#$&*
course phy 201
11-13-11I have more of this done on paper but I have been feeling really sick this weekend. Sorry, I just can't sit in front of the computer any longer.
Physics I Class 111031
Your data for the first two experiments should be submitted promptly. Your analysis should be submitted within a week, as should the rest of the assignment.
Balancing dominoes experiment:
Report all relevant data from the experiment, being sure to clearly identify the quantities you report and what they mean.
****
moment arms: 11.5cm, 4cm, 14cm
beam is 30.5cm total
force is 15g or 15000dynes for each domino
#$&*
Find the torque produced by the weight of each domino on the first beam, about the point of rotation. You may assume domino weights of 15 grams each.
****
T1=14cm*15g=210g cm pr 210000dyne cm
T2=4cm*15g=60g cm or 60000dyne cm
T3=11.5cm*15g=172.5g cm or 172500dyne cm
Tnet=210g cm+60g cm+172.5g cm=442.5 g cm or 4422500 dyne cm
#$&*
@&
The first mass is on the opposite side of the fulcrum as the other two. So if T1 is positive, the other two are negative and if T1 is negative, the other two are positive.
It's your choice whether T1 is positive or negative, but in any case the signs of the torques you add won't all be positive, and the magnitude of your result will be much less than the magnitude of any of the torques.
*@
Candy bar experiment:
How many oscillations did the candy bar complete in a minute?
****
150 oscillations in 1 minute
#$&*
What was the length of the rubber band chain when supporting 4 dominoes?
****
60.5 cm
#$&*
What was the length of the rubber band chain when supporting 8 dominoes?
****
41 cm
#$&*
@&
The rubber band wouldn't have gotten shorter; however if you have reported the marks on the meter stick, which would then have been upside down, you would get the correct difference between the lengths.
*@
Through how many radians did the reference point move during the 1-minute timing (it moved through a complete circle for every cycle you counted)?
****
1 oscillation=2pir
3.14 rad*150 oscill=471.24 radians in 1 minute
or 7.85 radians/sec
@&
2 pi is 2 * 3.14, so you would have double this number, of about 15.7 rad / sec.
*@
#$&*
What therefore was the angular velocity omega of the reference point?
****
change in angular postion/change in clock time
?
#$&*
@&
That would be your 15.7 rad / sec.
*@
How far was it, on the average, from the lowest point to the highest point in the candy bar's oscillation (you didn't measure this; just visualize the motion and make an estimate)?
****
the rubber band chain is 27 cm with no weights on it. i think that the weight of the candy bar and 4 dominos are close because the oscillation is close. if this guess is correct, the lowest point is about 33cm. and the highest point is about 26 or 27 cm.
#$&*
The diameter of the reference circle is equal to the estimate you made for the preceding question. How fast was that reference point moving around the arc of that circle?
****
approx 6 cm diameter= 3 cm radius
2pi3=18.85 cm is circumfrance
18.85=1 oscill
150 oscill in 1 min so, 2827.5 radian in 1 min or 47.125 radains/sec
#$&*
@&
The reference point was moving at about 15.7 rad / sec.
A radian on this circle corresponds to arc distance 3 cm, so the point is moving at
15.7 rad * (3 cm / rad) ) / sec = 47 cm / sec.
*@
What is the slope of the force vs. length graph for your rubber band chain? If you measured the domino stack then you can use the fact that for every millimeter of height the stack has mass 1.9 grams. If not just assume a mass of 15 grams.
****
0g---27cm
30g---28cm
60g---30.5cm
90g---35cm
120g---41cm
150g---50.5cm
#$&*
Your last answer is your estimate of k, and your count resulted in your previous answer for omega. What therefore is m?
****
4domino=60g so, candy bar=60g
#$&*
@&
Check the given solutions on these last two.
*@
Experiment with 3 rubber bands
The 'initial point' of each of your rubber band chains will be the end nearest the central paper clip, and the 'terminal point' will be the end further away from that clip. The length vector for each chain is the vector from its initial point to its terminal point.
What are the x and y components of each of the three length vectors?
****
xcomp ycomp
RB1=53.5cm-18.5cm=35cm RB1=24cm-37cm=-13cm
RB2=53.5cm-44cm=9.5cm RB2=24cm-37cm=-18.5cm
RB3=81cm-57cm=27cm RB3=8.5cm-21.5cm=.13cm
#$&*
What are the magnitudes of each of the three length vectors?
****
vector1= sqrrt(35cm)sqrd+(-13cm)sqrd=37.34cm
vector2=sqrrt(9.5cm)sqrd+(-18.5)sqrd=20.8cm
vectpr3=sqrrt(24cm)sqrd+(-13cm)sqrd=27.3cm
#$&*
Divide the x and y components of each of these vectors by its length. What are your results? The vectors you get here are called 'unit vectors', because if you divide a vector by its length you get a vector of length 1 (i.e., a vector of unit length).
****
x y
RB1 35cm/37.3cm=.94 -13/37.3=-.35
RB2 9.5cm/20.8cm=.46 -18.5/20.8=-.89
RB3 27/27.3=.88 -13/27.3=-.48
i'm not sure on my units
#$&*
According to your calibration of the colored calibrating chain, what was the tension in each of the rubber bands?
****
2 dominos=30g 5 dominos=75g
T1=33cm*75g=2475g cm
T2=28cm*30g=840 g cm
T3=32.8cm*75g=2460g cm
#$&*
@&
This last step doesn't appear to be based on your calibration of the colored chain.
