A 1.50 kg hoop of radius 0.160 m is released from rest at point A in the figure below, its center of gravity a distance of 1.80 m above the ground. The hoop rolls without slipping to the bottom of an incline and back up to point B where it is launched vertically into the air. The hoop then rises to its maximum height hmax at point C. HINT A 1.80 m v=0 A (a) At point B, find the hoop's translational speed v₂ (in m/s). m/s (b) At point B, find the hoop's rotational speed we (in rad/s). rad/s (c) At point C, find the hoop's rotational speed we (in rad/s). rad/s C B 10.300 m hmax (d) At point C. find the maximum height of the hoop's center of gravity (in m). m D

Is there any simpler solution to this? Without having to calculate so many variables?

(89) X 0 馄.all 89% 12:40 PM Homework 1 PHYs 2100-09-LAB, SPRING 2019 NAME CIN Question 1. A bird flies a distance of d- 150+3 m during a time 20.0+1.0 s The average speed of the bird is u-d/t = 7.5 m/s. what is the uncertainty of u? Question 2. Suppose that we're able to eliminate the uncertainty in our time mea- surement so that t = 20.0 ±0.0 s. What is the new uncertainty of u?

In the answer, it gives a formula f-ff=m(0) and afterwards simplies into 120 - ff = 0 and further into ff=120N. My question is how do you not have to divide by 0 to finish the equation if that makes sense. To simply further why are we able to ignore the m(0) portion?