Open Question
(a) Compute the torque developed by an industrial motor whose output is 150 kW at an angular speed of 4000 rev/min.
13. Rotational Inertia & Energy
Energy of Rolling Motion
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A hollow sphere of mass M and radius R rolls without slipping on a horizontal surface with angular speed W. Calculate the ratio of its linear kinetic energy to its total kinetic energy.
A
KL/Ktot =0.15
B
KL/Ktot = 0.60
C
KL/Ktot = 0.71
D
KL/Ktot = 1.67
Verified step by step guidance1
First, understand that the total kinetic energy of a rolling object is the sum of its translational kinetic energy and its rotational kinetic energy. For a hollow sphere, the moment of inertia I is given by \( I = \frac{2}{3}MR^2 \).
The translational kinetic energy (linear kinetic energy) \( K_L \) is given by \( K_L = \frac{1}{2} M v^2 \), where \( v \) is the linear velocity of the center of mass of the sphere.
The rotational kinetic energy \( K_R \) is given by \( K_R = \frac{1}{2} I \omega^2 \), where \( \omega \) is the angular speed. Since the sphere rolls without slipping, \( v = R\omega \).
Substitute the expression for \( I \) and \( v = R\omega \) into the rotational kinetic energy formula: \( K_R = \frac{1}{2} \times \frac{2}{3}MR^2 \times \omega^2 = \frac{1}{3}MR^2\omega^2 \).
The total kinetic energy \( K_{tot} \) is the sum of \( K_L \) and \( K_R \). Calculate the ratio \( \frac{K_L}{K_{tot}} \) by substituting the expressions for \( K_L \) and \( K_R \) into the equation \( K_{tot} = K_L + K_R \), and simplify the expression to find the ratio.
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