A marble {solid sphere, I = (2/5)MR2}, of mass M = 0.025kg, and R 1.2cm, is roll

Question

A marble {solid sphere, I = (2/5)MR2}, of mass M = 0.025kg, and R 1.2cm, is rolling down the following track without slipping. If H = 10 m, and the radius of the round part of the track is 2.0m, find; a) The speed of the center of mass of the marble at point Q b) The highest point that it reaches after it leaves point Q. c) The angular momentum of the marble at the point of part b

Explanation / Answer

mgH= 1/2 *mv^2 +1/2 Iw^2.....and w =v/r mgH =0.5(mv^2 +2/5 mr^*2v^2/r^2) hence v= = 11.83 m/s after this point it doesnt roll hence max height attained h is given by 1/2 mv^2 =mgh v=11.83 hence h =7.14 m hence as h>4 and marble is in the loop max height attained is equal to diameter =4m hence v at this point is given by 1/2 mv^2 =1/2 m11.83^2 -mg4 hence v = 7.84 m/s hence angular momentum about the center of the loop is mvr =0.025*7.84*2 =0.392 kgm^2/sec

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