[Super L]

is 500kg & radius is .5m. Wats KE? Avg angular Acc? well, I dont understand this question on my physics hw. I've come up with some answers but I rounded up a lot of equations and subbed in a lot of assumed unknowns so I don't know if I did it right. Thank you in advnace for your help!alright! thank you thus far! I appreciate it so much! For avg ang accel. the wheel spins down to a complete stop in 4 hours.

[Bekki B]

KE = 1/2 I (omega)^2They give you omega (convert rpm to radians per second).You can calculate I using mass and radius. Look up in your book the moment of inertia of a wheel (solid disk). There should be a table. It's mass times radius squared times some number I forget.You don't have enough information there to say anything about average angular acceleration.Edit:To get the average angular acceleration, divide change in omega (which is just your omega minus zero) by that time. Don't forget to convert from hours to seconds and from rpm to radians per second and all that.

[catarthur]

First you need the moment of inertia of the wheelfor a thin solid disk of mass m and radius r,I = mr^2/2so in your caseI = 500kg * (.5m)^2/2 = 62.5kg.m^2Kinetic energy of a rotating body isKe = 1/2 I w^2 whereI is moment of inertiaw is angular speed in rad/s300 rpm = 300 * 2Pi rad/60s = 31.4 rad/s soKe = 1/2*62.5kg.m^2 * (31.4rad/s)^2Ke = 30 842 kg.m^2/s^2) = 30 842 (kg.m/s^2).m = Ke = 30 842 N.m = 30 842 JSo kinetic energy is 30 842JIf the rotation speed is 300 rpm, than angular velocity is constant and the angular acceleration is 0 rad/s^2.