Recent content by pbnj

It's for a Coursera course, and it tells me it's incorrect. I've tried incrementing/decrementing my answer by 5 a few times, I tried using 2 sigfigs for everything, I tried assuming "diameter" meant "radius," but no luck. On the discussion forum there are no questions about this particular...

First, we calculate the maximum height using the first equation, noting that at maximum height, the velocity is purely horizontal with speed ##v\cos\theta##, and with initial vertical speed ##v\sin\theta##: $$ \begin{align} v_f^2 &= v_i^2 - 2g(h_f - h_i) \\ 0 &= (v\sin\theta)^2 - 2g(h_f - h_i)...

I've been using lean4 as my calculator, here's my source code: def g := 9.81 def m := 30.0 def M := 70.0 def L := 6.0 def h := 2.0 def rm := (0.5*L - h).abs def Fm := m*g def FM := M*g def τm := rm*Fm def rM := τm/FM def τM := -rM*FM #eval rM def rM' := rM + 0.1 def τM' := -rM'*FM def τnet...

I figure since we're only given the length and mass of the plank, it should be ##\frac{1}{12}mL^2##. Then I should use the parallel-axis theorem to get the moment of inertia of the plank about the pivot, which is ##I_m = \frac{1}{12}mL^2 + mr_m^2##. So moment of inertia of the plank and human...

Supposing L = 6m, m = 30kg, M = 70kg, h = 2m. Also set the origin at the leftmost edge of the plank. Free body diagram description: The center of mass of the plank is at ##\frac{L}{2}##, so considering it as a point mass, it feels the force of gravity ##F_m = mg##. The person is at some point...