ic_class_091007

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Class 091007

The first 6 questions are about the experiment we did

Monday, allowing a ball to back and forth on a pair of oppositely-inclined ramps

until it comes to rest.  Most of the remaining questions involve sketches,

estimates and explanations.  The last problem is a review of something

everyone should now know (but it ain't always so).

`q001.  Give your raw data for the experiment

conducted Monday, with a ball traveling down one incline then up another, until

it finally comes to rest.

Report you numbers in the form of an easily-read table,

including an explanation before or after the table of what the numbers mean.

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`q002.  How much distance did the ball cover between

the instant of release and the instant at which it came to rest?  Give the

distance, then explain how you got it from the data you reported in the

preceding.

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Typically the 25 mm ball will have traveled perhaps 50 cm

when rolling down one 30-cm incline and up another, then will travel about 40 cm

the other way, maybe 32 cm on the next down-and-up roll, etc..  The total

distance usually reported is between about 1.5 m and 3 m.

If you used a smaller ball, as did some groups, the total

distance will be significantly less.

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`q003.  The distance down the first ramp is about 30

cm, and the domino supporting it has a thickness of a little less than 1 cm. 

So the distance the ball traveled on the first ramp alone is more than 30 times

the distance of its vertical descent.

Between release and coming to rest (after rolling back and

forth for a time) the ball sometimes descends and sometimes ascends, but it ends

up lower than it starting point by a vertical distance equal to the thickness of

a single domino.

How many times the distance of its vertical descent is the

total distance traveled by the ball?

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The vertical descent is about a centimeter.

A 200 cm total distance would be about 200 times the

vertical descent.

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`q004.  When the ball descends, does its potential

energy increase or decrease?  Does its kinetic energy increase or decrease? 

When it ascends, does its potential energy increase or

decrease?  Does its kinetic energy increase or decrease? 

Sketch a graph you think might represent the potential

energy of the ball vs. clock time, from the instant of release to the instant it

comes to rest.

Describe your graph:

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The ball gets lower as it descends, so its gravitational

PE decreases.  (a fuller explanation:  the component of the

gravitational force along the incline is in the same direction as the

displacement so gravity does positive work on the ball; the change in PE being

equal and opposite to the work done ON the ball by gravity, the PE therefore

decreases)

Its speed increases as it descends, so its KE increases.

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`q005.  On the same graph, sketch the kinetic energy

vs. clock time.

Describe this graph, and describe how the two graphs

compare.

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KE is up when the ball reaches the bottom of the ramp,

down at the ends.

The ball reaches the 'low' point with less speed each time

than before, so the 'up' KE gets smaller and smaller each time.

The ball takes less and less time to reach the 'low'

point, so the 'up' KE occurs more and more frequently (the horizontal distance

between the 'ups' will get closer and closer to one another).

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`q006.  Sketch two more graphs, one representing the

ball's velocity vs. clock time, and another representing the ball's position vs.

clock time.

Describe your graphs.

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no 'given solution' will be provided for this question,

but we will likely look at this in class

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`q007.  The sketch you copied from the board in class

depicts an incline rising from left to right at angle alpha with horizontal. 

The x axis is directed up the incline.  The weight vector is vertical, and

makes angle beta with the y axis.  Theta is the angle from the positive x

axis to the weight vector, as measured in the counterclockwise direction.

Sketch the same figure, but adjust your angles so that the

angle of the incline with horizontal is 15 degrees.  (If you divide a 90

degree angle into three equal angles, then take half of one of these angles, you

will have a 15 degree angle.  Using this method you should be able to

hand-sketch a 15 degree angle within a couple of degrees.)

Sketch the projection lines for the weight vector, and

estimate the lengths of the arrows representing the x and y components of this

vector as percents of the length of the arrow representing the weight vector.

Give the following numbers below, in the specified order,

with one number to a line:

1. the angle alpha in degrees
2. the angle beta in degrees
3. the angle theta in degrees
4. the angle, in degrees, between the x axis and the
   horizontal direction
5. the angle, in degrees, between the x axis and the
   vertical direction
6. the angle, in degrees, between the y axis and the
   horizontal direction
7. the angle, in degrees, between the y axis and the
   vertical direction
8. the x component of the weight as a percent of the
   weight
9. the y component of the weight as a percent of the
   weight

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For the 20-degree incline sketched on the board we would

have the following values:

alpha = 20 deg

beta = 20 deg

theta = 250 deg

between the x axis and the

horizontal direction the nearest angle is 20 deg, with the angle from the

axis to the horizontal direction directed clockwise; in the counterclockwise

direction the angle is 160 deg

between the x axis and the

vertical direction the nearest angle is 70 deg, with the angle from the axis

to the horizontal direction directed counterclockwise; in the clockwise

direction the angle is 110 deg

between the y axis and the

horizontal direction the nearest angle is 70 deg, with the angle from the

axis to the horizontal direction directed counterclockwise; in the clockwise

direction the angle is 110 deg

between the y axis and the vertical direction the nearest

angle is 20 deg, with the angle from the axis to the vertical direction directed

clockwise; in the counterclockwise direction the angle is 160 deg

the x component is in the neighborhood of 40% of the

weight

the y component is in the neighborhood of 90% of the

weight

Based on these results you should be able to figure the

results for a 15 degree angle.

