Recent content by GrimesA

In that case it'll be A/(A+B)

That's why I was confused because they won't always be the same. If the radius of the sphere is greater then it'll have more rotational, if velocity is greater then translational will be greater

If I look at the rotational vs the translational it'll be half. Correct?

That I am completely clueless on

(.5)mv^2 + +(.5)*inertia*rotational velocity

That's my 2mgh*.96=mv^2 + (2/5)mr^2 * (v/r)^2 From there only mv^2 is the translational

I'm lost on why the 5/7 is there. All I know is 2/5*m*r^2 is the energy used to rotate

Sweet! I greatly appreciate the guidance! I think I know where to go from your advice.

My dilemma was that I was unsure what the initial y velocity is because it IS in fact moving in the y direction. Would it be Vf(incline)*sin(Height from table to ball/length of table to ball) ?

What an elementary mistake I made. The distance from the floor to the table was approx 140 cm but again I must find algebraic solutions.

The angle was approximately 9 so I thought the angle of the trajectory would be 81. And the length will vary because it won't necessarily be dropped from the top of the plane.

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color] So the problem that I have been assigned has formulas of rotational energy, momentum, trajectories, inertia, and inclined planes. A solid sphere is rolling down an inclined plane (that is...

Wow... Just wow... That's what I get for borrowing my girlfriend's calculator...My calculator never switches to radians. Thanks a ton to the both of you!

The sine(35.8) is -.9 something and cosine(35.8) is almost -.2 so that gives me negative in the parenthesis which makes everything else negative.

This turned out to still give me a wrong number when I put it into my spot.