Question

suppose a rolling solid sphere has a mass of 1.50 kg and a linear average speed of 0.410 m/s. provided the sphere does not slip, find the total kinetic energy of the sphere. multiple choice 0.177 j 0.105 j 0.704 j 0.430 j

Explanation:

Step1: Recall total kinetic energy formula for rolling without slipping

For a rolling solid sphere (moment of inertia \( I = \frac{2}{5}MR^2 \)) and no - slip condition (\( v = \omega R \)), the total kinetic energy \( K_{total}=K_{trans}+K_{rot}=\frac{1}{2}Mv^{2}+\frac{1}{2}I\omega^{2} \). Substitute \( I = \frac{2}{5}MR^2 \) and \( \omega=\frac{v}{R} \) into the rotational kinetic energy term:
\( K_{rot}=\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v^{2}}{R^{2}})=\frac{1}{5}Mv^{2} \)
Then \( K_{total}=\frac{1}{2}Mv^{2}+\frac{1}{5}Mv^{2}=\frac{5 + 2}{10}Mv^{2}=\frac{7}{10}Mv^{2} \)

Step2: Substitute the given values

Given \( M = 1.50\space kg \), \( v = 0.410\space m/s \)
\( K_{total}=\frac{7}{10}\times1.50\times(0.410)^{2} \)
First, calculate \( (0.410)^{2}=0.1681 \)
Then \( \frac{7}{10}\times1.50 = 1.05 \)
Then \( K_{total}=1.05\times0.1681\approx0.1765\space J\approx0.177\space J \)

Answer:

0.177 J (the option corresponding to 0.177 J)