please answer all of the questions in the imagine I submitted. thank you! **Circular Motion 1: Physics Phor Phun****Directions:** Answer the following questions as best you can.1. A 0.5kg cup of water is swung around in a vertical circle just fast enough so the water can't fall out. a) What's the least amount of acceleration this water and cup would have on them? b) If the string tied to the cup is 0.6m long, what would the velocity of the cup and water have to be in order to have this acceleration? c) What force would the string have to apply to the cup to get this to happen?2. A space station *(looking like a space donut)* has a radius of 50m. What would its velocity have to be in order to simulate the Earth's gravity? b) How long would it take for the space station to make one complete rotation at that speed?3. Presently the weight of an object is more than enough to keep something from "falling off" the surface of the Earth. If the Earth has a radius of 6,400,000m, how fast would something have to go *(or the Earth have to spin)* in order to no longer be kept in the circular path around the center of the Earth? *(You can pick the mass of any object you want.)*---©1998 Science Joy WagonPage 87

Given data: Mass of the cup is, m=0.5 kg Formula used: During the vertical motion, centripetal force is balanced by the force due to gravity, Therefore, Fc=Fgmv2R=mg Tension in the rope is, T= mv2R-mgCalculation: (a) When body move in the vertical circle, centripetal force and force due to gravity balances each other for safe move and without falling out. Therefore minimum acceleration of the cup and water is the centripetal acceleration. amin = gamin =9.8 m/s2 (b) During the vertical motion, centripetal force is balanced by the force due to gravity, Therefore, Fc=Fgmv2R=mgv2R=g Where, R is the radius of the vertical circular path, i.e. Length of circular path Give value for R is, R= 0.6 m v2R=gv=Rgv=0.6 m9.8 m/s2v= 2.4248 m/s (c) At the highest point in the vertical motion, Centripetal force is balanced by the gravitational force. Therefore, at highest point, tension in the rope is, T= 0 N At lowest point, Tension in the rope is, T= mv2R-mg In above case, minimum velocity and acceleration discussed. Therefore, for minimum condition, tension in the rope is, T= 0 N Answer: Therefore, (a) Minimum acceleration is, 9.8 m/s2 (b) Minimum velocity is 2.4248 m/s (c) Minimum tension in the rope is, 0 N