Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I am having some trouble with a few of the test questions for my upcoming tests and was hoping you could clarify a few things:
1)A Ferris wheel with radius 24 meters is moving fast enough that at the top of its arc a 150-lb person at the rim of the wheel has an apparent weight of only 63 pounds.
• How fast does a point on the rim of the wheel move?
• What will be the apparent weight of the same person at the bottom of the arc?
• What will be the apparent weight when the person is at the same height as the center of the wheel?
For this questions I understand that observed weight will be less at the top and more on the bottom due to centripital forces combination with normal force. I do not understand how to solve it however. Since observed weight is given in lbs, should i just convert this to Newtons and then solve for solve for velocity using F=v^2/r(m). If so would I use 150lbs converted to kg for mass and what would i use for force?
Then how would i get observed weight on the bottom?
The person's apparent weight is the force exerted on the person by the seat.
The net force is the sum of the force of the seat and the force of gravity (i.e., the person's weight).
The net force is equal to the centripetal force -m v^2 / r.
So we could write
F_seat + F_gravity = -m v^2 / r, so that
-v^2 / r = F_seat / m + F_gravity / m.
You could of course convert the forces to metric units and solve for v, and you should probably do so.
A more elegant solution is to note that F_gravity / m = -m g / m = -g, the negative of the acceleration of gravity (we implicitly have assumed that the upward direction is positive, which we declare by noting that we have done so).
The magnitude of F_seat is 63/150 the magnitude of the weight, and is directed upward. Thus
F_seat / m = 63/150 * m g / m = 63/150 g.
Thus
-v^2 / r = 63/150 g - g
so that
v^2 = r * (87/150) * 9.8 m/s^2 and
v = sqrt( 24 m * 87/150 * 9.8 m/s^2) = 12 m/s, approx.
2)A gun fires a bullet of mass 14 grams out of a barrel 23 cm long, from which it exits at 304 m/s. Assuming that the bullet accelerates uniformly from rest along the length of the barrel
• How long does it take the bullet to exit the barrel after the powder ignites?
• Using the Impulse-Momentum Theorem determine the average force exerted on the bullet as it accelerates along the length of the barrel.
• Using Newton's Second Law determine the average force exerted on the bullet as it accelerates along the length of the barrel.
I understand how to use the impulse momentum theorum but not here. I am confused as how to apply it. Could you help me out with this one?
Find the average velocity of the bullet in the barrel (assume uniform acceleration, which is a little unrealistic but anything else would be beyond the scope of this course).
Then you can find how long it takes to exit.
Its momentum changes from 0 to its exit momentum in the time interval you found. So you can solve F_ave `dt = `d(m v) for F_ave.
Of course you can (and should) also find the acceleration of the bullet and multiply its acceleration by its mass. You should get the same result.
See my notes, and if you aren't sure of yourself feel free to insert your solutions into a copy of this document and submit it.
Phy 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Professor, could you please send test 2 for phy 201 and test 3 for phy 202 to Cindy Hill at the Northern VA comm college testing center. I will be taking the tests this thursday. Thanks.
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I've sent the tests, and in addition a follow-up note. You might want to double-check to be sure they have been received and that they will be available on your intended schedule.