Recent content by jaewonjung

I see. shifting the axis from the center of the sphere to the other end of the rod is ML^2. Thanks!

Sorry for not including a picture. The axis is perpendicular to the rod, and is placed at the end of the rod that doesn't have the sphere attached to it.

Moment of inertia for a rod rotated around one end is 1/3ML^2 using the parallel axis theorem Moment of inertia for a sphere is 2/5 MR^2 Itotal=Isphere+Irod=2/5MR^2+1/3ML^2 However, the answer is 2/5MR^2+1/3ML^2+ML^2 Why is there an extra ML^2 added to the moment of inertia?

I solved it! The torque of the Tension is just Tr, not Trcostheta. Thanks for the help!

Forces in the x direction: static friction left=Tcostheta to the right Forces in the y direction: W= Normal force + Tsintheta Net torque: rTcostheta-RFs=0 Fs=Tcostheta, rTcostheta-R(Tcostheta)=0 rTcostheta=RTcostheta r=R Please help

Thanks for the reply! I see. If the angle was approaching 0, the period would approach infinity, since the pendulum would almost be completely horizontal and gravity would have a negligible effect. I understand why C is the answer and not D, but I still don't understand how to algebraically...

Well, I know that A is wrong since it is independent of the angle. The other answer choices have an angle component.

Since gravity is acting downward, I found the gravity component parallel to the plane, which was g/sin60. I substituted g/sin60 for g into that equation and got D, but the answer should be C.

I tried to substitute 1/2T in for T, because 1/2 T is when the ball is at H. With that, I found Vo, which was (H-1/8gT^2)/(1/2T). I plugged in Vo into h=vot+1/2gt^2 and put 1/4 T for t, but I could not get an answer in terms of H