Recent content by J Hann

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...

Doesn't (r - r cos x) imply that the vine breaks at x = 90 deg.? I think the problem assumes the initial angle from the vertical is 90 deg.

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?

Your solution looks OK at the surface of the sphere. The problem stated "near" the surface of 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...

Suppose the strings were attached to spring scales instead of fixed supports. What would you expect to happen to the scale readings during a cycle?

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).