Recent content by rschaefer2

Awesome, thank you! my new answer is (-10/14)mg, with the correct signs. Any confirmation?

Rolling up the ramp, the ball is rotating clockwise. Since ω is decreasing, the torque acting on the sphere must be in the opposite direction, counter clockwise. So the positive direction of rotation is counterclockwise If ay is positive, then the angular speed should also increase...

EDIT: From previous, I've changed torque to only static friction in the positive direction, as mg acts through the axis. Also, the friction force needs to counteract the clockwise movement, therefore needing to be positive (counterclockwise). So I'm pretty sure I've worked this out correctly. I...

My attempt: Equation 1 expands to: mg+fs=(2/5)mr2(ay/r) plugging the right side of this equation into equation 3: ay=(2/5)mray/m and this is where i get stuck, because the acceleration terms cancel out.

at=ay Okay, thanks! I'll try to work them out.

1) \sumτ=I(alpha) 2)v=ωr 3)fs=μFn 4) alpha=aT/r 5)aT=(-mg-fs)m Fn= -20mg/7 I=2/5mv2/r2 I can't seem to relate it using these, I'm probably missing one of the equations.

Yes, because v=ωr

Definitely slowing down, because v will be decreasing

The ball itself? Well angular acceleration will be the derivative of ω with respect to s? EDIT: s as in seconds, from g

Actually, static correct? Because the object is rolling.

Well, i guess there is friction to? But there wasn't any coefficients given for kinetic friction

Homework Statement A solid sphere of mass m and radius r rolls without slipping along the track shown below. It starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, much larger than r. (Consider up and to the right to be the positive...