course 201
10/25 8pm
Class 091021In class we observed a steel ball rolling down an incline with slope approximately .03, colliding with a stationary marble, with both objects then falling 90 cm to the floor.
The steel ball, when unimpeded by the marble, traveled 8.5 cm while falling.
The marble (which was initially stationary), after being hit by the ball, traveled 16 cm while falling.
The steel ball after hitting the marble traveled 5.5 cm while falling.
You should be able to figure out the time of fall, assuming the initial vertical velocity to be close to zero.
Approximating the time of fall by .4 sec we conclude that the ball reached the end of the ramp at a speed of about 20 cm/s, the marble traveled at 40 cm/s after being struck, and the steel ball was moving at about 12 cm/s after striking the marble.
`q001. Using the above figures:
What is the change in the ball's velocity between the instant before and the instant after it strikes the marble?
What is the change in the marble's velocity the instant before and the instant after the collision?
What do you therefore conjecture to be the ratio of the ball's mass to that of the marble?
****
The change in the balls velocity is 20 cm/sec-12 cm/s= 8 cm/sec
The change in the marbles velocity is 40 cm/sec- 0 cm/sec= 40 cm/sec
40 cm/sec / 8 cm/sec= 5, which illustrates that the marbles velocity changed five times as much as the balls velocity, which therefore means that the balls mass is five times as big as the marble.
&&&&
`q002. Using your own data for the same experiment, what are the following:
the unimpeded steel ball's velocity at the end of the ramp
the velocity of the glass marble the instant after being struck by the ball and
the velocity of the ball immediately after striking the marble?
What therefore should be the ratio of the ball's mass to that of the marble?
Be sure to include your raw data and explain how you get from the raw data to your conclusions.
****
When steel ball was allowed to roll down the ramp and fall to the floor on its own, it traveled 13.4 cm in the horizontal direction.
When the steel ball was let down the ramp with a marble resting at the end of the ramp, the steel ball traveled 10.6 cm in the horizontal direction and the marble traveled 18 cm in the horizontal direction.
Assuming a .4 second drop time based on previous experiments, we can calculate the horizontal velocity of each ball in each situation
Velocity of steel ball alone= `ds/`dt=13.4 cm/.4 sec= 33.5 cm/sec
Velocity of steel ball with marble= 10.6 cm/.4 sec= 26.5 cm/sec
Velocity of marble= 18 cm/.4 sec= 45 cm/sec
The steel ball changed velocity when it struck the marble.
33.5 cm/sec- 26.5 cm/sec= 7 cm/sec
The marble also had a change in velocity because it started from rest:
45 cm/sec- 0 cm/sec= 45 cm/sec
To find the ratio of the balls mass to the marbles mass:
45 cm/sec / 7 cm/sec= 6.4, indicating that the steel ball has 6.4 times the mass of the marble
&&&&
Let's assume that we can neglect forces exerted by the ramp on the glass marble and the steel ball, and that we can in fact neglect all forces except the forces exerted by each ball on the other. We can therefore regard the two balls, during the time of collision, as a closed system. We have draw our conclusions about the velocities involved in the collision from the projectile behavior of the objects after collision.
From F_net `dt = `dp and the fact that the forces exerted by the two objects are equal and opposite, we can draw the conclusion that the momentum change `dp of one object should be equal and opposite to that of the other. (recall that p = m v, so `dp = p_f - p_0 = m vf - m v0)
Since momentum changes are equal and opposite, we conclude that the total momentum of the system is the same immediately after collision as it was immediately before.
`q003. Let's let m1 and m2 be the two masses, let v1 and v2 be their velocities before the collision, and let u1 and u2 be their velocities after collision.
What are the values of v1, v2, u1 and u2 for your trial?
****
V1= 33.5 cm/sec
V2= 0 cm/sec
U1= 26.5 cm/sec
U2= 45 cm/sec
&&&&
`q004. Using the symbols m1, m2, v1, v2, u1 and u2, give the symbolic expression for each of the following:
The momentum of the steel ball before collision.
