Your post is difficult to understand. DeMorgans shows how the negation distributes over the conjunction and disjunction.
For example
~(A & B) <--> (~A V ~B)
This makes sense. The left hand side says
"It is not the case that both A and B are true."
The right hand side says,
"Either A is...
Right, as long as the planet isn't also rotating. You can consider it rotating about an axis through the Sun. Then you have
K=\frac{1}{2}mv^2=\frac{1}{2}mR^2\left(\frac{v^2}{R^2}\right)=\frac{1}{2}I\omega^2
But if in addition the planet spins, it has an additional kinetic energy due to its...
Ok, so in conclusion,
W_{net,real}=\frac{1}{2}\left(I\omega_f^2+mV_f^2\right)-\frac{1}{2}\left(I\omega_0^2+mV_0^2\right)
And
W_{net,real}=W_{trans}+W_{rot}
where
W_{trans}=\int_{\vec{x}_0}^{\vec{x}_f}\vec{F}(x)\cdot d\vec{x}and the limits of integration are the final and initial...
Ok, so I'm still kinda confused. What is the cylinder was only able to rotate. Then
W=\int_{0}^{L/R}FRd\theta=FL=\DeltaKE
In this case, the change in KE is only due to rotational motion.
What if the cylinder could only move translationally.
Then if the force acts over a distance L...
I figured it as such:
The rope has length L and the disc radius R. So the angle over which the torque acts will be (L/(2piR))(2pi)=L/R. So taking ω0=0,
\omega_f=\sqrt{\frac{2}{I}\int_{0}^{L/R}FRd\theta}=\sqrt{\frac{2FL}{I}}
I did this by imagining the disc as a pulley. I then wanted...
I was a bit confused at first as to why you asked the same question twice and got different results. I think I figured it out. You meant
The question "what is the probability of exactly three sixes?" is, as Doug said, 1 out of 216:
\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \approx...
Are you aware that the force for gravitation is
\vec{F}=-\frac{Gm_1m_2}{r^2}\hat{r}
You have two masses m and 4m and some unknown mass m_2, so the magnitude of the gravitational attraction between m and m_2 is
F_1=\frac{Gmm_2}{r_1^2}
where r_1 is the distance between m_2 and m. The...
"a third object would experience no gravitation force."
what they mean is, no net gravitation force
imagine a particle C lying on the line connecting A and B - what is the force that C feels from A? what is the force that C feels from B?
What is the net force?
Fr=I(a_1-a_2)/r
F=ma_2
(ma_2)r^2=I(a_1-a_2)
a_2(m+I)r^2=Ia_1
a_2=Ia_1/(m+I)
I believe this to be the correct solution, and this is in the rest frame of the ground.
I'm not quite sure what to put in for I, though.
Can you give specific example and work it both ways? That is, first considering the quantities are representing fractions of the total work and then the quantities representing the whole work. I can see what you're saying, but I don't see how to apply it to a problem.
I asked this...
Are these quantites:
W=\int_{\theta_0}^{\theta_f}{\tau}{d}\theta=\frac{1}{2}I(\omega_f^2-\omega_0^2)
W=\int_{r_0}^{r_f}Fdr=\frac{1}{2}m(v_f^2-v_0^2)
the same or different?
Is "v" in the second equation the speed of the center of mass? IOW, does the bottom integral give the change in...
I know. I've found at least four in the last two chapters, and I've verified these with my professor, so it's not just stupid Stephen being less smart than the author. The book is PHYSICS by Ohanian 2ed.
You must consider torques about the axis through the center.
The net external torque is
m_1R - m_2r=I_{net}\alpha
so
\alpha=\frac{m_1R - m_2r}{I_{net}}
where
I_{net}=\frac{1}{2}MR^2+\frac{1}{2}mr^2+m_1R^2+m_2r^2
you know r=1/2R, so you should be able to find the relation between m and M...
The book and I aren't getting along tonight. Maybe you can help.
A rope of length L has a tension T. Someone pushes on the rope with a force F at its midpoint and deflects the rope by a distance d. What is T is terms of L,d and F.
This is so simple I won't even explain my work...
A 24m nylon rope has a tension of 1.3x104N. The total mass of the rope is 2.7kg. If a wave pulse starts on one end, how long does it take to reach the other end.
I get 0.07s
v=\sqrt{\frac{T}{\mu}}
t=d/v=d\sqrt{\frac{\mu}{T}}
\mu=\frac{m}{l}=\frac{2.7kg}{24m}...