For the 15 degree angle, the x and y components will be

about 30% and something over 90%, the latter perhaps as great at 95%.

Your estimates may well have varied somewhat from the

estimates used here.

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`q008.  Repeat the preceding exercise for angle alpha

= 10 degrees.  You should identify all the quantities you did before, but

report here only the x component of the weight as a percent of the weight:

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No detailed solution is provided here.

However the x component of the weight will be in the

neighborhood of 20% of the weight, and the y component will be 95% or more of

the weight.

Your estimates may well have varied somewhat from the

estimates used here.

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`q009.  Repeat the preceding exercise for angle alpha

= 5 degrees.  You should identify all the quantities you did before, but

report here only the x component of the weight as a percent of the weight:

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No detailed solution is provided here.

However the x component of the weight will be in the

neighborhood of 10% of the weight, and the y component will be nearly equal to

the weight (just a bit less than 100%).

Your estimates may well have varied somewhat from the

estimates used here.

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`q010.  Explain how your pictures show why an

increasing slope implies and increasing acceleration.

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The ball accelerates along the ramp, which if the axes are

rotated to the specified position, is in the x direction.

The x component of the weight was seen above to decrease

with decreasing slope, and will therefore increase with increasing angle of

slope.

The ball is accelerated by the x component of the weight,

so increasing slope implies increasing acceleration.

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`q011.  Suppose the ball had a weight of 8 Newtons.

Using the percents you estimated previously:

What would be the x component of its weight on the 15

degree ramp?

What would be the x component of its weight on the 10

degree ramp?

What would be the x component of its weight on the 5

degree ramp?

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Assuming that the respective x components are 30%, 20% and

10% of the weight, we would conclude that the components are

.30 * 8 N = 2.4 N (15 deg ramp)

.20 * 8 N = 1.6 N (10 deg ramp)

.10 * 8 N = .8 N (5 deg ramp).

Your estimates may well have varied somewhat from the

estimates used here.

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`q012.  Suppose the ball had a mass of 3 kilograms. 

What would be its weight, in Newtons?

Using the percents you estimated previously:

What would be the x component of its weight on the 15

degree ramp?

What would be the x component of its weight on the 10

degree ramp?

What would be the x component of its weight on the 5

degree ramp?

Using Newton's Second Law, assuming that the net force on

the ball is equal to the x component of its weight, what then would be its

acceleration on the 15 degree ramp?

What would be its acceleration on each of the other ramps?

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If released from rest gravity would accelerate the ball at

9.8 m/s^2, so it is exerting force

F = m g = 3 kg * 9.8 m/s^2 = 29 N

on the ball. 

Weight being the force exerted by gravity, this is

therefore the weight of the ball.

On a 15 deg ramp the x component of the weight would be

30% of the weight, according to the estimate being used here, and would

therefore be

.30 * 29 N = 9 N .

On a 10 deg ramp the x component of the weight would be

20% of the weight, according to the estimate being used here, and would

therefore be

.20 * 29 N = 6 N .

On a 5 deg ramp the x component of the weight would be 10%

of the weight, according to the estimate being used here, and would therefore be

.10 * 29 N = 3 N .

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`q013.  Assume that the weight of the ball is 8

Newtons, as on one of the preceding problems.

Based on your estimates of the x component of the weight:

How much work would be done on the ball by this force if

it rolled 5 meters down the ramp?

Assuming it started from rest, how fast would it be going

after rolling 5 meters down the ramp?

Assuming it started from rest, what would be its kinetic

energy after rolling 5 meters down the ramp?

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On the 15-degree ramp the x component of the weight would

be 2.4 N, according to the estimates used here.

The x component would be directed down the ramp, so would

be in the same direction as the displacement.

work = force * displacement would therefore be positive,

and we would have

• work = 2.4 N * 5 meters = 12 N * meters = 12 N * m or
  12 kg m^2 / s^2.

 

The question of how fast the ball is moving is probably

premature at this point.  Theoretically you know enough to figure this out,

but if you did manage it you're doing exceptionally well.

One way to figure out the speed is to notice that the x

component is about 30% of the gravitational force, so if that's the only force

acting in the x direction the acceleration must be 30% that of gravity. 