The momentum of the glass marble before collision.
The momentum of the steel ball after collision.
The momentum of the glass marble after collision.
The total momentum of the two objects before collision.
The total momentum of the two objects after collision.
****
The momentum of the steel ball before collision.
M1*v1
The momentum of the glass marble before collision.
M2*v2
The momentum of the steel ball after collision.
M1*u1
The momentum of the glass marble after collision.
M2*u2
The total momentum of the two objects before collision.
M1*v1 + m2*v2
The total momentum of the two objects after collision.
M1*u1 + m2*u2
&&&&
`q005. Write the symbolic equation which states that the total momentum of the two objects before collision is equal to their total momentum after collision. (your equation will set your answers the last two expressions in the preceding question equal to one another)
****
M1*v1 + m2*v2= M1*u1 + m2*u2
&&&&
`q006. Substitute the values you observed for v1, v2, u1 and u2 into your equation. Your equation will still have the symbols m1 and m2, but the other symbols will be replaced by the various velocities you observed.
M1*v1 + m2*v2= M1*u1 + m2*u2
M1(33.5 cm/sec) + m2(0 cm/sec)= m1(26.5 cm/sec) +m2(45 cm/sec)
`q007. Rearrange your equation so that all the quantities containing m1 are on the left-hand side and all the quantities containing m2 are on the right-hand side (your equation should have been m1 v1 + m2 v2 = m1 u1 + m2 u2).
Factor m1 out of the left-hand side and m2 out of the right-hand side.
Solve the equation for m1. The left-hand side should end up just as m1; the right-hand side will include the symbol m2.
Give the steps of your solution:
****
M1(33.5 cm/sec) + m2(0 cm/sec)= m1(26.5 cm/sec) +m2(45 cm/sec)
M1(33.5 cm/sec) + 0= m1(26.5 cm/sec) +m2(45 cm/sec)
M1(33.5 cm/s)-m1(26.5 cm/s)=m2(45 cm/s)
M1(33.5 cm/sec- 26.5 cm/sec)=m2(45 cm/sec)
M1= m2(45 cm/sec)/(33.5 cm/s-26.5 cm/s)
You can simplify this to get m1 = 6.4 m2, identical with your previous conclusion about the masses.
&&&&
The foam wheel was observed to rotate through 8 revolutions, coming to rest after about 20 seconds. We want to find its acceleration. We follow the usual procedure for this situation: find the average velocity, use this along with the final velocity to to determine the initial velocity, then find the rate of change of velocity with respect to clock time.
Its position is measured in revolutions.
The average rate of change of position with respect to clock time is therefore
ave rate = change in position / change in clock time = 8 rev / 20 sec = .4 rev / sec.
Revolutions are a measure of angle or angular displacement. The 8 revolutions therefore represent a change in angular position. The rate of change of angular position will be called angular velocity. Angular velocity is represented by the symbol omega (the Greek symbol that looks something like a W but should never be called 'double-yew' or written as 'w'; write it as 'omega').
So we say that
ave angular velocity = .4 rev / sec, or
omega_Ave = .4 rev / sec.
This is the 'average velocity'. To find the acceleration we have to find the initial velocity.
A linear velocity vs. t trapezoid tells us that since final vel is zero, the initial vel is double the average.
Our conclusion:
the initial angular velocity is .8 rev / sec, double the average angular velocity
the average rate of change of angular velocity with respect to clock time is therefore
ave rate = (0 rev / sec - .8 rev/sec) / (20 sec) = -.4 rev / sec^2.
This is what we call acceleration. Since it's the average rate of change of angular velocity, we call this the angular acceleration. Angular acceleration is represented by the symbol alpha.
Thus our conclusion is that
angular acceleration = alpha = `dOmega / `dt = -.04 rev/s^2.
