1)A uniform solid ball of radius r=1.00 cm starts from rest at a height h and ro

Question

1)A uniform solid ball of radius r=1.00 cm starts from rest at a height h and rolls without slipping along a loop-the-loop track of radius R=0.50 m. What is the smallest value of h for which the sphere will not leave the track at the top of the loop? (Note that the center of mass of the ball travels a loop of radius R-r, i.e., the height it reaches is 2R-r). Use energy conservation.)

2)A uniform solid ball of radius r=1.00 cm starts from rest at a height h and slips without rolling along a loop-the-loop track of radius R=0.50 m. What is the smallest value of h for which the sphere will not leave the track at the top of the loop?

Explanation / Answer

to go in a circle at the top mg = mv^2/(R-r) v^2 = g (R-r) conservation of energy E initial = E final mg(h+r) = 1/2 mv^2 + 1/2 I w^2 + mg(2R-r) for a ball I = 2/5 Mr^2 mg (h+r) = 1/2 m v^2 + 1/2 2/5 m r^2 v^2 /r^2 + mg (2R-r) g(h+r) = (1/2 + 1/5) g(R-r) + g (2R -r) (h+r) = (1/2 + 1/5)(R-r) + (2R -r) solve for h, h=1.323 m 2) if slipping no 1/2 I w^2 term mg(h+r) = 1/2 mv^2 + mg(2R-r) mg (h+r) = 1/2 m g (R-r) + mg (2R-r) (h+r) = 1/2 (R-r) + (2R-r) solve for h, h=1.225

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