(So acceleration would be .30 * 9.8 m/s^2 = 3 m/s^2, approx.; `ds is 5 meters

and v0 is 0, and the fourth equation of unif accel motion would give us vf = 5.4

m/s.)

• Another way is to use the 8 N weight to figure out
  the mass, then divide the 2.4 N x-component of the force by the mass.
• Since the gravitational force is the only force
  acting partly in the direction of motion, the work calculated here would be
  the work done by the net force.  So the kinetic energy of the ball
  would change by 12 kg m^2 / s^2.  (weight = m g so m = weight / g = 8 N
  / (9.8 m/s^2) = .8 kg, approx.. so acceleration would be a = F_net / m = 2.4
  N / (.8 kg) = 3 m/s^2, and the conclusion would again be vf = 5.4 m/s).

 

We could reason out the KE of the ball using the

work-kinetic energy theorem, or we could just calculate it using the velocity:

• Work-KE Theorem (work done by net force = change in
  KE): 

Since no other forces act in the x direction, and

since the net force in the y direction is zero (normal force is equal and

opposite to the vertical component of the gravitational force), the net

force is the x component of the gravitational force, which we have estimated

to be 2.4 N.

Since the ball started from rest, its initial KE is

zero. 

The work 12 kg m^2 / s^2 was done by the net force, so

is equal to the change in KE. 

We could therefore conclude that its final KE is 12 kg

m^2 / s^2.

• Alternatively, we could simply calculate 1/2 m v^2
  with m = .8 kg and v = 5.4 m/s.  Within roundoff error we would get 12
  kg m^2 / s^2, in agreement with the work-KE theorem.

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`q014.  Assume that the weight of the ball is 8

Newtons, as before.

Based on your estimates of the x component of the weight:

How much work would be done on the ball by this force if

it rolled 3 meters up the ramp?

Assuming it started with a speed of 6 meters / second, how

fast would it be going after rolling 3 meters up the ramp?

Assuming it started with a speed of 6 meters / second,

what would be its kinetic energy after rolling 3 meters up the ramp?

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`q015.  Sketch the tension force for a pendulum which

is displaced to an angle of 15 degrees with vertical.

Using an x-y coordinate system with the x axis horizontal,

estimate the x component of the tension as a percent of the tension.

Repeat for angles of 10 degrees and 5 degrees.

Give your three estimates below:

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`q016.  Based on your answers to the preceding, would

you expect the acceleration of the pendulum to increase, decrease or remain the

same as it swings back to its equilibrium position?  Be sure to explain

your reasoning.

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`q017.  Explain why it would or would not be a good

idea to use the equations of uniformly accelerated motion to model the motion of

a freely swinging pendulum.

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`q018.  You should be able to do these review

problems in a couple of minutes each.  If not, you still have some work to

do on projectile motion: 

If an object falls freely from rest through a vertical

distance of 2 meters, while traveling at a constant horizontal velocity of 250

cm/second, then how far does it travel in the horizontal direction as it falls?

If another object falls freely from rest through the same

distance, and in the process of falling travels 40 cm in the horizontal

direction, then what is its horizontal velocity?

Your solution:

 

 

 

 

 

 

 

 

Confidence rating:

 

Given solution:  Let downward be the positive

direction.

The vertical displacement is therefore 2 meters, the

vertical acceleration 980 cm/s^2, the initial vertical velocity zero.  We

can work in units of cm or meters, but should decide this at the beginning. 

Either express 2 meters as 200 cm, or 980 cm/s^2 at 9.8 m/s^2, then simply use

the equations.

The fourth equation gives us vf^2 = v0^2 + 2 a `ds. 

Substituting the quantities we get vf = 6.4 m/s, approximately.  We also

get a negative solution, but discard it since it is clear the final velocity is

downward.

The average vertical velocity is therefore vAve_y = (0 m/s

+ 6.4 m/s) / 2 = 3.2 m/s, and the time of fall is

`dt = `ds / vAve = 2 m / (3.2 m/s) = .63 sec, approx..

The horizontal velocity is constant, since gravity doesn't

act in that direction and we are neglecting air resistance and other possible

forces in the horizontal direction.

Thus the horizontal displacement is

`ds = vAve_x * `dt = 250 cm/s * .63 sec = 160 cm, approx.. 

This can also be expressed as 1.6 meters.

 

For the second question the initial vertical velocity and

displacement are the same as before so `dt is still .63 sec.

The horizontal displacement is 40 cm, so the constant

horizontal velocity is

vAve_x = 40 cm / (.63 s) = 60 cm/s, approx..

Self critique:

 

 

 

 

 

Self-critique rating:

 

 

 

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Very good. The annotated commentary above is for your reference.