A uniform sphere of mass 500 ; g rolls withou | vedclass

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(A) Given: Mass $m = 500 \; g = 0.5 \; kg$,velocity $v = 5.00 \; cm/s = 0.05 \; m/s$.For a solid sphere rolling without slipping,the moment of inertia

Vedclass · Accepted Answer

$8.75 	imes 10^{-4} \; J$

Vedclass · Answer

(A) Given: Mass $m = 500 \; g = 0.5 \; kg$,velocity $v = 5.00 \; cm/s = 0.05 \; m/s$.For a solid sphere rolling without slipping,the moment of inertia about the center of mass is $I = \frac{2}{5} mR^2$.The total kinetic energy $KE$ is the sum of translational and rotational kinetic energy:$KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$.Since $v = R\omega$,we have $\omega = v/R$.$KE = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{2}{5} mR^2) (\frac{v}{R})^2 = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2$.Substituting the values:$KE = 0.7 	imes 0.5 \; kg 	imes (0.05 \; m/s)^2 = 0.35 	imes 0.0025 \; J = 8.75 	imes 10^{-4} \; J$.

$A$ uniform sphere of mass $500 \; g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00 \; cm/s$. Its kinetic energy is

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