Our conclusions:
omega_Ave = .4 rev / sec
omega_f = .8 rev / sec
alpha = -.04 rev / s^2.
We could express these quantities using degrees instead of revolutions. Since 1 revolution = 360 degrees, we have
omega_Ave = .4 rev / sec = .4 (360 deg) / sec = 144 deg / sec
alpha = -.04 rev / s^2 = -.04 * 360 deg / s^2 = -14.4 deg/s^2.
Or we could express these quantities in terms of radians. You might not be familiar with radians, and if not you should note and remember that 1 revolution = 2 pi radians. Using this fact
omega_Ave = .4 rev / sec = .4 * 2 pi rad / sec = .8 *pi rad / sec and
alpha = -.04 rev / s^2 = -.04 * 2 pi rad / s^2 = -.08 pi rad / sec.
How would someone who measured the motion in cycles of a 9 cm pendulum calculate these quantities? Let's see:
We know that for this length 1 cycle = .2 sqrt(9) sec = .6 sec.
We know omega_Ave in rev / sec. To get this quantity in rev/cycle we need to know what to substitute for 'sec'.
If 1 cycle = .6 sec, then 1 sec = 1 / .6 cycle = 1.67 cycle. So we can substitute 1.67 cycle for 'sec', obtaining
omega_Ave = .4 rev / sec = .4 rev / (1.67 cycle) = .24 rev / cycle.
alpha = -.04 rev / sec^2 = -.04 rev / (1.67 cycle)^2 = -.014 rev / cycle^2.
`q008. If someone observed that the wheel went through 10 revolutions in 50 cycles of a pendulum of unspecified length, then in units of revolutions and cycles, reason out the values of omega_Ave and alpha.
****
`dtheta= 10 revolutions
`dt= 50 cycles
omegaAve= `dtheta/`dt=10 revolutions/50 cycles= .2 revolution/cycle
since omegaF is 0, we can assume that omega0 is 2*omegaAve or .4 revolution/cycle
alpha= `domega/`dt= .4 revolution/cycle/50 cycles= .004 revolution/cycle^2
arithmetic: .4 / 50 = .008, not .004
Otherwise fine.
&&&&
`q009. If we later find that the pendulum in the preceding has length .8 cm, then what would be the values of omega_Ave and alpha in terms of revolutions and seconds, in terms of degrees and seconds, and in terms of radians and seconds?
****
One cycle of a .8 cm pendulum would be .2*sqrt.8= .18 seconds
Substituting this into our values for omegaAve and alpha:
OmegaAve=.2 revolutions/cycle= .2 revolutions/(.18 seconds)= 1.1 revolution/sec
Alpha= .004 revolution/cycle^2= .004 revolution/(.18 seconds)^2=.004 revolution/.0324 sec^2= .123 revolution/sec^2
Knowing that one revolution is 360 degrees,
OmegaAve= 1.1 revolution/sec= 1.1 (360 degrees)/sec= 396 degrees/sec
Alpha=.123 revolution/sec^2= .123 (360 degrees)/sec^2= 44.3 degrees/sec^2
Knowing that one revolution is 2pi radians,
OmegaAve= 1.1 revolution/sec=1.1(2pi radians)/sec= 6.9 radians/sec
Alpha= .123 revolution/sec^2= .123(2pi radians)/sec^2= .77 radians/sec^2
&&&&
An Atwood machine consists of a 20 kg mass on one side and a 22 kg mass on the other. Let the positive direction be that in which the 20 kg mass descends. The system therefore experiences forces of 20 kg * 9.8 m/s^2 = 200 N (approx.) in the positive direction, and 22 kg * 9.8 m/s^2 = 220 N in the negative direction, so the net force is 200 N - 220 N = -20 N. The total mass is 20 kg + 22 kg = 42 kg, so the acceleration of the system is
a = F_net / m = -20 N / (42 kg) = -.48 m/s^2.
We come to this conclusion without considering the tension in the string connecting the two masses. To figure out this tension we could consider just the 20 kg mass. The forces on this mass are the tension T, acting upward, and the weight (200 N) of the 20 kg mass, acting downward. Letting downward be the positive direction (consistent with our choice in the original problem) we see that
F_net = 200 N - T.
We know that F_net = m a, so we can say
m a = 200 N - T.
We also know that a = -0.48 m/s^2. Of course we know that m = 20 kg. So we can solve the equation for T:
T = 200 N - m a = 200 N - 20 kg * (-0.48 m/s^2) = 200 N - (-10 N) = 210 N.
That is, the tension is 210 N. We have already assumed that the tension acts upward (we did this by saying that the tension force is -T), so the tension is 210 N in the upward direction.
This makes sense. The net force on the 20 kg mass is 200 N - 210 N = -10 N, i.e., 10 N in the upward direction. Its acceleration is therefore -10 N / (20 kg) = -.5 m/s^2, within approximation error of the more accurate -.48 m/s^2.
`q010. We have seen that the acceleration of the system is -.48 m/s^2. Use this fact to find the tension acting on the 22 kg mass. Your reasoning will be similar to that used above, but some of the details will be different.
****
Fnet=ma and fnet=220N-Tension, so we can say:
Ma=220N-Tension
We know that a=-.48 m/sec^2 and m=22 kg so
(22 kg)(-.48 m/sec^2)= 220 N-Tension
Tension=220 N-(22 kg)(-.48 m/sec^2)
Tension=220 N+10.6 N
Tension=231 Newtons
&&&&
A 20 kg mass on a 10 degree incline, with the incline running down and to the right, experiences a gravitational force of about 200 N.
If an x-y coordinate system is imposed on this system, with the x axis down and to the right parallel to the incline, then the downward gravitational force will lie at angle 280 deg, measured counterclockwise from the positive x axis.
From the figure on the board we estimated that the y component of the 200 N weight is about -.95% of the weight, while the x component is about +30% of the weight. According to our estimates we thus estimated
wt_x = .30 * 200 N = 60 N
wt_y = -.95 * 200 N = -190 N
(ball-park estimated values).
Using the sine and cosine we can make these calculations more accurate:
wt_x = 200 N * cos(280 deg)
wt_y = 200 N * sin(280 deg)
(precise values, which will correct our estimates).
`q011. Using the sine and cosine what do you get for the x and y components of the weight? How far off were our ballpark estimates?
****
WtX=35
WtY=-197
Our estimate for the Y component was close, but our guess of 30% for the x component was too much.
&&&&
`q012. What are the x and y components of the weight of a 40 kg mass on a 20 degree incline?
Sketch and estimate, give your estimated percents, and the resulting x and y components of the weight.
Use sines and cosines to get the precise values of the components.
****
40kg(9.8 m/s^2)=392N
WtX estimate=30%=.3(392N)=118 N
WtY estimate=-90%=-.9(392N)=-353 N
Theta=290 degrees
WtX=392N cos(290)=134N
WtY=392Nsin(290)=-368N
&&&&
`q013. Continuing the preceding, if the normal force is equal to the y component of the weight and the frictional force resisting motion is 15% of the normal force, then what is the net force on the block? What therefore is its acceleration?
****
Fnormal=WtY=-368 N
Ffriction=.15WtY=.15(-368)=-58N
Fnet=Fconservative+Fnonconservative
Force of gravity was found earlier to be 392 N
Fnet=392N+(-368N)+(-58N)=-34 N
392 N, -368 N and -58 N do not all act along the same axis. They are the magnitudes of vectors and/or components acting along different directions and have to be added as vectors.
F=ma
A=F/m
A=-34kgm/sec^2 / 40kg=-.85m/sec^2
&&&&
Homework:
Your label for this assignment:
ic_class_091021
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.
Work through and submit q_A_15 and q_A_16. These qa's are concerned primarily with vectors, and you should find them straightforward.
URL's of qa's 10-19:
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
Read through Chapter 2 of your text to familiarize yourself with the author's notation, and to familiarize yourself with his examples and methods of solution. Just about everything you see should be familiar to you. The author does a good job explaining a lot of the things we've seen in class, and you'll find the text to be a very useful reference.
You should take a good look at Introductory Problem Sets 3 and 4. The Phy 121 test, probably to be assigned the week after next, will consist mostly of problems from these sets. A number of these problems will also appear on the Phy 201 test.
Appended copy with solutions:
class 091021
Class 091021
In class we observed a steel ball rolling down an incline
with slope approximately .03, colliding with a stationary marble, with both
objects then falling 90 cm to the floor.
- The steel ball, when unimpeded by the marble,
traveled 8.5 cm while falling.
- The marble (which was initially stationary), after
being hit by the ball, traveled 16 cm while falling.
- The steel ball after hitting the marble traveled 5.5
cm while falling.
- You should be able to figure out the time of fall,
assuming the initial vertical velocity to be close to zero.
Approximating the time of fall by .4 sec we conclude that
the ball reached the end of the ramp at a speed of about 20 cm/s, the marble
traveled at 40 cm/s after being struck, and the steel ball was moving at about
12 cm/s after striking the marble.
`q001. Using the above figures:
- What is the change in the ball's velocity between the
instant before and the instant after it strikes the marble?
- What is the change in the marble's velocity the
instant before and the instant after the collision?
- What do you therefore conjecture to be the ratio of
the ball's mass to that of the marble?
****
The ball's velocity just before collision is its velocity
20 cm/s at the end of the ramp. After collision its velocity is 12 cm/s,
so its velocity changed by -8 cm/s.
The marble was initially at rest. As a result of the
collision it velocity changed to 40 cm/s.
The two objects exert equal and opposite forces on one
another during the time interval during which they are in contact. So they
have equal and opposite momentum changes.
For these two objects we therefore have
m1 `dv1 = -m2 `dv2 so that
m2 / m1 = - `dv1 / `dv2 and, letting object 1 be the
marble and object 2 the steel ball,
m2 / m1 = - (40 cm/s) / (-8 cm/s) = 5.</h3>
&&&&
`q002. Using your own data for the same experiment,
what are the following:
- the unimpeded steel ball's velocity at the end of the
ramp
- the velocity of the glass marble the instant after
being struck by the ball and
- the velocity of the ball immediately after striking
the marble?
What therefore should be the ratio of the ball's mass to
that of the marble?
Be sure to include your raw data and explain how you get
from the raw data to your conclusions.
****
<h3>Your analysis will be similar to that of the
example.</h3>
&&&&
Let's assume that we can neglect forces exerted by the
ramp on the glass marble and the steel ball, and that we can in fact neglect all
forces except the forces exerted by each ball on the other. We can
therefore regard the two balls, during the time of collision, as a closed
system. We have draw our conclusions about the velocities involved in the
collision from the projectile behavior of the objects after collision.
From F_net `dt = `dp and the fact that the forces exerted
by the two objects are equal and opposite, we can draw the conclusion that the
momentum change `dp of one object should be equal and opposite to that of the
other. (recall that p = m v, so `dp = p_f - p_0 = m vf - m v0)
Since momentum changes are equal and opposite, we conclude
that the total momentum of the system is the same immediately after collision as
it was immediately before.
`q003. Let's let m1 and m2 be the two masses, let v1
and v2 be their velocities before the collision, and let u1 and u2 be their
velocities after collision.
- What are the values of v1, v2, u1 and u2 for your
trial?
****
For the example, we have
- v1 = 0
- v2 = 20 cm/s
- u1 = 40 cm/s
- u2 = -8 cm/s.
&&&&
`q004. Using the symbols m1, m2, v1, v2, u1 and u2,
give the symbolic expression for each of the following:
- The momentum of the steel ball before collision.
- The momentum of the glass marble before collision.
- The momentum of the steel ball after collision.
- The momentum of the glass marble after collision.
- The total momentum of the two objects before
collision.
- The total momentum of the two objects after
collision.
****
<h3>
- The momentum of the steel ball before collision is m2
v2.
- The momentum of the glass marble before collision is
m1 v1.
- The momentum of the steel ball after collision is m2
u2.
- The momentum of the glass marble after collision is
m1 u1.
- The total momentum of the two objects before
collision is m1 v1 + m2 v2.
- The total momentum of the two objects after
collision is m1 u1 + m2 u2.
</h3>
&&&&
`q005. Write the symbolic equation which states that
the total momentum of the two objects before collision is equal to their total
momentum after collision. (your equation will set your answers the last
two expressions in the preceding question equal to one another)
****
<h3>The equation is
m1 v1 + m2 v2 = m1 u1 + m2 u2.</h3>
&&&&
`q006. Substitute the values you observed for v1,
v2, u1 and u2 into your equation. Your equation will still have the
symbols m1 and m2, but the other symbols will be replaced by the various
velocities you observed.
****
<h3>For the given example the equation would be
m1 * 20 cm/s + m2 * 0 = m1 * 12 cm/s + m2 * 40 cm/s.</h3>
&&&&
`q007. Rearrange your equation so that all the
quantities containing m1 are on the left-hand side and all the quantities
containing m2 are on the right-hand side (your equation should have been m1 v1 +
m2 v2 = m1 u1 + m2 u2).
Factor m1 out of the left-hand side and m2 out of the
right-hand side.
Solve the equation for m1. The left-hand side should
end up just as m1; the right-hand side will include the symbol m2.
Give the steps of your solution:
****
<h3>m1 * 20 cm/s + m2 * 0 = m1 * 12 cm/s + m2 * 40 cm/s
simplifies to
m1 * 20 cm/s = m1 * 12 cm/s + m2 * 40 cm/s. Dividing
both sides by the unit cm/s we get
m1 * 20 = m1 * 12 + m2 * 40, or
20 m1 = 12 m1 + 40 m2. Rearranging we get
20 m1 - 12 m1 = 40 m2 so that
8 m1 = 40 m2 and
m1 = 40 m2 / 8 or
m1 = 5 m2.</h3>
&&&&
The foam wheel was observed to rotate through 8
revolutions, coming to rest after about 20 seconds. We want to find its
acceleration. We follow the usual procedure for this situation: find
the average velocity, use this along with the final velocity to to determine the
initial velocity, then find the rate of change of velocity with respect to clock
time.
Its position is measured in revolutions.
The average rate of change of position with respect to
clock time is therefore
- ave rate = change in position / change in clock time
= 8 rev / 20 sec = .4 rev / sec.
Revolutions are a measure of angle or angular
displacement. The 8 revolutions therefore represent a change in angular
position. The rate of change of angular position will be called angular
velocity. Angular velocity is represented by the symbol omega (the Greek
symbol that looks something like a W but should never be called 'double-yew' or
written as 'w'; write it as 'omega').
So we say that
- ave angular velocity = .4 rev / sec, or
- omega_Ave = .4 rev / sec.
This is the 'average velocity'. To find the
acceleration we have to find the initial velocity.
A linear velocity vs. t trapezoid tells us that since
final vel is zero, the initial vel is double the average.
Our conclusion:
- the initial angular velocity is .8 rev / sec, double
the average angular velocity
- the average rate of change of angular velocity with
respect to clock time is therefore
- ave rate = (0 rev / sec - .8 rev/sec) / (20 sec) =
-.4 rev / sec^2.
This is what we call acceleration. Since it's the
average rate of change of angular velocity, we call this the angular
acceleration. Angular acceleration is represented by the symbol alpha.
Thus our conclusion is that
- angular acceleration = alpha = `dOmega / `dt = -.04
rev/s^2.
Our conclusions:
- omega_Ave = .4 rev / sec
- omega_f = .8 rev / sec
- alpha = -.04 rev / s^2.
We could express these quantities using degrees instead of
revolutions. Since 1 revolution = 360 degrees, we have
- omega_Ave = .4 rev / sec = .4 (360 deg) / sec = 144
deg / sec
- alpha = -.04 rev / s^2 = -.04 * 360 deg / s^2 = -14.4
deg/s^2.
Or we could express these quantities in terms of radians.
You might not be familiar with radians, and if not you should note and remember
that 1 revolution = 2 pi radians. Using this fact
- omega_Ave = .4 rev / sec = .4 * 2 pi rad / sec = .8
*pi rad / sec and
- alpha = -.04 rev / s^2 = -.04 * 2 pi rad / s^2 = -.08
pi rad / sec.
How would someone who measured the motion in cycles of a 9
cm pendulum calculate these quantities? Let's see:
- We know that for this length 1 cycle = .2 sqrt(9) sec
= .6 sec.
- We know omega_Ave in rev / sec. To get this
quantity in rev/cycle we need to know what to substitute for 'sec'.
- If 1 cycle = .6 sec, then 1 sec = 1 / .6 cycle = 1.67
cycle. So we can substitute 1.67 cycle for 'sec', obtaining
- omega_Ave = .4 rev / sec = .4 rev / (1.67 cycle) =
.24 rev / cycle.
- alpha = -.04 rev / sec^2 = -.04 rev / (1.67 cycle)^2
= -.014 rev / cycle^2.
`q008. If someone observed that the wheel went
through 10 revolutions in 50 cycles of a pendulum of unspecified length, then in
units of revolutions and cycles, reason out the values of omega_Ave and alpha.
****
omega_Ave is ave. angular velocity, which is ave rate of
change of angular position with respect to clock time, so
- omega_Ave = (change in angular position) / (change in
clock time) = 10 revolutions / (50 cycles) = .2 rev / cycle.
Assuming uniform angular acceleration, since the system
ends up at rest its initial angular velocity will be double its average angular
velocity so
- omega_0 = .4 rev / cycle.
Angular acceleration is ave rate of change of angular
velocity with respect to clock time, so
- alpha = (change in angular velocity) / (change in
clock time) = (0 - .4 rev/sec) / (50 cycles) = -.008 rev / cycle^2.
&&&&
`q009. If we later find that the pendulum in the
preceding has length .8 cm, then what would be the values of omega_Ave and alpha
in terms of revolutions and seconds, in terms of degrees and seconds, and in
terms of radians and seconds?
****
A pendulum of length .8 cm has period .2 sqrt(.8) sec =
.18 sec.
A revolution corresponds to 360 degrees, or to 2 pi
radians.
So the average angular velocity is
omega_Ave = .2 rev / cycle = .2 rev / (.18 sec) = 1.1 rev
/ sec, which in turn can be written
1.1 rev / sec = 1.1 (360 deg) / sec = 400 deg / sec, or as
1.1 rev / sec = 1.1 ( 2 pi rad) / sec = 2.2 pi rad / sec.
2.2 pi rad / sec can be approximated as 2.2 * 3.14 rad /
sec = 6.9 rad / sec.
&&&&
An Atwood machine consists of a 20 kg mass on one side and
a 22 kg mass on the other. Let the positive direction be that in which the
20 kg mass descends. The system therefore experiences forces of 20 kg *
9.8 m/s^2 = 200 N (approx.) in the positive direction, and 22 kg * 9.8 m/s^2 =
220 N in the negative direction, so the net force is 200 N - 220 N = -20 N.
The total mass is 20 kg + 22 kg = 42 kg, so the acceleration of the system is
- a = F_net / m = -20 N / (42 kg) = -.48 m/s^2.
We come to this conclusion without considering the tension
in the string connecting the two masses. To figure out this tension we
could consider just the 20 kg mass. The forces on this mass are the
tension T, acting upward, and the weight (200 N) of the 20 kg mass, acting
downward. Letting downward be the positive direction (consistent with our
choice in the original problem) we see that
We know that F_net = m a, so we can say
We also know that a = -0.48 m/s^2. Of course we know
that m = 20 kg. So we can solve the equation for T:
- T = 200 N - m a = 200 N - 20 kg * (-0.48 m/s^2) = 200
N - (-10 N) = 210 N.
That is, the tension is 210 N. We have already
assumed that the tension acts upward (we did this by saying that the tension
force is -T), so the tension is 210 N in the upward direction.
This makes sense. The net force on the 20 kg mass is
200 N - 210 N = -10 N, i.e., 10 N in the upward direction. Its
acceleration is therefore -10 N / (20 kg) = -.5 m/s^2, within approximation
error of the more accurate -.48 m/s^2.
`q010. We have seen that the acceleration of the
system is -.48 m/s^2. Use this fact to find the tension acting on the 22
kg mass. Your reasoning will be similar to that used above, but some of
the details will be different.
****
For this mass, F_net = m a = 22 kg * .48 m/s^2 = 10 N,
approx..
The weight of the 22 kg mass is 220 N, approx..
- net force = weight + tension, so that
- 10 N = 220 N + tension. We easily rearrange
this to obtain
- tension = 10 N - 220 N = -210 N.
&&&&
A 20 kg mass on a 10 degree incline, with the incline
running down and to the right, experiences a gravitational force of about 200 N.
If an x-y coordinate system is imposed on this system,
with the x axis down and to the right parallel to the incline, then the downward
gravitational force will lie at angle 280 deg, measured counterclockwise from
the positive x axis.
From the figure on the board we estimated that the y
component of the 200 N weight is about -.95% of the weight, while the x
component is about +30% of the weight. According to our estimates we thus
estimated
- wt_x = .30 * 200 N = 60 N
- wt_y = -.95 * 200 N = -190 N
- (ball-park estimated values).
Using the sine and cosine we can make these calculations
more accurate:
- wt_x = 200 N * cos(280 deg)
- wt_y = 200 N * sin(280 deg)
`q011. Using the sine and cosine what do you get for
the x and y components of the weight? How far off were our ballpark
estimates?
****
We get
- wt_x = 200 N * cos(280 deg) = 200 N * .17 = 34 N
- wt_y = 200 N * sin(280 deg) = 200 N * (-.98) = -196
N.
&&&&
`q012. What are the x and y components of the weight of a 40 kg
mass on a 20 degree incline?
Sketch and estimate, give your estimated percents, and the
resulting x and y components of the weight.
Use sines and cosines to get the precise values of the
components.
****
The weight is 40 kg * 9.8 m/s^2 = 400 N, approx..
If the figure is sketched in a manner analogous to the
preceding, the weight vector makes angle 270 deg + 20 deg = 290 deg. The
components are therefore
- wt_x = 400 N * cos(290 deg) = 400 N * .34 = 140
N and
- wt_y = 400 N * sin(290 deg) = .400 N * (-.94) = - 380
N (both final results very approximate)
&&&&
`q013. Continuing the preceding, if the normal force is equal to the y component of the
weight and the frictional force resisting motion is 15% of the normal force, then what is the net force
on the block?
What therefore is its acceleration?
****
The y component of the weight is -380 N so the normal
force will be + 380 N in the positive y direction.
The normal force is therefore 380 N.
15% of the normal force is 380 N * .15 = 57 N, and is
directed opposite the motion.
Assuming the motion to be down the incline, then
frictional force is in the negative x direction and the net force, which is is
in the x direction, will be
- F_net = wt_x + f_frict = 140 N + (-57 N) = 90 N,
approx.
The acceleration will be
- a = F_net / m = 90 N / (40 kg) = 2.25 m/s^2.
&&&&
Good work overall. See my notes and the appended document for comments and details.