A hollow sphere of radius 0.20 m, with rotational inertia I = 0.070 kg m2 about
ID: 2006980 • Letter: A
Question
A hollow sphere of radius 0.20 m, with rotational inertia I = 0.070 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 23° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 80 J.(a) How much of this initial kinetic energy is rotational?
J
(b) What is the speed of the center of mass of the sphere at the initial position?
m/s
Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?
J
(d) What is the speed of its center of mass now?
m/s
Explanation / Answer
Given that the radius of the hallow sphere is R = 0.20 m The moment of inertia is I = 0.070 kg .m2 The angle of inclination is = 230 The initial kinetic energy is K = 80 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 80 J = (1/2)I( U /R)2 + (1/2)mU2 80 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s Given that the radius of the hallow sphere is R = 0.20 m The moment of inertia is I = 0.070 kg .m2 The angle of inclination is = 230 The initial kinetic energy is K = 80 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 80 J = (1/2)I( U /R)2 + (1/2)mU2 80 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s Given that the radius of the hallow sphere is R = 0.20 m The moment of inertia is I = 0.070 kg .m2 The angle of inclination is = 230 The initial kinetic energy is K = 80 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 80 J = (1/2)I( U /R)2 + (1/2)mU2 80 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s The moment of inertia is I = 0.070 kg .m2 The angle of inclination is = 230 The initial kinetic energy is K = 80 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 80 J = (1/2)I( U /R)2 + (1/2)mU2 80 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s The angle of inclination is = 230 The initial kinetic energy is K = 80 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 80 J = (1/2)I( U /R)2 + (1/2)mU2 80 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s 80 J = (1/3)m U2 + (1/2)mU2 80 J = (5/6)m U2 -----------(a) mU2 = 6* 80 J / 5 = 96 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 80 J - ( 96 J /2) = 32J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 2.625 kg From the equation (1) mU2 = 96 J U = ( 96J / m )1/2 = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s = 9.79 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 K1 + P.E1 = K2 + P.E2 K2 = K1 + P.E1 - P.E2 = 80 J +(0 - mgh) = 80 J - mg(1.0m*sin) = 80 - 10.05 = 69.94 J This is the final total kinetic energy (d) From the equation (a) the total kinetic energy is K2 = (5/6)m V2 V = [ 6*K2* / 5m ]1/2 = 5.62m/s V = [ 6*K2* / 5m ]1/2 = 5.62m/s = 5.62m/sRelated Questions
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