class 091028

Class 091028

STUDENT QUESTION

Could you tell me how to solve for the 'dPE, and could you give it to me in equation form if possible, or just tell me which qa gives the info.

INSTRUCTOR RESPONSE

The answer to your question depends on the situation. I'm giving you a general answer here; after working to understand this, if you have additional questions for a specific situation, I'll be glad to help you see how to apply these ideas. Just be sure to include a copy of this document, as well as the specific situation you want to address and your best thinking about that situation.

Any time you think about potential energy, you should immediately remind yourself of the definition of potential energy:

`dPE for a system is equal and opposite to the work done by the conservative force on the system for the interval in question.

This definition involves work, so you should also remind yourself of the definition of work:

Work is the product of force and displacement in the direction of the force.

The definition of work involves force and displacement.

So if you know the conservative forces and the initial and final position of the system, you can find `dPE:

Identify the displacement `ds for the interval and multiply it by the conservative force.

Be careful about the directions of the force and the displacement. The displacement you use is either in the same direction as the force (in which case the force and displacement have the same sign and the work is positive), or opposite the direction of the force (in which case the force and displacement have opposite signs and the work is negative).`

You can also determine `dPE if you know the work done by nonconservative forces and the change in KE:

The work-energy theorem tells us that `dW_NC_ON = `dPE + `dKE.
It follows immediately that `dPE = `dW_NC_ON - `dKE.

`q001. ** One-step-at-a-time questions:

If the velocity of a 5-kg object changes from +3 m/s to -2 m/s, what is the change in its velocity, and what is the change in its momentum?

****

Initial velocity is +3 m/s, final velocity is -2 m/s.

Change in velocity is

`dv = vf - v0 = -2 cm/s - 3 m/s = -5 m/s.

Initial momentum is 5 kg * 3 m/s = 15 m/s; final momentum is 5 kg * (-2 m/s) = -10 m/s.

Change in momentum is

`dp = p_f - p_0 = -10 kg m/s - 15 kg m/s = -25 kg m/s.

&&&&

A 12-kg object experiences a velocity change of +2 m/s during a time interval of .01 second. What average force was exerted on it during that interval?

****

The impulse-momentum theorem applies to a force exerted for a given time interval:

F_net `dt = `d(m v); when m is constant this can be written

F_net `dt = m `dv.

Thus

F_net = m `dv / `dt = 12 kg * 2 m/s / (.01 s) = 2400 kg m/s^2 = 2400 N.

&&&&

What is the resultant of two displacement vectors, the first having magnitude 8 meters directed at angle 53 degrees, the second with x component -3 meters and y component -4 meters?
****

The magnitude of the first displacement vector is 8 meters, the angle is 53 deg, so its components are
- x comp of 1st vector = 8 m * cos(53 deg) = 8 m * .6 = 4.8 m
- y comp of 1st vector = 8 m * sin(53 deg) = 8 m * .8 = 6.4 m
To find the resultant vector R we add the two vectors. To do this we add their components, obtaining

R_x = 4.8 m + (-3 m) = 1.8 m and

R_y = 6.4 m + (-4 m) = 2.4 m.

The magnitude of the resultant vector is found using the Pythagorean Theorem:

magnitude = sqrt( (R_x)^2 + (R_y)^2 ) = sqrt( (1.8 m)^2 + (2.4 m)^2 ) = sqrt ( 3.24 m^2 + 5.76 m^2) = sqrt(9 m^2) = 3 m.

The angle of the resultant is

theta = arcTan(R_y / R_x) = arcTan ( 2.4 m / (1.8 m) ) = arcTan(1.33) = 53 deg.

&&&&

Two objects collide, one of mass 5 kg and the other of mass 12 kg. The momentum of the 5 kg object changes by -40 kg m/s. What is the momentum change of the 12 kg object, and what is the change in its velocity?

****

The forces exerted on the particles by one another are equal and opposite, and the act for the same time interval. So the values of the impulse F `dt will be equal and opposite.

Since F `dt = `d(m v), the changes in momentum will be equal and opposite.

The momentum change of the first object is -40 kg m/s, so the momentum change of the second is +40 kg m/s.

The mass of the second object is 12 kg. If `dv is the change in its velocity we have

m `dv = `dp so that

`dv = `dp / m = +40 kg m/s / (12 kg) = 3.33 m/s.

&&&&

What is the direction of a displacement vector whose x and y components are +5 m and - 4 m? What are the components of a force vector of magnitude 80 N in the same direction?

****

The direction of the vector is

theta = arcTan(-4 m / (5 m) ) = arcTan(-.8) = -39 deg, approx..

