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.