Question

A small solid porcelain sphere, with a mass m and radius r, is placed on the inclined section of the metal track shown below, such that its lowest point is at a height h above the bottom of the loop. The sphere is then released from rest, and it rolls on the track without slipping. In your analysis, use the approximation that the radius r of the sphere is much smaller than both the radius R of the loop and the height h. (Use the following as necessary: MR, and g for the acceleration of gravity.) Solid sphere of mass m and radius r<< R. (a) What is the minimum value of h (in terms of R) such that the sphere completes the loop? for Rạr (b) What are the components of the net force on the sphere at the point P if h = 3R? Use a coordinate system where the +x-direction is to the right and the +y-direction is up. (Assume point P has a height R above the horizontal section of the track.) Fx = [F, =

Answer 1

A)

Forces acting on the sphere at the top of the loop,

N + mg = mv^2/R

Since, force N at the top of the loop is zero,

mg = mv^2/R

Velocity, v = sqrt(gR)

Angular velocity, w = sqrt(gR)/r

Using law of conservation of energy,

mgh = 0.5 mv^2 + 0.5 Iw^2 + mg (2R)

gh = 0.5 gR + (1/5) gR + 2 gR

h = 2.7 R

B)

Using law of conservation of energy,

3 mgR = 0.5 mv^2 + 0.5 Iw^2 + mgR

Moment of inertia, I = (2/5) mr^2,

Angular velocity, w = v/r

2gR = (7/10) v^2

Velocity, v = sqrt(20 gR/7)

Torque, T = I alpha = (2/5) mr^2 alpha

Also, T = fr

Equating both,

Alpha = 5g/7r

Net force in vertical direction,

Fy = - (5/7) mg

Net force in horizontal direction,

Fx = - mv^2/R = - m((20/7) gR)/R = - (20/7) mg

Fx = - (20/7) mg

Fy = - (5/7) mg

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