A hollow sphere of radius 0.25 m, with rotational inertia I = 0.054 kg m2 about
ID: 1965892 • Letter: A
Question
A hollow sphere of radius 0.25 m, with rotational inertia I = 0.054 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 70 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 = 0.054 kg .m2 The angle of inclination is = 230 The initial kinetic energy is K = 70 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 70 J = (1/2)I( U /R)2 + (1/2)mU2 70 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 70 J = (1/3)m U2 + (1/2)mU2 70 J = (1/3)m U2 + (1/2)mU2 70 J = (5/6)m U2 -----------(a) mU2 = 6* 70 J / 5 = 84 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 70 J - ( 84 J /2) = 28 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3*0.054/2*(0.25)^2 = 1.296 kg From the equation (1) mU2 = 84 J U = ( 84 J / m )1/2 = 8.05 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 = 70 J +(0 - mgh) = 70 J - mg(1.0m*sin) = 20.37 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 = 4.34 m/s The angle of inclination is = 230 The initial kinetic energy is K = 70 J --------------------------------------------------------------------------- The intiial kinetic energy is K = (1/2)I2 + (1/2)mU2 70 J = (1/2)I( U /R)2 + (1/2)mU2 70 J = (1/2)(2/3)mR2* U2 / R2 + (1/2)mU2 70 J = (1/3)m U2 + (1/2)mU2 70 J = (1/3)m U2 + (1/2)mU2 70 J = (5/6)m U2 -----------(a) mU2 = 6* 70 J / 5 = 84 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 70 J - ( 84 J /2) = 28 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3*0.054/2*(0.25)^2 = 1.296 kg From the equation (1) mU2 = 84 J U = ( 84 J / m )1/2 = 8.05 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 = 70 J +(0 - mgh) = 70 J - mg(1.0m*sin) = 20.37 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 = 4.34 m/s 70 J = (1/3)m U2 + (1/2)mU2 70 J = (5/6)m U2 -----------(a) mU2 = 6* 70 J / 5 = 84 J ------------(1) Then the rotational kinetic energy is (1/2)I2 = K - (1/2)mU2 = 70 J - ( 84 J /2) = 28 J But the moment of inertia is I = (2/3)mR2 m = 3I / 2R2 = 3*0.054/2*(0.25)^2 = 1.296 kg From the equation (1) mU2 = 84 J U = ( 84 J / m )1/2 = 8.05 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 = 70 J +(0 - mgh) = 70 J - mg(1.0m*sin) = 20.37 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 = 4.34 m/s = 8.05 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 = 70 J +(0 - mgh) = 70 J - mg(1.0m*sin) = 20.37 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 = 4.34 m/s V = [ 6*K2* / 5m ]1/2 = 4.34 m/s = 4.34 m/sRelated Questions
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