting with Fg = Fc, we get: . Rearranging, and solving for the velocity, we end up with a final expression for the orbital speed of , or .

Question 2. What assumption was made when deriving the above equation? Why is it justified? What does the cancellation of little mass m imply?

In the above derivation, we gave a centripetal force for a small object of mass m as it orbits the larger celestial body of mass M in a circular path. In truth, the larger body also moves a little, preserving the center of mass of the system. However, since M is much larger than m, we can derive our orbital speed equation in peace by assuming that M doesn’t move at all in comparison to m.

Speaking of m, it tends to cancel out all the time, just like it did in our derivation of g. Little masses hate being anywhere, apparently. Looking at our equation again, , we can see that the orbital speed only depends on the larger celestial body’s mass, M, and the distance at which the smaller body is orbiting, r. This means a huge spaceship of a few tons and a poor astronaut in orbit at the same distance away from the planetary center would move at the same orbital speed.

Question 3. The above equation tells us how to find the orbital speed of a smaller object around another of larger mass. How would we use this information to come up with an expression for the orbital period T, the time to complete one revolution?

Answer: The circumference of a circle is C = 2πr. This may be used to find the distance the Moon travels in one full orbit of the Earth. Since velocity is equal to distance divided by time, . We also have very cleverly derived that , or . Substituting our expression for v into this last equation, we get . Solving for T, the orbital period of a circular orbit is . The orbital time depends only on the radius and the orbited mass, not the one orbiting.Here, the units are time.

Question 4. We know today that although planetary orbits around the Sun are elliptical, as discovered by Johannes Kepler in the 17th century’.

Assuming Earth’s orbit around the Sun to be circular, show that its period T is approximately 365 days, or one year. Take the Sun’s mass to be M = 1.99 × 1030 kg and the average mean distance from the Sun to be r = 1.51 × 1011 m.

Answer: Starting with , we plug in our average mean distance, the mass of the Sun, and G as . After some calculator magic (and what did people like Isaac Newton do before calculators?), this leads to T = 3.20 × 107 s = 370 days. Pretty good.

The five-day difference can be accounted for by the fact the Earth doesn’t really make a circular orbit around the Sun.

Question 5. What are all the forces in a force diagram for a person in an elevator as it accelerates upwards?

Answer: There are only two forces on the person: the force of gravity, which is essentially a constant anywhere near the surface of the Earth, and the normal force, which increases during the time of acceleration.

If that person were to stand on an old-fashioned dial scale, he’d see that the scale records the normal force and not the force of gravity because the weight recorded on the scale would increase as the elevator accelerates.

Question 6. Describe the net forces on the person in the elevator as it accelerates, travels at a steady rate, and then decelerates.

Answer: The steady rate equates to no net forces, because if there’s no acceleration, then there’s no net force as we learned in Newton’s laws. As the elevator accelerates upward, the net force on the person is upward, of which the magnitude is the difference between the normal force and the force of gravity. And lastly, as the elevator decelerates, the normal force lessens, so the net force is downward and the difference between the force of gravity and the now-small normal force.

If only we had this person’s mass and the acceleration rates for the elevator, we could calculate these forces precisely. Too bad, so sad.

Experimentally, we could work out the person’s mass and the acceleration rates from force measurements on a scale within the elevator.

Question 7. A person standing on a scale in an elevator records his weight as the elevator accelerates upwards as 160 pounds, his weight while the elevator travels at constant speed (or at rest) as 140 pounds, and lastly his weight while the elevator decelerates as 120 pounds. What is the acceleration of the elevator?

Answer: Our first step? Destination: SI units. There are 2.2 pounds per kilogram on planet Earth, so this-person-who-shall-remain-nameless has a mass of , which means his weight from F = ma is 63.6(9.8) = 623.6 N. Similarly, the recorded “weight” during acceleration is 712.7 N and during deceleration is 534.5 N.

As we learned in #5, this “weight” is actually the value of the normal force. Its magnitude is equal to the weight during times of constant or zero velocity. During acceleration, the net force is 712.7 N – 623.6 N = 89.1 N = ma. We learned the mass is 63.6 kg earlier, so .

For deceleration, the magnitude of the net force is the same but the sign changes, so as the elevator slows down.

Question 8. A person standing on a scale in an elevator records his weight as the elevator accelerates downwards as 120 pounds, his weight while the elevator travels at constant speed (or at rest) as 140 pounds, and lastly his weight while the elevator decelerates as 160 pounds. Why aren’t the signs of the acceleration the same as when the elevator traveled upward?

Answer: The short answer is that acceleration and force are both vectors. For the elevator to accelerate downward is to decrease the normal force on the person, not increase it like it did going the other direction. The acceleration is then negative because it’s in the negative (downward) direction.

As the elevator decelerates while traveling in the negative direction, the vector points the opposite of the direction of motion, so in the positive direction.

Question 9. Gandalf seems to defy the laws of gravity by catching a sword that was dropped 16 seconds before he falls himself, and exactly 21 seconds before he catches it. Assuming the wizard uses his amazing magical powers to accelerate faster than , at what constant acceleration must he fall to catch his sword in time to show the Balrog demon who’s the boss?

Answer: Let’s use one of our equations of motion, . If we take yi to be the top of the cliff and set that equal to 0, and set vi = 0, then we can calculate how far the sword has fallen in 21 seconds under the usual acceleration, because that’s the time associated with the distance Gandalf overcomes to catch up.

We have y = -0.5(9.8)(21)2= -2161 m. This is where the sword is some 5 seconds after Gandalf lets go, urging the fellowship to go on without him.

Gandalf reaches -2161 m with an acceleration much greater than in order to catch it in a mere 5 seconds from the time of his fall. All we need to do is use the same equation, but solve for a instead. How long we want Gandalf to take is up to us. Rearranging the equation, we write , which nearly (but not quite) 20 times as fast of an acceleration as the acceleration from gravity.

Question 10. We learned three equations of motion for one -dimensional situations with constant acceleration, such as free fall. Use the definitions of velocity and acceleration to derive them as written below.

Answer: Let’s consider an object initially, with t = 0, v = vi and ax = a. By the way, let’s agree to drop the x subscripts since we’re only considering one dimension anyway. The same object t seconds later will have the following properties: t = t, v = vf and a= a, since a is constant. Knowing that , we can write and solve for vf as vf = vi + at.

Goodie. One down, two to go.
On to the next one.

The average velocity between vf and vi is . Velocity is defined by . Equating the two means . Yep, we’re left with an algebra problem.

First we rearrange to , after which we add xi to both sides, and use the equation vf = vi + at. This leads to . We then get, with our quick minds and sharp brains, the second equation of motion, .

Both these equations include time as a variable. To get to our third equation of motion without time will take more time, no pun intended, and is a bit messier. Long story short: solve for t in one of the equations we derived and then substitute it into the other equation.

Solving for t in our first equation of motion as easier than the second, we have . Next, we plug that into and simplify away. Yes, it takes awhile.

Whew! We made it. We are algebra champs.

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