In the figure here, a solid brass ball of mass 0.309 g will roll smoothly along

Question

In the figure here, a solid brass ball of mass 0.309 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.19 m, and the ball has radius r muchlessthan R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop (b) If the ball is released at height h = 6R, what is the magnitude of the horizontal force component acting on the ball at point Q? (a) h = the tolerance is +/-2% (b) F = N

Explanation / Answer

As r<<R , we can consider ball as point mass.

a) Let vt be the velocity at the top of the loop when ball is in the verge of leaving the track.

Then ,   

weight of the ball=centrifugal force

or, mg=mvt²/R

or, vt= (g*R)= (9.81*0.19) = 1.36 m/s

velocity at bottom, vb

(1/2)mvb² = (1/2)mvt²+mg(2R)

or, vb²= vt² + 4gR = 5vt²

or, vb= 5* 1.36 = 3.04 m/s

height, h= vb²/(2g) = 0.47 m

b) Here, h=6R=6*0.19= 1.14 m

velocity at bottom,

vb=(2gh) = (2*9.81*1.14) = 4.73 m/s

let velocity at Q be vQ

(1/2)mvb² = (1/2)mvQ²+mgR

or, vQ²= vb²-2gR= 4.73² - (2*9.81*0.19) = 18.64 m/s

horizontal force at Q, F = mvQ²/R = 0.309*10^(-3)*18.64/0.19 = 0.03 N

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