A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0505 kg·m2 abou

Question

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0505 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 32.6° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 22.0 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.00 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Explanation / Answer

Let w = omega the angular velocity of the sphere I = 2/3 m R^2 the moment of inertia of a spherical shell (I) KE = 1/2 m v^2 + 1/2 I w^2 = 1/2 m v^2 + 1/3 m R^2 w^2 KE = (1/2 + 1/3) m v^2 Total energy = 5/6 m v^2 (II) (1/3) / (5/6) = 2/5 KE = 2/5 * 22 = 8.8 J is rotational m = 3 I / R^2 = 3 * .0505 / .04 = 1.89 kg from (I) 5/6 m v^2 = 22 v = [132 / (5 * 1.89)]^1/2 = 3.74 m/s After 1 m m g h = 1.89 * 9.8 * sin 32.6 = 9.98 J energy lost KE = 22 - 9.98 = 12 J after traveling 1 m v = [6 * 12 / (5 * 1.89)]^1/2 = 2.76 m/s using (II)

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