His test is tomorrow and I am stuck in helping him. (High School senior.)
2000 kg car, speed 30 m/s around a horizontal circular track, radius 100m. What coefficient of friction is necessary to allow the car to follow the curve of the track? The rest of the question we got, including the Normal force is 19600 N.
"Newton’s cannon is placed on a platform 19.6 m high at the equator of the earth (Re= 6.38 x 10 to the 6), G=6.67 x 10 to the -11 Nm2/kg2, and Me= 5.98 x 10 to the 24 kg) and is used to fire a 10.0 kg cannonball into orbit. Ignore air resistance.
a) how fast horizontally would you need to fire the cannonball so it stays at a height of 19.6 m above the earth’s surface?
b) What would be the period of this orbit?
c) What is the mechanical energy of the cannonball at this height?"
We have the right answers given in the study guide* but can’t figure out how to do the problems.
*1) 0.918
2a) 7900 m/s b) 5070 s c) 3.13 x 10 to the 8 J
Thanks for talking me through this so I can explain it to him!
depends a little on what exactly he’s expected to know. One way is to equate the force of gravity with the “fictitious” centripetal force
Force of gravity is G * Me * 10 (mass of cannonball) / (Re + 19.6)^2
Centripetal force is 10 (ditto) * v^2 / (Re + 19.6)
equate and solve for v squared then v
b) Orbital period is (distance traveled) / v
distance traveled is 2 * pi * (Re + 19.6)
c) Based on the numerical answer give, I assume mechanical energy means the kinetic energy of the cannonball (not it’s potential energy He should know the difference between these two) Kinetic energy is 0.5 * 10 * v^2 where v is your first answer and 10 is the cannonball mass.
The teacher probably expects him to know about significant digits. For example adding 19.6 meters to the radius of the earth makes no difference since the radius of the earth isn’t given to that accuracy.
Nitpick: Centripetal force is not in any sense fictitious. It’s just any force, of any nature, that happens to be causing some object to change its direction of motion. It points towards the center of curvature of the trajectory. Centrifugal force, by contrast, is the label given to the “fictitious” force which, in a rotating reference frame, pulls objects out away from the center. Despite the “fictitious” label, it’s still perfectly fine to use it, as long as you’re in the rotating reference frame.
The way my kid’s textbook puts it though is to remember that there actually is no force pushing away from the center: there is inertia that would have you continue at a tangent to the curve (i.e. in a straight line forward) if not for the centripetal force being applied that kept the object traveling in the curve.
Just woke him up early to review these two with him. Thanks and “Ah!” from him as well. I’m sure he’ll do fine on the test. His response to my “Damn Newton was smart” line however was “Yeah, but he was an asshole.”
Number 3 son … his younger sister is 6 years his junior and we’ll see if I have to relearn physics again in a few years! (I do hope some money for her college tuition appears out of the quantum foam.)
Tell your son to writ: “F = ma” at the top of his test paper, and whenever he gets stuck, look at that equation. I only half-kid about that. You’d be surprised how many times that works!
Now the task is helping him deal with getting the dreaded waitlist response from three schools over the past 24 hours … he’s already been accepted to three, including one that I am pretty sure he’d be very happy to go to and happy at, and he has his very top choices yet to give their responses. I am going to try to get him to understand that it’s a crapshoot and one school’s answer is not very predictive of what the other ones he has yet to hear from will say and that even if he is rejected from all the other programs he applied to he’s got a great choice for him in the bag; his ego will have to deal.
