OPENQA24

#$&*

course PHY 121

4.18.11 at 11am

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

v^2/r

=(3m/sec)^2/5

=1.8m/sec^2

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 Newton's Second Law, must therefore be

Fcent = 12 kg * 1.8 meters/second ^ 2.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

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

Newtons?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

CA = v^2/r

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 Newton breaking force, then we have

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):

I was thinking that r would be half of the 0.7m?

------------------------------------------------

Self-critique rating:

@& The length of the sting is the radius, which is half the diameter of the circle.

However we use the radius when calculating centripetal force.*@

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Accel is 1.8m/sec^2

Velocity to break it must be 18.7m/sec

18.7m/sec / 1.8m/sec^2 = 10.4sec

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):

I'm not really sure what I solved and what it would even represent, if anything.

@& You divided a velocity by an acceleration, which is not generally meaningful.

If you divide a change in velocity by an acceleration, you get the time required for that velocity change. Not a bad thought, but it doesn't help with the question here. *@

------------------------------------------------

Self-critique rating:

*********************************************

Question: `q004. Explain in terms of basic intuition why a force is required to keep a

mass traveling any circular path.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Less force would be required to keep an object on a linear path as opposed to a circular path. Some force would have to interfere to keep motion circular.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#Good responses. See my notes and let me know if you have questions. &#