Recent content by tomstringer

Isn't it true that the bob's velocity tangent to the circular path will become constant at the moment the box starts its freefall? I'm taking "relative to the box" to mean the same as "relative to the ceiling of the box". I see the bob continuing in its circular path at a constant speed, after...

I agree with Sdtootle. The key idea here is to note that the time it takes to travel from A to D is the same time that the car's "shadow" moves from B to D. t(AD) = t(BD). Then 100/v = 100(sq root 2). Solve for v. Then t(total journey) = D/v .

Gravitational force ends but centripetal force, Fc = mv^2/L, remains where Fc is independent of g. Hence, the bob continues, in the absence of friction from the pivot, with constant velocity v. Fnet = T + mgsin(Theta) = ma = mv^2/L = Fc where L is length of pendulum. When g = 0, Fnet = Fc

Done. Thanks.

θ, the angle from the horizontal, was used to simplify the solution. Since the question asks, "what is the angle from the verticle" I added 45°.

Homework Statement A rod of neligeable mass is released from the horizontal position. As a ball at the end of the rod falls, it reaches a point at which the tension, T, in the rod equals the ball's weight. At what angle from the verticle does this occur. I am not getting the same answer...

Thanks voko, I got it! Boy was I stuck on stupid. Thanks for being patient.

Right. The vertical displacement is simply T x (V sin theta) - gT^2 and this works for so many problems of this sort, but not this one. P.S. tan and cos don't like each other well enough to make this work so far as I can tell.

We have D = VT or 50 = Tcos theta for the horizontal and 1/2gT2 sin theta for the vertical components, but while the horizontal motion equation could provide us with T, the vertical equation does not because the apogees of the two parabolic paths are not relevant to the question. I will work on...

Thanks but in order to get the time parameters one needs to know θ which is the unknown. Also, the two angles in question each give us a flight time that is different. I agree with you that working the path equation backward to find the θ's is hard, but I think this is the only way.

The end point of the trajectory is the landing point(beyond the goal post) and may be determined with the range equation. But the question is not this--rather it is asking for the maximum and minimum angle the kicker must kick the ball in order to clear the post. We know the hight and distance...

Sorry. We are asked to find the max and min angles between which the kicker must kick the ball in order to barely clear the goal post.

Homework Statement This problem out of Halliday and Resnick is stumping me. A kicker can kick the ball with an initial velocity of 25 m/sec. The ball is 50 m from a 3.44 m high goal post. Homework Equations I have the range equation and the Path equations for a trajectory. The...

A foot ball is kicked with a velocity of 10 m/sec. 50 meters in front of a 10 meter goal post. What is the minimum and maximum angle needed to clear the goal post?