A small solid marble of mass m and radius r will roll without slipping along the

Question

A small solid marble of mass m and radius r will roll without slipping along the loop-the-loop track shown in Fig. 12-34 if it is released from rest somewhere on the straight section of track. For the following answers use m for the mass, r for the radius of the marble, R for the radius of the loop-the-loop and g for the acceleration due to gravity.

(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?

Explanation / Answer

a)
minimum speed at the top of the loop, v = sqrt(g*R)

let h is the minimum height required.

Apply conservation of energy

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

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

m*g*h = 0.5*m*g*R + 0.2*m*v^2 + 2*m*g*R

m*g*h = 0.5*m*g*R + 0.2*m*g*R + 2*m*g*R

m*g*h = 2.7*m*g*R

h = 2.7*R <<<<<<------Answer

b) let v is the speed at point Q.

Apply conservation of energy

m*g*h = 0.7*m*v^2 + m*g*R

m*g*(6*R) = 0.7*m*v^2 + m*g*R

0.7*m*v^2 = 5*m*g*R

v^2 = (50/7)*g*R

Horizontal componet of force exrted on the marble = m*v^2/R

= m*(50/7)*g*R/R

= 7.14*m*g <<<<<<------Answer

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