A 4.42 kg, 62.5 cm diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. What is the absolute value of the cylinder‘s initial angular acceleration?
There was one going on proper now. I was once so worn out and it woke me up and would not ******* turn off. If I had standard garments on, i might've long gone in the market and shot the tires out. Lol. Neighbors should ******* pay attention to their ******* auto.
i think this question requires you to calculate the torque produced when the supporting force is released, and the weight force causes torque. T I x (alpha) where I is moment of inertia, and alpha is angular acceleration The torque produced is a result of the weight force, and produces the angular acceleration about the axis. F(weight) 4.42 x 9.81 43.36 N this acts at the centre of gravity, which i assume to be 1/2(0.625) 0.3125 m from the edge joined to the axle This allows us to find the torque produced T Fd 43.36 x 0.3125 13.55 Nm we equate our original equation to this T I x (alpha) I x (alpha) 13.55 moment of inertia of a cylindrical disk is 1/2 Mr^2 I 1/2 x (4.42) x 0.3125^2 0.215 kg.m^2 hence alpha 13.55 / 0.215 63.02 rad/s^2 these answers depend on whether or not the centre of mass is one radius length from the fixed point. the answer doesnt seem right to me, but i wont redo it, maybe my working will lead you to the right answer, im sure you have to use torque to relate the weight force to the angular acceleration.
