We have recently started studying rotational and rolling motion, Ive struggled with this more than anything, Im pretty lost on these problems: 1) Consider a thin rod of mass 3.4 kg, length 6.8m, and uniform density. The rod is pivoted at one end on a frictionless horizontal pin. The rod is initally held in horizontal position but eventually allowed to swing down. What is the angular acceleration at the instant it makes a 48 degree angle with the horizontal? So far Ive done this. Line density of rod = 3.4kg/6.8m =0.5 kg / m Moment of inertia = Integrate over 0 to 6.8 ((X^2) * 0.5 dx) = (0.5/3) * 6.8^2 =52.4 kg m^2 Then using torque: (3.4) * (9.8) * (0.5 * 6.8) * sin(48) = 52.4a (where 'a' is angular acceleration) This comes out to 1.6066 m/s which converted to radians /s = 1.6066 m/s / 6.8 = .23626 rad/s^2 but this is wrong, what is going wrong in this calculation? A 0.13kg basketball has 0.145m diameter and may be approximated as a thin sphereical shell. The moment of inertia is I = (2/3)m(R^2) and coefficient of friction 0.33. Starting from rest how long will it take a basketball to roll, without slipping, 6.18m down an incline that makes and angle of 30 degrees with the horizontal? Ive tried many different ways of going about this problem but Im still lost. Ive been trying to somehow solve with: KE=(1/2)mv^2 + (1/2) I*w^2 I= (lambda)*m*r^2 Im having no luck with these equations, and Im having a hard time just grasping the concepts of the problem, especially how to take friction into account, can anyone offer any insight? Thanks for any help.