Measured counterclockwise from the positive x axis, this is 360 deg - 39 deg = 341 deg.

A force vector of magnitude 80 N in the same direction has the same angle, so the components of this force vector are

F_x = 80 N cos(341 deg) = 80 N * .78 = 62 N, approx.

F_y = 80 N sin(341 deg) = 80 N * .62 = 50 N, approx..

&&&&

A rubber band chain exerts force 4 N when its length is 50 cm, and 12 N when its length is 80 cm. As the chain is extended from the first length to the second, what average force is exerted by the rubber band (assuming that the force vs. length is linear)? How much work is done by the chain during this interval?

****

The average force, assuming linearity of F vs. length, is

F_ave = (4 N + 12 N ) = 8 N.

As the chain is extended, the ends of the rubber band must be moved in the direction opposite the tension force, so the work done by the chain is negative. The force is exerted through a displacement of 80 cm - 50 cm = 30 cm, and the displacement is along the line of the force.

Thus the work done by the rubber band is

`dW_by = - (8 N * 30 cm) = -240 N * cm = -240 N * (.01 m) = -2.4 N * m = -2.4 Joules.

&&&&

A 200 kg mass is subjected to a net force of 1000 Newtons for half a second. What is its acceleration? What therefore is its change in velocity? What therefore is the change in its momentum? What is the impulse of the force on this interval?

****

The acceleration is

a = F_net / m = 1000 N / (200 kg) = 5 m/s^2.

The change in velocity is

`dv = a `dt = 5 m/s^2 * .5 s = 2.5 m/s.

The change in momentum is

`dp = m `dv = 200 kg * 2.5 m/s = 500 kg m/s.

The impulse of the force is

F_net `dt = 1000 N * .5 s = 500 N * s = 500 kg m / s^2 * s = 500 kg m/s.

The impulse is seen to be equal to the change in the momentum.

&&&&

`q002. Give your data for the experiment in which you attempted to release the pendulum the point that would maximize its horizontal range. Include all data that would be necessary to replicate your trials.

Attempt to organize your data in such a way that it is easy to read at a glance.

Include a description how you conducted the experiment.

****

You should have received individual feedback on your data submission.

&&&&

`q003. How much force would it take to accelerate an Atwood machine consisting of two 50 kg masses at 9.8 m/s^2?

What is the total weight of these two masses?
If the addition of a single dollar coin to one side of the system results in an acceleration of 5 cm/s^2, then what is the mass of the coin? What is the weight of the coin?
How many coins would it therefore take to match the mass of the system?
(Assume that the friction in the pulley is negligible).

****

The system has a total mass of 100 kg, so this would require a force of

F_net = 100 kg * 9.8 m/s^2 = 980 N.

The total weight of the system is therefore 980 N.

When in balance, the system experiences net force zero in its direction of motion.

When the coin is added, the net force on the system is

F_net_coin = 100 kg * .05 m/s^2 = 5 N.

This force comes from the gravitational force on the coin, so we have

wt_coin = 5 N.

The mass of the coin is its weight divided by the acceleration of gravity:

m = wt / g = 5 N / (9.8 m/s^2) = .5 kg, approx..

That's a pretty massive dollar coin.

It's worth noting that this is .5% the mass of the system, and that the 5 m/s^2 acceleration is .5% the acceleration of gravity. This connection is worth thinking about.

&&&&

`q004. Three rubber bands are attached to a small ring at the origin of an x-y coordinate system. The opposite ends of the rubber bands are respectively at the points (8 cm, 6 cm), (-10 cm, 0) and (0, -12 cm).

· For each rubber band, determine the angle of the displacement vector from the origin to its other end.

****

The lengths of the three rubber bands are respectively

L_1 = sqrt( (8 cm)^2 + (6 cm)^2 ) = 10 cm

L_2 = sqrt( (-10 cm)^2 + (0 cm)^2 ) = 10 cm and

L_3 = sqrt( (0 cm)^s + (-12 cm)^2) = 12 cm.

The lengths could be used with force vs. length graphs for the three rubber bands, to find the force exerted by each. This won't be necessary here, since the forces are simply given. However, as we say in class, this is what we would ordinarily do.

The three angles are

theta_1 = arcTan(6 cm / (8 cm) ) = arcTan(.75) = 37 degrees.

theta_2 = 180 deg (this vector is clearly directed along the negative x axis) and

theta_3 = 270 deg (this vector being clearly directed along the negative y axis).

&&&&

Using a graph of force vs. length for each rubber band, we find that they exert forces of 5.0 N, 3.8 N and 3.2 N, respectively.

· Find the x and y components of each each force.

