I have the working for part d) but I don't understand certain parts, so my problems lie with part d), f) and g) 1. The problem statement, all variables and given/known data 3. A solid wooden disk is suspended as shown: it has mass 7.70 kg and radius 0.219 m . It is rotated from rest about its vertical axis, reaching an angular velocity of 3.69 rad/s in 15.1 s. http://i41.tinypic.com/2u9mgzs.jpg Calculate (a) the average angular acceleration (b) the angular momentum. (c) the rotational kinetic energy A 0.17 kg metal block is gently placed on the disk 0.140 m from the rotational axis. (d) Show that the new angular velocity is 3.62 rad/s (e) Calculate the new rotational kinetic energy. The wooden cylinder is very slowly accelerated, and when the new angular velocity is 4.02 rad/s the block is just about to slide. Calculate (f) the frictional force on the block. (g) the coefficient of static friction μs between the block and the disk. 2. Relevant equations a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t I(Disk) = MR^2/2 L=I[itex]\omega[/itex] Krot = 1/2I[itex]\omega[/itex]^2 W2 = I/I2 x [itex]\omega[/itex] 3. The attempt at a solution (a) a = [itex]\Delta[/itex][itex]\omega[/itex]/[itex]\Delta[/itex]t = 3.69/15.1 = 0.244 rad/s^2 (b) I(Disk) = MR^2/2 = (7.70x0.219^2)/2 = 0.185 kg m^2 L=I[itex]\omega[/itex] = 0.185x3.69 = 0.681 kg m^2 s^-1 (c) Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.185)(3.69)^2 = 1.257J (d) I know L1 = L2, can we prove that this is the case if we didn't have given the value of angular velocity? [itex]\tau[/itex]NET = 0 (Not sure how I can prove this) I2 = I1 + MR^2 (Inertia of block) I2 = 0.185 + 0.17(0.14)^2 = 0.188 kg m^2 [itex]\omega[/itex]2 = (I1/I2)x[itex]\omega[/itex] = (0.185/0.188)x3.69 = 3.62 rad/s (e) Krot = 1/2I[itex]\omega[/itex]^2 = 1/2(0.188)(3.62)^2 = 1.23J (f) & (g) no idea how to work out, can anyone please give me a hand?