But if it spins about an axis, it must spin about any parallel axis. That's what's throwing me off. The angular acceleration should be the same about any parallel axis. If I use the CM as a axis, then I get F=ma, FR=Ia/R, eliminating F, I get maR=Ia/R, but then a drops out!
Ok, I am still interested in knowing how you so easily knew what the projections must be without any apparent calculation. I got these same results, but I had to calculate the CM one at a time and then add the results together to find the projections.
A bowling ball sits on a level floor of a subway car. If the car has a horizontal acceleartion a, what is the acceleration of the ball wrt the ground? Ball rolls w/o slipping.
The forces that act on the ball are its weight, a normal force, and static friction. The weight and normal offset...
This question makes no sense. A ball doesn't travel with force.
I agree with Jason's answer. The 8 should go off at a 45 degree angle to the initial velocity of either ball with speed v*sqrt2 where v is the initial speed of the 6 ball.
Five identical wood blocks of sides L and thickness H are shifted in one direction to form a leaning tower of the maximum protrusion. How should you stack the blocks to achieve the maximum protrusion? What is the maximum protrusion? What if you had an infinite number of blocks?
For any...
by cons of p,
m_1v_{1i}=m_1v_{1f}-m_2v_{2f}
where v_2f is reckoned as negative
By the energy conditions,
m_1v_{1i}^2=\frac {1}{2}(m_1v_{1f}^2+m_2v_{2f}^2)
which gives two equations with two unknowns.
Yes that helps a lot. Do you use - on the botton when they are moving apart and + on the top when they are moving apart?
Also, can you clearly define f and f'?
f is the frequency as observed by the source?
f' is the frequency as observed by the detector?
Also, these speed v_d and v_s are...
My book derives two formulas for Doppler shift. One for when the source moves and one for when the observer moves.
What about when both are moving?
I tried deriving it myself... but I couldn't :(
If you want, you could just give me a hint on how to derive it.
I still don't see how a driving force without damping won't cause the amplitude to keep increasing. Are you not continually doing positive work on the system and thus adding energy to the system?
No, I'm just given that a driving force is acting on the system. In my book it talks about driving forces in the context of couteracting the effects of damping. In class (where the above equation came from), we talked about driving forces without damping. But without damping the amplitude...
It is very hard for me to understand something if I cannot derive it. I'm going to have to dig through my calc book to do this one.
I think in any event my professor is more concerned with the qualitative features of a driving force.
This is very confusing.
Is F always acting to pull the...
If I understand #2 correctly
Consider only x positive.
A=bh=(2x)(y)=(2x)(6-x2)=12x-2x3
dA/dx=12-6x2=0
12=6x2
x=sqrt2
This is the only solution (if x were negative, you'd get the same area) so it's probably a max. Check just to be sure.
d2A/dx2=-12x<0
yup a max
(by the way x=-sqrt2...
If you solve the Diff Eq with either an inital displacement or velocity, in addtion to the driving function, then you shold see some terms representing non driven oscillations.
Ok, That's relieving. So it does reduce.
How would you write the equation of motion with an initial...
For forced oscillations we have
\frac {d^2x} {dt^2} = -\omega_N^2x+\frac {F_0} {m} cos\omega_Ft
The solution is
x(t)=\frac {F_0} {m(\omega_N^2-\omega_F^2)}cos\omega_Ft
This doesn't seem to reduce to oscillations where F0=0. Shouldn't it?
Draw a picture! It should be clear. What does h(x)=-h(-x) mean? What does a definite integral represent if h(x) is positive?
Think of h(x)=x3 for instance. This should make the answer clear.
So its acceleration relative to the table is zero. And since the initial velocity of that point was zero, at all successive instants, the point in contact with the surface should be at rest relative to the table. And thus no slipping because slipping would imply movement relative to the table...
Hi, I'm trying to understand what's going on here.
When you write F=Ma, a is the acceleration of the center of mass. When you write F=I α=I a/R, a is the linear acceleration of a point at a distance R from the axis of rotation (as if the ball were not moving at all). So these are not...
This advice isn't entirely correct. I didn't take into account the pulley's mass.
Try solving these for a.
m1a=m1g-T1
m2a=T2
(T2-T1)R=Ia/R=(1/2)MRa
DocAl's approach might work also, but I would avoid that direction simply because I don't feel comfortable using it.