Question

7. (100 points) In Figure 1, we have a solid sphere of mass m and radius r that is at rest at the top of an incline. At the bottom of the incline, there is a loop. (a) What is the minimum value of h such that the sphere just barely makes it around the loop? (b) What is the total force acting on the sphere at point P, ir 3R? Difficulty: 4/5

PLEASE GIVE ANNOTATIONS EXPLAINING THE CONCEPTS BEHIND EACH STEP.

Answer 1

Please Remember → Total Energy remains constant through out motion, KE + P. E = constant K.F - kinetic Energy PE a potential Energy (a) o miruso t su, height=o (TE) i = (T.E)F mgh + Ilm (0) ² = 1 mb ² + mg cos nigh = /ho - gh = 42 Now 2V Centripital acceleration = 12 TR-r) N-mga mo? At Height print, n=0 for min mg = mora 102= g(R-1) R-H)

Total Energy at lowest print - Įmos " " " Heightest "smo F mg 21R-D) I mo ² = 1 {mv ² + 2 mg (R-r) Koho za tim g (R-1) + 20ng (R-1) x 2 bez sikt) Now, ghe van ons that SPR=) 'min (Rt) (b) if h= BR mg h= 1 mua ng (3R) = 1 mua Tw=6gr/ Now, Energy conservation for loop Ima ² = 10 + mg (R-1) m& gr = 1 mo't mgr ş for simplicity consider 3mg R-mg R = 1 m 2 I R-ro R 3 angr = 1 m ² 10²=49R

imig u = mx 4g R N = mx HQ R = 4mg = 4mg Nek force Net For Net Force = Jimgjet (ings 2 = mgJ1+16 If net = JF mg]