A 2.30 kg thin, spherical shell of radius 0.160 m is released from rest at point

Question

A 2.30 kg thin, spherical shell of radius 0.160 m is released from rest at point A in the figure below, its center of gravity a distance of 1.80 m above the ground. The spherical shell rolis without slipping to the bottom of an incline and back up to point B where it is launched vertically into the air. The spherical shell then rises to its maximum height hmax at point C HINT 1.80 m imax 0.350 m (a) At point B, find the spherical shell's translational speed Ve (in m/s). m/s (b) At point B, find the spherical shell's rotational speed wg (in rad/s). rad/s (c) At point C, find the spherical shell's rotational speed (in rad/s). rad/s (d) At point C, find the maximum height hmax of the spherical shel's center of gravity (in m).

Explanation / Answer

let

h1 = 1.8 m
h2 = 0.35 m

a) Apply conservation of energy

gain in kinetic energy = loss of potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*(h1 - h2)

(1/2)*m*v^2 + (1/2)*(2/3)*m*R^2*w^2 = m*g*(h1 - h2)

(1/2)*m*v^2 + (1/3)*m*v^2 = m*g*(h1 - h2)

0.8333*m*v^2 = m*g*(h1 - h2)

v = sqrt(g*(h1 - h2)/0.8333)

= sqrt(9.8*(1.8 - 0.35)/0.8333)

= 4.13 m/s

b) w = v/R

= 4.13/0.16

= 25.8 rad/s

c) 25.8 rad/s

because, when it leaves the track the shell rotates with consatnt angular velocity.

d) h_max = h2 + v^2/(2*g)

= 0.35 + 4.13^2/(2*9.8)

= 1.22 m

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