The components are

F1_x = L1 cos(theta_1) = 5.0 N cos(37 deg) = 4.0 N

F1_y = L1 sin(theta) = 5.0 N sin(37 deg) = 3.0 N.

F2_x = 3.8 N cos(180 deg) = -3.8 N

F2_y = 3.8 N sin(180 deg) = 0 N

F3_x = 3.2 N cos(270 deg) = 0 N

F3_y = 3.2 N sin(270 deg) = -3.2 N.

· Use your results to find the x and y components of the net force, and

The components of the net force are found by adding the appropriate components of the individual forces:

F_net_x = 4.0 N + (-3.8 N) + 0 N = 0.2 N

F_net_y = -3.0 N + 0 N + (-3.2 N) = -0.2 N.

· find the magnitude and direction of the net force.

The magnitude of the net force is

F_net = sqrt( (F_net_x)^2 + F_net_y)^2 ) = sqrt( (0.2 N)^2 + (-0.2 N)^2 ) = sqrt(.08 N^2) = 0.28 N, approx.

and its angle is

theta = arcTan(-0.2 N / (0.2 N)) = arcTan(-1) = -45 deg, which when measured counterclockwise from the positive x axis is

theta = 360 deg - 45 deg = 315 deg.

****

&&&&

`q005. A ball starts rolling along the x axis at point x = x0, with velocity v0. Its acceleration is a. This continues through time interval `dt.

By how much does the velocity of the ball change?
What is the ball's displacement during the interval?
What therefore is its final position along the x axis?

****

During the time interval the ball's velocity changes by

`dv = a `dt,

ending up with velocity

vf = v0 + `dv = v0 + a `dt.

Its average velocity, assuming linear v vs. t (i.e., constant acceleration) is

vAve = (vf + v0) / 2 = (v0 + a `dt + v0) / 2 = (2 v0 + a `dt) / 2 = v0 + 1/2 a `dt.

Its displacement is therefore

`ds = vAve `dt = (v0 + 1/2 a `dt) * `dt = v0 `dt + 1/2 a `dt^2.

Its final position along the x axis is therefore

x_f = x_0 + `ds = x_0 + v0 `dt + 1/2 a `dt^2.

Our solutions are therefore

vf = v0 + a `dt

`ds = v0 `dt + 1/2 a `dt^2 and

x_f = x0 + v0 `dt + 1/2 a `dt^2.

You should recognize these expressions as being similar to expressions given in your text. If we assume that x0 and v0 are position and velocity at clock time t_0 = 0, then the time interval up to t = t_f is just `dt = t_f - t_0 = t_f = 0 = t_f. Our equations would then become

vf = v0 + a * t_f

`ds = v0 * t_f^2 + 1/2 a * t_f^2 and

x_f = x0 + v0 * t_f + 1/2 a t_f^2.

If we agree that vf, x_f and t_f can just be represented by v, x and t, our equations are

v = v0 + a t

`ds = v0 t + 1/2 a t^2 and

x = x0 + v0 t + 1/2 a t^2.

These equations specify the velocity v and position x at clock time t.

v and x can now be understood as functions of t.

The authors of your text present the four equations of uniformly accelerated motion as

v = v0 + a t

x = x0 + v0 t + 1/2 a t^2

(these two are identical to those given above)

vAve = (v + v0) / 2 (the familiar statement that avererage velocity is the average of initial and final velocity, assuming acceleration to be constant) and

v^2 = v0^2 + 2 a ( x - x0 ).

The last is identical to our equation vf^2 = v0^2 + 2 a `ds, except that `ds is expressed as x - x0.

The author's equations have the advantage that v and x are now expressed as functions of t.

They have the disadvantage of having six symbols (v0, v, x0, x, a, t) rather than our five (`ds, v0, vf, a, `dt), and we can no longer say that 'given three of the five quantities, we can find the other two'.

&&&&

Homework:

Your label for this assignment:

ic_class_091028

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form. Answer the questions posed above.

Note that you should be preparing to take Test 1 sometime next week.

You have been assigned the first six problems, and should now learn the rest of the problems in Set 5 of the Introductory Problem Set. A link that should work is at http://vhmthphy.vhcc.edu/ph1introsets/default.htm .

You should also read Chapter 6 in your text, which can serve you in the future as a good organized reference. By Monday you should have read Chapters 2, 4 and 6. You should make note of some of the differences in notation (e.g., my use of `dW_NC_ON where the book uses just W_NC).

Work through and submit q_A_18, q_A_19 and q_A_20

URL's of qa's 10-20:

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_11.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_12.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_13.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_14.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_15.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_16.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_17.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_18.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_19.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_20.htm