If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
024. Centripetal Acceleration
Question: `q001. Note that this assignment contains 4 questions.
. Note that this assignment contains 4 questions.
When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration.
If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?
Your solution:
Confidence rating:
Given Solution:
The centripetal acceleration of the object is v^2 / r = (3
meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2.
The centripetal force, by
Self-critique (if necessary):
Self-critique rating:
Question:
`q002. How fast must a 50 g mass at the end of a
string of length 70 cm be spun in a circular path in order to break the string,
which has a breaking strength of 25
Your solution:
Confidence rating:
Given Solution:
The centripetal acceleration as speed v will be v^2 / r,
where r = 70 cm = .7 meters. The
centripetal force will therefore be
m v^2 / r, where m is the 50 g = .05 kg mass.
If F stands for the 25
m v^2 / r = F, which we solve for v to obtain
v = `sqrt(F * r / m). Substituting the given values we obtain
v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.
Self-critique (if necessary):
Self-critique rating:
Question: `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?
Your solution:
Confidence rating:
Given Solution:
The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.
STUDENT COMMENT:
I read through the solution but still
wouldn't be able to solve this.
INSTRUCTOR RESPONSE
The question comes down to this:
At 18.7 m/s (the result found in the preceding), how many times will the mass travel around a circle of radius .7 meters in 1 second?
The circumference of the circle is about 4.4 meters, so at 18.7 m/s the object will go around the circle a little over 4 times in 1 second.
Self-critique (if necessary):
Self-critique rating:
Question:
`q004. Explain in terms of basic intuition why a
force is required to keep a mass traveling any circular path.
Your solution:
Confidence rating:
Given Solution:
We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed.
If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle.
In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.
STUDENT RESPONSE (good intuition but statement isn't quite right)
Something has to keep the momentum going
for anything in a circular path to continue. Otherwise, it will fly off in a
vector.
INSTRUCTOR CRITIQUE
Nothing is required to keep something
moving in a straight line; in the absence of a force it will maintain its
momentum, in the same direction as the original.
The force is required to cause the object to deviation from its 'natural'
straight-line motion.
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
Question: `q004. The centripetal force on a 0.2 kg mass, as it travels around a circle of radius 80 cm, is 12 Newtons. How fast is it moving?
Your solution:
Confidence rating:
Self-critique (if necessary):
Self-critique rating: