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Topic: HW5a #2
Jill of the Jungle swings on a vine 7.1 m long. What is the tension in the
vine if Jill,
whose mass is 61 kg, is moving at 2.5 m/s when the vine is vertical?
Student A: This is the one where the girl is swinging on the vine and you
need to find
the tension on the rope given the length of the rope, speed when she's
vertical and her mass.
I know you need to solve for the acceleration, but I'm not sure how to do it
or what
I'm doing wrong. Does the problem require separating x and y components?
Dr. Man: Good question: Does the problem require separating x and y
components?
It is always good practice to do things step by step. When the string is
vertical,
1, identify all the forces on the girl
2, chooce x,y directions for that position.
3, find forces and force components if any.
4,Solve Fnetx = m ax;
Fnety =m ay
Just notice that for circular motion, what we really care most now is the
total force in the
radius direction pointing to the orbit center.
We call it Fnet_r or Fcp and we know the total foce in the radius direction
pointing to
the center = m acp
Or, Fnet_r = m v^2 /r
When you solve this eqation you get the answer.
But you need an correct equation. That includes, finding the correct total
force in radius
direction pointing to the center, and using the correct v and r.
Notice that when the rope is vertical, the direction pointing to the center
of the orbit is the
same direction as vertical y direction.
Student B: So are you saying we simply plug the numbers into Fnetr = mv^2/r??
because
that's what I did and did not get the right answer.. it seems to make the
most sense to
me because the mass, velocity and radius are given. It kept saying I had an
order of
magnitude issue.
Dr. Man: Your answer is way off then. Keep in mind that Fnetr is not tension.
Always identify ALL the forces on the object,
find forces and force components in radius direction.
Your Fnetr needs to include not only tension but all other forces (gravity
etc.) or their
components along the r direction.
from ur drawing u see clearly Fnetr is the sum of which forces.
then solve unknown forces.
Topic: hw 5b #2
Mass A (3 kg) and mass B (2 kg) sit on the ground, 1 m apart.
a.) What is size of the the gravitational force:
on A from B?
on B from A?
Student A: For this particular problem, I tried using the equation
Fg= G (m1m2/r^2) and am not getting the right answer. Is there something I am
missing here with the first part?
Dr. Man: Check the lecture notes and your text book, chapter 12.
G is an extremely small number.
G is not g.
You know that the gravity attraction between the huge earth and 2kg object is
only 19.6 newtons
We expect Fg between 3kg and 2 kg to be extremely small. Less than 10^(-9)
Newtons.
Real lecture notes one more time.
Student B: I used exercise 12-1 (pg.380) as a guide to doing this problem but
I'm
still not getting the answer. I got part B already but am still having
trouble finding part A
Dr. Man: Follow the lecture notes.
This problem is completely solved in the lecture notes already.
Part A is straight forward, the gravitational force between mass 1 and mass 2
is Fg= G
(m1m2/r^2)
Use right number for G, r, m, m2 and don't forget r square.
Topic: HW5a #5
A hockey puck of mass m is attached to a string that passes through a hole in
the
center of a table, as shown in Figure 6-46. The hockey puck moves in a circle
of
radius r. Tied to the other end of the string, and hanging vertically beneath
the table,
is a mass M. Assuming the tabletop is perfectly smooth, what speed must the
hockey
puck have if the mass M is to remain at rest? (Use g for acceleration due to
gravity,
and m, r, and M as necessary.)
Dr. Man: In this question, you answer an expression with g, m r and M in it.
Basically, you need to set equation for the mass on the table.
Fnet r =T= mv2/r.
You need to find out what T is by looking at the forces on the big M under
the table.
Keep in mind that the big M remains at rest, so that the total force on it
must be ____?
Then you know Tension.
Then you solve v. v=Blahblah m g r M..., just write the right side in the
box.
Topic: HW5b #1
two masses a distance R apart attract each other with a gravitational force
F. For each of the
cases below, give the factor by which F is changed if:
a) if R is multiplied b 2, then F new = _____ x F old
b) if R is multiplied by 2.6, then F new = _____ x F old
c) if R is multiplied by 3, then F new = _____ x F old
Student A: I don't seem to understand how to answer the question.. I'm sure
that
its much simpler than what I am doing but I don't know how to go about
answering this.
Dr. Man: This question is also solved in lecture notes and during the
classes.
Basically, between mass1 and mass2, is the distance is r, the gravitational
attraction force is
Fg_old= G m1 m2 / R^2
Suppose that m1 and m2 do not change, but the distance between them is
multiplied by 2,
Now, the new Fg becomes how many times of the old Fg?
The new Fg should be more or less than the Old Fg?
If more how many times?
If less, by a factor of what?
Fnew = a number * Fold, You just need to find that number
Topic: HW 5b #3
In this problem, be especially wary of round-off; keep 2-3 (non-zero) digits
in each answer and
throughout your calculations.
Based on the following data about planet X (which orbits around the Sun):
Planet X's distance from Sun = 4.4*1012 m
Planet X's radius = 1*106 m
Planet X's mass = 6.8*1022 kg
a.) Find gx, the size of the acceleration due to gravity on the surface of
Planet X.
b.) What is the weight of a 10 kg mass on the surface of Planet X?
(How does this compare to its weight on Earth?)
c.) How long would it take for a ball dropped from a height of 4 m to hit the
ground?
(How does this compare to the time it would take on Earth?)
d.) At 1 of Planet X's radii above the planet's surface, what is gx?
e.) Find the orbital speed of Planet X around the Sun.
f.) How long is a year on Planet X? Express your answers in both seconds and
Earth years:
Topic: Hw 5b #3 Part A
Student A: Find g_x, the size of the acceleration due to gravity on the
surface of planet X.
For this one, I used the equation of universal gravity, plugged in the masses
of both planets
, constant G, and for the r-- the radius of both planets plus the distance
between them.
Therefore, I got F_g. To get acceleration, that means i have to use F=ma and
plug in F_g
as F and m as the mass of planet x to get a, acceleration. But I got it
wrong.
Am I doing something wrong? I'll try to look at the lecture notes again.
Dr. Man: To find the gravitational force Fg between the sun and planets, you
use the
mass of the sun and the planet and the distance between the sun and the
planet.
(no need to add the radius of planet or sun), It is too small comparing with
the distance
between the planet and the sun. When the distance between the sun and the
planet is given
it already means the center to center distance. To find the gravitational
force Fg between
the planet and a object m1 on the planet, you use the mass of the PLANET and
m1, and the
distance between the mass center of the planet and the object, which is the
radius of the
planet. g_x by definition is the gravitational force between the planet and
m1, divided by m1
Fg (between planetX to m1) = m1 *g_x
g_x=Fg (between planetX to m1)/m1
You then find that m1 cancels and for different m1, you get the same g_x for
one planet.
Try this for Earth first, See lecture notes for detailed calculations, you
should expect
Earth g_x = 9.8 N/kg
Then do it for the other planet
Student B: I keep getting the same answer for this problem but webassign says
its wrong.
When asked what is g_x, i use the universal gravity multiplied by the mass of
planet x,
and then divide by the distance between the sun and the planet and square it.
Dr. Man: think twice which distance you should use.
read my above explanation and lecture notes again.
Fg (between planetX to m1) = m1 *g_x
Does Fg (between planetX to m1) has anything to do with the sun? and the
distance to the sun?
How did we get g_earth= 9.8 N/kg on earth?
Which r to use?
Student C: I am having a really hard time with this problem....I am using the
equation (G*mass of sun*mass of planet)/(distance from the sun^2) what am
I doing wrong???
Dr. Man: Depending on which part of the question you are talking about.
Considering the force between the sun and the planet (how much does the sun
attracts
the planet, you use G*mass of sun*mass of planet)/(distance from the sun^2.
But considering an object m1on the surface of planet X, what is the force the
planet
acts on the object? That force divided by m1, is the g in that planet. On
earth, 1 kg object
weights 9.8 N. The earth attracts the 1kg mass with F=m * g_earth
g_earth =9.8N/kg =9,8m/s^2. Where did this g_earth come from? Actually The
earth
attracts the mass m with
F=G* m_earth*m/R^2 = m *g_earth
Notice that m cancels in both sides.
So g_earth=F/m= G* m_earth/R^2
Similarly another planetX the planet attract the mass m with F = ???? =m g_X
If you write the force between the planet nd mass m correctly, you can find
g_X easily.
All these have been explained during the lecture. And are also written in the
lecture notes
Topic: hw 5b #3 Part E
Student A: I tried to set the force of gravity between the sun and the planet
(G*m1*m2)/r^2
to the centripetal force (v^2/r) as it seems like these would beequal and
opposite but I
keep getting the velocity wrong. Is there a different way to approach this?
Dr. Man: there is no equal and opposite.
Total force on the planet pointing to the center of the orbit IS EQUAL to
gravity between the
sun and the planet. (there is no another force going along the opposite
direction)
Total Fr =m v^2/r
2, Don't forget m in the above equation?
which m do you use
which r do you use?
Student A:
G = gravitational constant
m1 = mass of planet
m2 = mass of sun
r = distance between sun and planet
v = speed of planet?
I meant that they were equal, not opposite sorry.
Dr. Man: good this is right. don't forget that you have
mass *v^2/r on your other side of the equation.
Part F
Student A: I’ve looked at the lecture notes several times and went to a help
session.
I am still getting the wrong answer.
Is T=2pi(distance from planet to sun) /orbital speed ?
Is T already in seconds, and how do you convert it to Earth years?
Dr. Man: Good question. Yes. T=2pi(distance from planet to sun) /orbital
speed
And then one earth year =365 days
1 day =24 hr
1hr =3600 s
You then know how to convert your seconds into earth year
Student A:
have a strong feeling i am getting the answers right for question 2 and 3,
but how i am entering them
the computer is not accepting. for an example does the system understand
1.23e17
1.23x10^5
1.23*10^5
does webassign understand this notation? or am i do the problem wrong, for
part b of 2 and and part of
3, iam getting answers that require this notation, is that correct, i am
using the universal gravity
equation, with g times mass of sun, mass of planet x, divided by difference
between the 2 plants squared
is that right?, that for g_x. I am also using that equation to find out the
2b also.
Dr. Man:
yes. as the first assignment Webassign introduction showed, webassign
understand 1.23e17 , but not
1.23*10^17
If it is 1.48*10^-10 you should write 1.48e-10
However, your answer were really wrong.
You get huge numbers, maybe because you used the huge mass of the earth.
Don't forget the /R^2
R is the distance between mass center of the earth to the mass center of the
object, in this case it is the
radius of the earth.
See lecture notes.
On the other hand, previously, we already learned that a 1kg object is
attracted to earth for 9.8 newtons.
mass m is attracted to earth for mg newtons.
When you get numbers like e17, for gravity force on earth, think about
whether it is possible.
Remember we talked about estimate order of magnitude in chapter one.
Similarly, if g-x in another
planet is not around 9.8, but e5, how fast things are going to fall in that
planet? is that possible?
BTW, start earlier and allow enough time to understand HW
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