Test 2 Review

course Phy 201

I am not going to do good on this test, I can do all the work in the queries but when it comes to the test they seem extremely different. I am having extreme difficulty with any orbit questions.

Tr}f}׾~Veɘassignment #003

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Pxĕxvci

Liberal Arts Mathematics I

08-08-2008

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11:28:02

GOOD STUDENT ANSWER: We calculate the potential energy of the system when the rubber band is fully stretched, and compare with the F `ds total as the rail slides across the floor to see if all the potential energy was dissipated against friction.

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11:31:51

Problem Number 2

An Atwood machine consists of 34 paper clips, each of mass .4 grams, suspended from each side of a light pulley.

If 4 clips are transferred from one side to the other, what will be the total gravitational force on the system?

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9.8m/s^2 * .0152 = .15N

9.8m/s^2 * .012 = .12N

Total of .27N

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11:33:07

GOOD STUDENT SOLUTION: Well Im assuming that there are 34 paper clips on each side. So;

34 * .0004 kg= .0136 kg on one side when even

4 * .0004 kg = .0016 kg less on the side that has 4 removed

so that gives you .0152 kg on one side and .012 kg on the other

now to get the gravitational force on the system you multiply both by 9.8 m/sec^2 and that will give you the force acting on each one, which are: .14896 N and .1176 N respectively.

Now to get the net gravitational force just subtract and that will give you a net gravitational force of

GravFnet = .03136 N

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11:35:38

If the frictional force exerted by the pulley is .05 times the total weight of the system, then what is the net accelerating force?

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.05*.03kg = .0147N

.03N - .0147N = .0153N

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11:37:52

well take the weights given above and combine them, then multiply by the .05, that will give us:

[(.0136 kg + .0016 kg) * 9.8 m/sec^2] * .05= .007448 N for the frictional force

now subtract GravFnet Ff = Fnet, or

0.03136 N - .007448 N = .023912 N net force.

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11:38:28

Divide net force by the total mass of the system to get the acceleration of the system

.023912 N / .0152 kg = 1.573 m/sec^2

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11:48:54

STUDENT ATTEMPT: Alright Im not very good at this YET, so I could use some help!!!!!!!!

I know what is given in the problem. From this I can figure: angular velocity = 2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and

because 1m* omega = linear velocity is 20. 944 m/sec

AND THE REST I GET LOST ON, IM HAVING PROBLEMS WITH ALL THESE RELATIONSHIPS!!!!!!!!!!!!!!!!

INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing you're forgetting here.

This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal force would therefore be mass_ball * 440 m/s^2. **

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Ok, so what is the Tension???

If the ball is being held in its circular path by a light string, then the tension in the string is the centripetal force on the ball.

To get the centripetal force you multiply the mass by the centripetal acceleration.

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X{ű֫~O叶

assignment #003

Pxĕxvci

Liberal Arts Mathematics I

08-08-2008

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12:00:33

STUDENT ATTEMPT: Alright Im not very good at this YET, so I could use some help!!!!!!!!

I know what is given in the problem. From this I can figure: angular velocity = 2pi*(f) where f = 3.33333 rev/sec so angular velocity = 20. 94359102 sec^-1 and

because 1m* omega = linear velocity is 20. 944 m/sec

AND THE REST I GET LOST ON, IM HAVING PROBLEMS WITH ALL THESE RELATIONSHIPS!!!!!!!!!!!!!!!!

INSTRUCTOR NOTE: ** Centripetal acceleration is v^2 / r. That's the key thing you're forgetting here.

This gives you aCent = (20.94 m/s)^2 / (1 m)^2 = 440 m/s^2. The centripetal force would therefore be mass_ball * 440 m/s^2. **

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I can't learn anything from this because I can't understand it. I don't know what is a typo, and what is not a typo. There are too many typo's on these assignments and I don't know what I'm doing. Some of them I can catch, but it still leaves me thinking ""Did I do it wrong?"".

If centripetal acceleration is v^2/r, then why do you have (20.94m/s)^2 / (1m)^2??? Shouldn't the denominator just be 1m, not squared.

Is centripetal force the tension?

The square on the 1 meter is not correct. The centripetal acceleration is correct.

