Have a hoop (15kg) with a radius of 1.5m. Angular speed is 5 rad/s. What is the total KE possessed by the hoop? The hoop begins to climb an incline which is at a 15 degree angle. Measured along the incline, how far up the incline does it rool before stopping if the incline has a coefficient of friction of .15?
Circumference=pi*2r = 3.14 * 3 = 9.42 m.
m = 15 kg.
V = 5rad/s * 9.42m/6.28rad = 7.50 m/s.
KE = 0.5m*V^2 = 0.5*15*7.50^2 = 421.9 J.
Fp = mg*sinA = 15*9.8*sin15 = 38.05 N. = Force parallel to the incline.
Fn = mg*cosA = 15*9.8*cos15 = 142 N. =
Normal = Force perpendicular to the incline.
Fk = u*Fn = 0.15 * 142 = 21.3 N. = Force
of kinetic friction.
a = (Fp-Fk)/m = (38.05-21.3)/15 = 1.12
m/s^2.
d = (V^2-Vo^20/2a = (0-7.5^2)/2.24=25.1m
Up the incline.