Please check my setups and kindly tell me the correct method of finding the answers if possible. 1) The radius of a 3.0 kg wheel is 6.0 cm. The wheel is released from rest a point A on a 30 degree incline. The wheel rolls without slipping and moves 2.4 m to point B in 1.20 s. The moment of inertia of the wheel is closest to: a.0.0057 kg*m^2 b.0.0048 kg*m^2 c.c. 0.0051 kg*m^2 d.0.0054 kg*m^2 e.0.0060 kg*m^2 At first, I thought it would be easy by just using I = 0.5*m*r^2 = 0.0054 kg*m/s^2. Then, I thought of the conservation of energy. mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r h = L*sin(30) = 1.2 m, where L = 2.4 m Is v = 2.4 m/ 1.2 s = 2m/s? w = (2 m/s)/(.06 m) = 33.33 rad/s I have all values so isolate I. I = (2*29.28 J)/ (w^2) = (2*29.28 J)/ (33.33 rad/s)^2 = 0.0521 kg*m^2 This doesn’t match any of the choices. What did I do wrong? 2) A hoop is released from rest at the top of a plane inclined at 24 degrees above horizontal. How long does it take the hoop to roll 24.0 m down the plane? a. 3.27 s b. 4.91 s c. 4.48 s d. 7.70 s e. 3.13 s I used conservation of energy for this part. I = M*r^2 for a hoop. If L, length of incline, is 24 m then hoop must travel, height = L*sin (24) = 9.76168 m. mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r mgh = 0.5*mv^2 + 0.5(m*r^2)*(v/r)^2 mgh = 0.5m*v^2 + 0.5*m*v^2 mgh = m*v^2 v = sqrt(g*h) ??? If I do that, v = sqrt(9.8 m/s^2*9.76168 m) = 9.78998 m/s 24 m/9.78998 m/s = 2.4514 s What did I do incorrectly?? Thank you.