One way might be to place the center of the triangle at the origin and orient the triangle such that
the instantaneous velocity of Vc is along the y-axis.
Then resolve the velocities into their x-y components.
Note that then VBy = VAy, VCx = 0, and VAx = - VBx at that instant in time
if the...
I get h / L = .0174 for the vertical distance fallen..
If that is the case, then what does 3.2 m have to do with the problem
unless that is used to determine the starting angle.
I missed the fact that the vine is 18 meters long so you could find the initial angle
using 18 meters for the length of the vine and the vertical drop from that
point is 3.2 m to the bottom of the swing.
(Then, of course, he really wouldn't drop 3.2 m if the vine breaks
before he reaches the...
I agree with this answer.
Your method of balancing torques is interesting - I mean the term (Tsin30⋅4).
Usually, one would include the force at the pivot point of the bar and then eliminate this
force when balancing the translational forces.
You have apparently done this indirectly.
The moment of inertia I of a body about any axis is equal to the moment of inertia
I CM of the body about a parallel axis through its center of mass plus the mass M of the
body times the square of the perpendicular distance L between the axes:
I = Icm + M L^2.
For a cylinder about an edge I =...
Taking torque about the bottom point is a convenient way of eliminating
the frictional force.
Are you familiar with the parallel axis theorem?
You can also apply Newton's equations about the center of mass
to eliminate the frictional force.
Static electricity has a way of dissipating.
Try to avoid any sharp points (pointed ends of connecting wires, etc.)
Alligator clips should work OK.
You might try folding the edges of the foil strips to get rid of the sharp edges.
Also, the bowl would have to be clean to prevent leakage between...
You solved for Q using D/2 and V.
Why can't you just use E = k Q / r^2 where r is the distance from the
center of the sphere to a point external to the sphere?
Maybe it would help to resolve the issue if you calculated the "contracted" distance that
each ship travels and then the time to traverse this distance.
Use the 100 min. that you calculated as measured by Liz.
Consider the number of wavelengths of the light as it passes thru the glass
as compared to the number of wavelengths of light as it passes thru
an equivalent thickness of air.
In my earlier post I implied that use of a spring scales would show discontinuities in the motion.
Kinematics seems to indicate that this is not the case and a scales would be
of little or no use.
As indicated in post #8 other factors must be at work here such as
(work in bending the cords...
What if you replaced photon with "2 protons with speeds of say .999999 c"?
It seems that is what the question implies ( reference frames moving at speed c).
The moment of inertia for a solid cylinder of radius R2 is 1/2 m R2^2.
If you remove a cylinder of radius R1 from the solid cylinder then you are
left with 1/2 m (R2^2 - R1^2) which is what I assumed was meant by a "thick walled cylinder".
You wrote: Then the moment of inertia for a thick walled cylinder =
Don't you mean : 1/2m(r2² - r1²)?
Also, you could solve the appropriate equations in terms of I = M k^2
where k is known as the radius of gyration.
Then you could conveniently compare solutions for various moments of inertia...
But it seems if you place one end of the chain at the origin then you have to divide by n (infinity) to
get the potential energy per ion. But if you place the middle of the chain at the origin then you have the
difference of two infinite series that proceed as:
1/2 + 1/4 + 1/6 ........ 1/2n and...
What if you take one charge out of the infinite chain and then consider the work done
in adding the charge back into the chain?
For the first two adjacent charges : W1 = -2kq / d
and then for the next 2 charges W2 = 2 k q / (2 d)
then you get an infinite series of terms like (2 k q / d) *...
What seems to be missing is the implication that the train speeds back to 10 m/s in 10 sec.
Work (J) = Change in KE = Force * Distance
Power (Watts) = Work / time
One useful fact to keep in mind is that in an (linear) elastic collision of two objects is that
the relative speed of approach equals the relative speed of separation after the collision.
The file you attached is not in Excel format.
Were you given the actual spreadsheet., if so you can check the result cells for
the formulas used.
As an example in cell H1 there must be a formula such as: =B1*D1