Recent content by BrettF

Ahh this makes sense and is very clear, thank you. In the calculation I just did I used d just as the distance from the mass to the pivot point and got the right answer. Thank you so much for your help!

What I'm thinking is that d is the distance from the center of mass, but the pivot point is the end of the rod. I guess I'm just confused, would you be able to rephrase the question please? Thank you for your help.

Ahh, that makes sense, the moment of intertia for the weight is not mL^2 since it's not at the end of the rod, it should be m(d+L/2)^2 right? So the inertia should be I = (1/3)ML^2 + m(d+L/2)^2 [I think] Except I solved for d using d = Iπ^2/((M+m)g) and substituting the I from above and got the...

Homework Statement You are at a furniture store and notice that a grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows donw; when it is moved upward, the pendulum swings faster...

I have been trying this problem for a while and can't seem to figure it out: A bicycle wheel has a radius R = 32.0 cm and a mass M = 1.82 kg which you may assume to be concentrated on the outside radius. A resistive force f = 137 N (due to the ground) is applied to the rim of the tire. A force...