*@
Multiply the tension by the x and y components of the corresponding unit vector. Your result for each rubber band chain will be the components of its tension vector.
****
x y
T1= 2475g cm*3cm=86625 2475g cm*.13cm=-32175
T2= 840g cm*9.5cm=7980 840g cm*-18.5=-15540
T3= 2460g cm*24cm=59040 2460g cm*-13=-31980
i dont know about the units of measurement on this either
#$&*
Add up the x components of your three tension vectors. What do you get?
****
86625+7980+59040=153645
again ? on units of measure
#$&*
Add up the y components of your three tension vectors. What do you get?
****
-32175+(-15540)+(-31980)=-79695
? units
#$&*
The sum of your x components is the x component of the resultant vector; the sum of the y components is the y component of the resultant vector. What therefore is the magnitude of your resultant vector?
****
153645-79695=73950
? units
#$&*
The resultant vector represents the total effect of the three forces acting on the paper clip in the middle. The paper clip has zero acceleration, so the net force is actually zero. One measure of how accurate this experiment might be is the comparison between your resultant and the ideal. A reasonable way to make the comparison is to compare the magnitude of your calculated resultant with the largest of the three tension forces.
What is the magnitude of your resultant, as a percent of the magnitude of the largest of the three tension forces?
****
greatest tension=86625
resultant vector=73950
#$&*
@&
You're off track on this solution. There is a very similar problem in one of the homework assignments, with a step-by-step given solution. When you get to that and receive the given solution, you can see where you went off track here.
*@
`q001. The mass of the Earth is about 6 * 10^24 kg. The mass of the Moon is about 8 * 10^22 kg. The two are separated by about 400 000 kilometers. G = 6.67 * 10^-11 N m^2 / kg^2.
What is the gravitational force between the Earth and the Moon?
****
m1=6*10^24kg m2=8*10^22kg
F=G*(m1*m2)/r^2
6.67*10^-1n(m^2/kg^2)*((6*10^24kg)*(8*10^22kg))/(2.0*10^8m)^2
=8.0*10^20N kg
@&
The separation is 4 * 10^8 m, which is what you should use for r in your denominator.
*@
#$&*
What is the acceleration of the Moon toward the Earth?
****
a=v^2/r
4pi^2(3.84*10^8m)/(2.36*10^6s)^2
= 0.00272m/s^2=2.72*10^-3m/s^2
#$&*
What is the acceleration of the Earth toward the Moon?
****
#$&*
Assuming that the Moon's path in its orbit is a circle (which is pretty much the case) of radius about 400 000 kilometers, and that it takes 28 days to complete one orbit (again pretty close to the actual time required), then how fast is it moving?
****
28days*(24hours/day)*(3600sec/hour)=2.4*10^6sec
a cent= v^2/r
4pi^2(4.0*10^8m)/(2.4*10^6s)^2
= 2.7*10^-3m/s^2
#$&*
What therefore its its centripetal acceleration?
****
#$&*
@&
You're on the right track throughout on this problem. Check the given solutions for some details.
*@
`q002. Summary of SHM:
If the force vs. length graph of an elastic object is linear then its slope is called its force constant.
If the tension force exerted this object is (in a certain way not yet specified) responsible for the net force on an object then that force will be of the form F_net = - k x, where x is the position of the object relative to its equilibrium position.
If the net force on a mass m at position x is - k x, then the mass will either remain in its equilibrium position (x = 0), or will oscillate in a manner modeled by the projection on a line through the origin of a reference point moving around a circle with angular velocity omega = sqrqt(k/m).. You should understand how to use the formula omega = sqrt(k / m), and you should understand motion around a circle which is characterized by a constant angular velocity omega. You can expect that it will take a few examples and more explanation for you to understand the part about the projection..
If A is the radius of the circle then the amplitude of the motion is A, the speed of the point about the reference circle is v = A * omega, the maximum KE of the object occurs at the equilibrium point and is equal to 1/2 m v^2, the total mechanical energy of the oscillation is 1/2 m v^2, and the potential energy at position x relative to the x = 0 position is 1/2 k x^2.
An object of mass 50 grams is suspended from a rubber band chain. When it supports a hanging mass weighing .5 Newtons the chain is 70 cm long. When it supports a hanging mass weighing .9 Newtons the chain is 90 cm long.
What is the average slope of the force vs. length graph for the given length interval?
****
m1=.5N chain L=70cm
m2=.9N chain L=90cm
(.9N-.5N)/(90cm-70cm)=0.4N/20cm
#$&*
Assuming that the force vs. length graph is linear (not really so for a rubber band chain, but close enough not to make a big difference in the oscillation of our object), what is the value of k for this chain?
****
0.4N/20cm=0.02Ncm
#$&*
What therefore should be the angular velocity of our reference point?
****
W=sqrrt(.02/50g)=0.02
#$&*
How long should it take the reference point to complete one circuit around the circle?
****
#$&*
@&
Right track here as well. Check the given soluitons.
*@
`q003. A vector `R of constant length 1 has its initial point at the origin. Its terminal point moves at a constant speed around the circle, so that the vector moves like a spinning dial. It should be clear that the circle has radius 1.
When the vector makes an angle of 15 degrees with the x axis, what are its x and y components?
****
x y
.97 .26
#$&*
Answer the same for angle 30 degrees.
****
x y
.87 .5
#$&*
Answer for angle 45 degrees.
****
x y
.71 .71"
@&
Very good on this last problem.
*@
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
_________________________