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12:07:27

The net force on the ball is always equal to the centripetal force Fcent = mass_ball * 440 m/s^2. Gravity exerts a force equal to mass_ball * 9.8 m/s^2. If the ball is at the top of its arc the net force consists of the tension plus the gravitational force, both of which act in the same direction, which is downward. So using downward as the positive direction we have

tension + mass_ball * 9.8 m/s^2 = mass_ball * 440 m/s^2.

We find that

tension = mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = mass_ball * 430.2 m/s^2.

You could substitute a reasonable mass (e.g., .4 kg) for the ball and obtain all quantities in Newtons.

Note that these calculations are not in fact accurate to 4 significant figures. The numbers shown here are intended to demonstrate how the tension differs from the centripetal force.

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12:10:45

What if the ball is at the bottom of its arc?

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Fg and Ft are in same direction, therefore we add them together,

437m/s^2 + 9.8m/s^2 =446.8m/s^2

Tension = 446.8m/s^2 * mball

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12:11:54

If the ball is at the bottom of its arc the net force consists of the tension plus the gravitational force, with tension up and gravity down. Again using downward as the positive direction and noting that the upward centripetal acceleration is with this assumption negative we have

tension + mass_ball * 9.8 m/s^2 = -mass_ball * 440 m/s^2.

We find that

tension = -mass_ball * 440 m/s^2 - mass_ball * 9.8 m/s^2 = -mass_ball * 449.8 m/s^2.

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12:13:19

What if the ball is at its halfway height?

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Fcent would be perpendicular to the Fgrav, therefore, Fcent would have no influence.

Tension = massball * 9.8m/s^2

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12:14:17

At the halfway height the centripetal force and gravitational acceleration are perpendicular and hence independent. The only force acting toward the center is the tension in the string, which must therefore supply the entire centripetal force. The tension is mass_ball * 440 m/s^2 toward the center.

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Ok, vice versa, Fgrav is downward, not affecting the tension, so the only force affecting it is Fcent.

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12:24:10

Setting the two expressions equal we have the equation

G M / r^2 = v^2 / r.

To solve for v we first multiply both sides by r to get

G M r = v^2,

Taking the square root of both sides and reversing sides of the equation gives us

v = sqrt(G M r).

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I don't think this is right.

G M / r^2 = v^2/ r, if you multiply both sids by r you get GM/r = v^2, Right????

Right. Should be v = sqrt(G M / r).

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12:39:21

If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?

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So do we ignore the disk since it has negligible mass, otherwise how do I calculate its moment of inertia without its mass.

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12:39:38

What then will be the speed of each of the masses?

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12:39:52

*&*& Add up the individual kinetic energies.

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12:39:58

*&*& The two quantities should be the same.

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12:44:44

How long does it take the bullet to exit the barrel after the powder ignites?

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vAve = (267m/s + 0)/2 = 133.5m/s

`ds = vAve`dt

.38m = 133.5m/s`dt

.0028s = `dt

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12:45:47

Well I know that v0= 0 and vf= 267 m/sec and `ds= .38 m and from there I can use vf^2 = v0^2 + 2a`ds so I figure a= 93801 m/ sec^2

From there I can find the change in time to be 2.846441948 * 10^- 3

** OK but it's easier to average init and final vel then divide into displacement (e.g., vAve = (0 + 267) / 2 m/s = 133.5 m/s so to move .38 meters requires .38 m / (133.5m/s) = .028 sec approx. **

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12:51:30

Using the Impulse-Momentum Theorem determine the average force exerted on the bullet as it accelerates along the length of the barrel.

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I think it is 8.8kgm/s

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13:05:55

By the impulse-momentum theorem Fave `dt = m `dv we get Fave = (m`dv)/ `dt so from there I can substitute in the values already determined and find the force to be 3095.4434 N

** Good work. Impulse-momentum is appropriate here. It would also have been possible to find acceleration and multiply by mass, but impulse-momentum is more direct. **

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I don't understand. The impulse theorem says

impulse = Fnet*`dt = `dp

I calculated the force by doing a*m.

`dp = m*(vf-v0) = .033kg*267m/s = 8.8kgm/s

Fnet*`dt = `dp

Fnet*.0028s = 8.8kgm/s

fnet = 3142N

This is consistent with the given result. Acceleration would be 267 m/s / (.0028 s); when multiplied by the mass this gives the same result as the impulse-momentum theorem.

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13:07:56

i dont know what to do here.

** Sure you do. You found the acceleration above when you didn't need it. You need it now. **

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What am I looking for, i have everything already.

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