A sphere of radius a and mass m rolls along a | vedclass

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(A) The total initial energy of the sphere is $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$.Since it rolls without slipping,$\omega_0 = \frac{v_0

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$\frac{7v_0^2}{10g \sin 	heta}$

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(A) The total initial energy of the sphere is $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2$.Since it rolls without slipping,$\omega_0 = \frac{v_0}{a}$.For a solid sphere,the moment of inertia is $I = \frac{2}{5}ma^2$.Substituting these into the energy equation: $E_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{2}{5}ma^2)(\frac{v_0}{a})^2 = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$.At the maximum height $h$,the sphere momentarily comes to rest,so its final kinetic energy is zero. By the law of conservation of energy,$E_i = E_f = mgh$.Therefore,$mgh = \frac{7}{10}mv_0^2$,which gives $h = \frac{7v_0^2}{10g}$.The distance $d$ traveled along the incline is related to height $h$ by $h = d \sin 	heta$.Thus,$d \sin 	heta = \frac{7v_0^2}{10g}$,which implies $d = \frac{7v_0^2}{10g \sin 	heta}$.This matches option $A$.

$A$ sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping,how far up the sphere will travel?

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