Student Learning Objectives
- To understand the basic ideas of circular motion.
- To understand the concept of acceleration in circular motion
- To solve quantitative kinematics problems involving circular motion and to interpret the results.
- To apply Newton's laws in the context of circular motion
Objectives
After studying the material of this chapter, you should be able to:
1. Calculate the centripetal acceleration of a point mass in uniform circular motion given the radius of the circle and either the linear speed or the period of the motion.
2. Identify the force that is the cause of the centripetal acceleration and determine the direction of the acceleration vector.
3. Use Newton's laws of motion and the concept of centripetal acceleration to solve word problems.
4. Distinguish between centripetal acceleration and tangential acceleration.
5. State the relationship between the period of the motion and the frequency of rotation and express this relationship using a mathematical equation.
6. Write the equation for Newton's universal law of gravitation and explain the meaning of each symbol in the equation.
7. Determine the magnitude and direction of the gravitational field strength (g) at a distance r from a body of mass m.
8. Use Newton's second law of motion, the universal law of gravitation, and the concept of centripetal acceleration to solve problems involving the orbital motion of satellites.
9. Explain the "apparent" weightlessness of an astronaut in orbit.
*11. Use Kepler's third law to solve word problems involving planetary motion.
12. Use Newton's second law of motion, the universal law of gravitation, and the concept of centripetal acceleration to derive Kepler's third law.
*13. Solve word problems related to Kepler's third law.
14. Identify the four forces that exist in nature.
PhyzGuide
HW
Pen Cast Example Page
what is the equation for centripetal force?
ReplyDeleteF(centripetal) = mass x velocity^2 / radius
DeleteThis may help
Deletehttp://hyperphysics.phy-astr.gsu.edu/hbase/cf.html
How do you go about solving the coefficient of static friction in number 5?
ReplyDeletei missed the due date for this assignment, is it next class?
ReplyDeleteYes pretty sure
Deletecan anyone explain 4 and 5?
ReplyDeleteFor 1-3 i got:
1) A. .0692 newtons
B. .1384 Newtons (tension was doubled)
C. .0346 Newtons (tension was halved
2) Acceleration doubles when radius is halved
3) Tension is 4 times as much when velocity is doubled
4)??
5)??
please completely ignore number 1 i made so many dumb mistakes
DeleteMarty, I got about 2.3 meters per second for the velocity in number 4 and number 5 I'm having trouble..
DeleteColin
i explained number 5 as best i could down below and i also got 2.3 meters/second for number 4
DeleteIn number 5, does the length of the second hand matter in the final answer? It's throwing me off...
ReplyDeleteDoes anyone know what 9 is about im confused
ReplyDeletethe homework is only section B 1-5 Haley
DeleteWhat's the equation to find tension in #1?
ReplyDeleteThis comment has been removed by the author.
Deletei think it is just F=ma
Deletealso a=v^2/r and v=2*pi*r/T (i think)
Deletecan anyone explain to me how to do #3?
ReplyDeletei mean #4
Deleteme too!! I just do not understand what the picture looks like that they are describing, so it is not making sense in my head
DeleteI'm having trouble getting any of these really. For 1a. I got 342 N B I got that it doubles and C It came out to be 1/8 of the original. How do we get the tension of the string? Do we use mv^2/r? For number 2 I got the the Acceleration was halved as well, for 3. being that V must be squared in the equation for Fc, the doubled V causes 4x the force to be amassed. For 4. I somehow found the V to be .000031 m/s and I am lost on number 5...what is the coefficient of static friction?
ReplyDeleteyour new 1 is correct, i had a stupid few mistakes above
Deletefor 2: a=v^2/r so if you halve r a would be double, not halved? by saying the period stays the same it means the velocity is also the same. 2/1 is half of 2/.5 if you wanted to plug numbers in, for example.
We agree on number 3.
i do not understand 4 or 5 at all so i'm going in at lunch
i agree with marty and the one fourth original for c is also what i got. Jon, if u want for two u can just plug in simple numbers and see the results
DeleteI just noticed I had to convert g to kg so I now have .325N for 1A, .68 for B and that it is a quarter the original for C.
ReplyDeleteThis may be obvious, but are force and tension the same thing?
ReplyDeleteI'm pretty sure they are
Deletei thought in a circle there is no velocity.
ReplyDeleteif it is moving there has to be a velocity, that is its speed, so there is
DeleteThe idea is that the average velocity is zero in the normal respect because its position reverts to the original but because it is moving rotationally it has to have a velocity.
DeleteThere is velocity, circular motion, the formula is alittle different though: V=2*Pie*r/T(period)
DeleteColin
i also calculated from centimeters to meters for number 1
ReplyDeletei completely forgot about that... my answers were two places off & not knowing why was killing me.
Deletethanks!
i think for number 4 you set the centripetal force equal to the force of gravity on the hanging mass then solve for V. So:
ReplyDeleteFc = mv^2/r = Fg = ma = 7.9(9.8) = 77.4
77.4=(2.8(x)^2)/.19
Solving for x gives you a velocity of 2.29 meters/second
For 5 i came across the answer online and i think you have to assume that the pebble's mass is 1kg. you would set uFnormal = Fc = mv^2/r
ReplyDeletesolve for Velocity using 2pi(r)/period = 2X3.13X.01/60 (seconds it takes for the second hand to spin one full time) = .00105 m/s
plug it in: 1 X .00105^2/.01 = 1.1X10^-4
Now set that equal to uFnormal and we know Fnormal has to be the same as the force of gravity so Fnormal = ma = 1X10. Solve for u
1.1X10-4 = 10u
1.1X10^-5
Can anyone tell me the equation you used for # 1 (the centripetal I believe)?
ReplyDelete(2πr)/T
Deletealso a=(v^2)/r and F=ma
is centripetal force the same as Fnet (but when it is going in a circle) ?
ReplyDeleteyeah-- http://upload.wikimedia.org/math/d/f/8/df86712e000fe347516b8f39b9490815.png
Deletefor #1, how would the radius be found?
ReplyDeletethey give it to you. 93. you're swinging the string in a circle so it's only half the diameter.
Deletehttp://www.ithaca.edu/faculty/mcsullivan/PH117/CTs/CT-AngularMomentum-IMAGES/AngMomentumBallWithForce.gif
(so just imagine the red dot is the stopper and the black line is the string. the dotted line would be the... orbit? i guess? it would be the circle made by the swing)
I cant figure out nnumbers 1, 4, &5. is there a relationship between tension and Fc
ReplyDeletethe tension would be equal to the centripetal force
DeleteFor the mass of the sun problem, based on the chart, I rounded but I got about 4*10^30 kilograms when I know it should be more like 2*10^30. Can someone give me the components needed to find the mass based on their chart, because I feel like mine or high.
ReplyDeleteColin