A hollow sphere of radius 0.15 m, with rotational inertia I = 0.082 kg m^2 about
ID: 2006896 • Letter: A
Question
A hollow sphere of radius 0.15 m, with rotational inertia I = 0.082 kg m^2 about a line through its center of mass, rolls without slipping up a surface inclined at 30° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 50 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?
Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?
(d) What is the speed of its center of mass now?
Explanation / Answer
The moment of inertia is I = 0.082 kg .m2 The angle of inclination is = 300 The initial kinetic energy is K = 50 J The intiial kinetic energy is K = (1/2)I2 + (1/2)mu2 50 J = (1/2)I( u /R)2 + (1/2)mu2 50 J = (1/2)(2/3)mR2* u2 / R2 + (1/2)mu2 50J = (1/3)m u2 + (1/2)mu2 50 J = (5/6)m u2 -----------(1) mu2 = 6* 50 J / 5 = 60 J ------------(2) b ) the rotational kinetic energy is (1/2)I2 = K - (1/2)mu2 = 50 J - ( 60 J / 2) = 20 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3 * 0.082 kg m^2 / 2 ( 0.15m)^2 = 5.46 kg From the equation (2) mu2 = 60 J u = ( 60J / 5.46 ) = 3.31 m/s This is the speed of the center of mass (c) if the sphere moves 1.0 m up then From the conservation of energy at intiial point and the final point K.E_11 + P.E_1 = K.E_2 + P.E_2 K2 = K1 + P.E1 - P.E2 = 50J +(0 - mgh) = 50J - mg(1.0m*sin) = 50 J - 5.46 kg * 9.8 m/s^2 ( 1.0 m sin 30) = 23.2J (d) From the equation (1) the total kinetic energy is K.E_2 = (5/6)m v2 v = [ 6*K.E_2* / 5m ] = 6 * 23.2 J / 5 * 5.46 ) = 2.25m/s The angle of inclination is = 300 The initial kinetic energy is K = 50 J The intiial kinetic energy is K = (1/2)I2 + (1/2)mu2 50 J = (1/2)I( u /R)2 + (1/2)mu2 50 J = (1/2)(2/3)mR2* u2 / R2 + (1/2)mu2 50J = (1/3)m u2 + (1/2)mu2 50 J = (5/6)m u2 -----------(1) mu2 = 6* 50 J / 5 = 60 J ------------(2) b ) the rotational kinetic energy is (1/2)I2 = K - (1/2)mu2 = 50 J - ( 60 J / 2) = 20 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3 * 0.082 kg m^2 / 2 ( 0.15m)^2 = 5.46 kg From the equation (2) mu2 = 60 J u = ( 60J / 5.46 ) = 3.31 m/s This is the speed of the center of mass (c) if the sphere moves 1.0 m up then From the conservation of energy at intiial point and the final point K.E_11 + P.E_1 = K.E_2 + P.E_2 K2 = K1 + P.E1 - P.E2 = 50J +(0 - mgh) = 50J - mg(1.0m*sin) = 50 J - 5.46 kg * 9.8 m/s^2 ( 1.0 m sin 30) = 23.2J (d) From the equation (1) the total kinetic energy is K.E_2 = (5/6)m v2 v = [ 6*K.E_2* / 5m ] = 6 * 23.2 J / 5 * 5.46 ) = 2.25m/s 50 J = (5/6)m u2 -----------(1) mu2 = 6* 50 J / 5 = 60 J ------------(2) b ) the rotational kinetic energy is (1/2)I2 = K - (1/2)mu2 = 50 J - ( 60 J / 2) = 20 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3 * 0.082 kg m^2 / 2 ( 0.15m)^2 = 5.46 kg From the equation (2) mu2 = 60 J u = ( 60J / 5.46 ) = 3.31 m/s This is the speed of the center of mass (c) if the sphere moves 1.0 m up then From the conservation of energy at intiial point and the final point K.E_11 + P.E_1 = K.E_2 + P.E_2 K2 = K1 + P.E1 - P.E2 = 50J +(0 - mgh) = 50J - mg(1.0m*sin) = 50 J - 5.46 kg * 9.8 m/s^2 ( 1.0 m sin 30) = 23.2J (d) From the equation (1) the total kinetic energy is K.E_2 = (5/6)m v2 v = [ 6*K.E_2* / 5m ] = 6 * 23.2 J / 5 * 5.46 ) = 2.25m/s v = [ 6*K.E_2* / 5m ] = 6 * 23.2 J / 5 * 5.46 ) = 2.25m/s = 6 * 23.2 J / 5 * 5.46 ) = 2.25m/sRelated Questions
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