A sphere of mass 2 , kg and radius 0.5 , m is | vedclass

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(C) For a sphere rolling up an incline without slipping,the acceleration $a$ is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.For a solid sp

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$0.57$

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(C) For a sphere rolling up an incline without slipping,the acceleration $a$ is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}}$.For a solid sphere,the moment of inertia $I = \frac{2}{5} mR^2$.Substituting this,$a = \frac{g \sin 	heta}{1 + \frac{2}{5}} = \frac{g \sin 	heta}{7/5} = \frac{5}{7} g \sin 	heta$.Given $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$ for approximation),$	heta = 30^{\circ}$,and $v_0 = 1 \, m/s$.Using $g = 9.8 \, m/s^2$: $a = \frac{5}{7} 	imes 9.8 	imes \sin(30^{\circ}) = \frac{5}{7} 	imes 9.8 	imes 0.5 = 3.5 \, m/s^2$.The time taken to reach the highest point is $t_{up} = \frac{v_0}{a} = \frac{1}{3.5} = \frac{2}{7} \, s$.The total time to return to point $A$ is $T = 2 	imes t_{up} = 2 	imes \frac{2}{7} = \frac{4}{7} \approx 0.57 \, s$.

$A$ sphere of mass $2 \, kg$ and radius $0.5 \, m$ is rolling with an initial speed of $1 \, m/s$ up an inclined plane which makes an angle of $30^{\circ}$ with the horizontal plane,without slipping. How long will the sphere take to return to the starting point $A$? (in seconds)

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