In this application, you will examine a small, box of mass 370 g sliding down a frictionless hemisphere With a radius of 5.3 m. It Will start at the top of the hemisphere at the point U with a negligibly small speed - Just enough to start the process of mowing. but not large enough to measure. It then slides down the sphere in process I. At the point 1. it Wlll go fast enough that it will lose contact with the sphere only to fall to point 2 in process II. Part 3 of 5 - AI Launch: Point 1 At the exact point where the box loses contact with the spherer the component of the weight in the radial direction is exactly the centripetal acceleration required to keep the box traveling in a circle This is depicted by the free body diagram at left (which is the limit N i 0 of the free body diagram that describes the motion along path segment I. To find the position and velomty of the box, you'll have to look at what's happening to the centripetal acceleration (which is proportional to u2) and the kinetic energy {also proportional to uz) at the same time. To do this, you should first lay out each set of equations in standard form. Each will have terms in the speed 2 and height H. (a) Firstr write the energy conservation equation at the point 1. K U E E(tl) = 10153 N x DimensionallyI incorrect. Please check the type or dimension of your unit. u2 =: H =|:| (b) Next. find the force equation in the normal direction. Applied Forces Net Force You now have two equations and two unknowns. Start by eliminating u2 to find the height at which the box loses contact, and then use that knowledge to find the velocity. (c) What is the height at which the box loses contact with the sphere? H = (d) What is the speed of the box when it loses contact? Now, turn these into vectors. (e) What is the position of the box when it leaves the sphere in physical form? Fltlle |