Recent content by Matt Raj

The friction from the rod and the friction from the plane on the cylinder should be the same due to torque equilibrium on the cylinder. If we let N_1 be the normal force on the rod and N_2 be the normal force on the cylinder from the plane, I expected µN_1 = µN_2. Looking at torque on the rod...

"l" is the length of the right stick. I mistakenly assumed that the length of the left stick was "L" when the problem stated that it was infinite. There's a normal force from the right stick that should yield N= "the torque from the weight of the right stick", but how do you get the side of the...

I attempted this problem trying to balance torques but I couldn't because the length of the left stick is unknown. From the right stick I got that mg/2* cos θ = F_f but from the left stick I got that Mg/2sin θ =N* r where r is the ratio of the length of where the sticks meet and the total length...

Sorry, I divided by cos(theta) instead of multiplying. The equations should be mv^2/R=mg*cos(theta) and mgR=mgRcos(theta)+1/2(mv^2). Solving this gave me 2/3=cos(theta), so the height that the ball leaves the hill is Rcos(theta)=2R/3.

Using the centripetal acceleration, I found that mv^2/R=mg/cos(theta), so v^2=Rg/cos(theta). Furthermore, I used conservation of energy to find that mgR=mgRcos(theta)+1/2(mv^2). I tried solving this system of equations but I got cos^2(theta)-cos^3(theta)=Rg and couldn't go anywhere after that.

Homework Statement There is an object at the top of a frictionless hemispherical hill with radius R. t time t=0, it is given a small impulse so that it starts sliding down the hill. Find the height from the ground where the ball becomes airborne. Express your answer in terms of R. Homework...