Thanks, I finally figured it out! Unfortunately, I got the wrong answer, so I'm approaching the actual physics problem wrong. At least I got some good geometry practice out of it :)
I am assuming you mean by dropping a perpendicular that is the height of the triangle, so it is split into two right triangles? I can't figure out what to do from there though because I don't know how the 52 m side is split.
Homework Statement
So I'm doing a physics problem, and I think I'd get the right answer if I solved this triangle. I don't know any angles. The base is 56 meters long, its height is 500 meters, and the difference between the other two sides is 4 meters. Can I figure out the sides based on this...
Ah. The clockwise torque should equal the counterclockwise torque. Would counterclockwise be MgL? And clockwise would be... the tension? That seems much too simple. :(
I know; I described it instead. It is a trapezoid with a longer top than bottom. I guess I should also say that θ is each outside (acute) angle that the beams make with the ground.
Homework Statement
This is actually a sound wave problem, but I think I'll be fine when I actually get to that part; my issue is that it is a cumulative problem that involves torque, which I haven't had practice with since the fall. It's embarrassing how little I remember how to do from just a...
Well, I don't have any equations with v for simple harmonic motion. All I know is F=-kx, which leads to a=(-k/m)x, which gives us ω=√(k/m). I just converted v back to ω, which puts R back into the picture. Is there a simpler way?
YES. I think so. If R=2∏x (which would cancel out the x's and R's and give me the factor of 2∏ that I want on the right side) then I've got it. Is that true? If so, can you explain why?
Oh, whoops, I'm sorry.
1/2kx^2 = 1/2mv^2 + 1/2Iv^2/R^2
Is I additive here? Would it simply be MR^2?
That would make the equation simplify to 1/2kx^2 = mv^2.
Homework Statement
) Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, massless rod. They are attached to a spring with force constant k using a frictionless ring around the axle. If the spring is pulled out and released, the cylinders...
Oh my goodness, you're right. I've been looking for the angular frequency when this question calls for regular frequency. Thank you so much! I can't believe I wasted so much time on such a silly oversight :P