In the figure here, a solid brass ball of mass 0.266 g will roll smoothly along

Question

In the figure here, a solid brass ball of mass 0.266 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.11 m, and the ball has radius 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 h7R, what is the magnitude of the horizontal force component acting on the ball at point Q? balis released et heghto what 's the nagniude the h frcte component acting on the ball t in Qe (a) h= (b) F=

Explanation / Answer

using law of conservation of energy

Total energy at the top of the verge = Total energy at the top of the loop

(m*g*h) = (m*g*2R)+(0.5*I*w^2)+(0.5*m*v^2)

at the top of the loop ,the speed of the sphere to continue its circular motion is v = sqrt(R*g)

I = (2/5)*M*R^2

(m*g*h) = (m*g*2R)+(0.5*(2/5)*m*R^2*(V/R)^2)+(0.5*m*V^2)

simplifuing we get

h = 2.7*R = 2.7*0.11 = 0.297 m

b) F = m*v^2/R

using law of conservation of energy

m*g*h = (m*g*R)+(0.5*(2/5)*m*R^2*(V/R)^2)+(0.5*m*V^2)

m cancels on both sides

9.8*7*0.11 = (9.8*0.11)+(0.7*V^2)
V = 3.03 m/s

F = m*V^2/R = (0.266*10^-3*3.03^2)/0.11 = 0.0222 N

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