An object of uniform density rolls up a curve | vedclass

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(D) By the law of conservation of energy, the initial kinetic energy (translational + rotational) is converted into gravitational potential energy at

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solid sphere

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(D) By the law of conservation of energy, the initial kinetic energy (translational + rotational) is converted into gravitational potential energy at the maximum height $h$.Initial kinetic energy $K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega^2$.Since the object rolls without slipping, $\omega = \frac{v_0}{R}$.Substituting the moment of inertia $I = kmR^2$, we get:$K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}(kmR^2)(\frac{v_0}{R})^2 = \frac{1}{2}mv_0^2(1+k)$.At maximum height $h$, the potential energy is $U_f = mgh$.Equating $K_i = U_f$, we have $\frac{1}{2}mv_0^2(1+k) = mgh$.Thus, $h = \frac{v_0^2(1+k)}{2g}$.Given $h = \frac{7v_0^2}{10g}$, we equate the two expressions:$\frac{v_0^2(1+k)}{2g} = \frac{7v_0^2}{10g} \Rightarrow 1+k = \frac{14}{10} = 1.4 \Rightarrow k = 0.4$.For a solid sphere, the moment of inertia is $I = \frac{2}{5}mR^2$, so $k = 0.4$.Therefore, the object is a solid sphere.

An object of uniform density rolls up a curved path with an initial velocity $v_0$ as shown in the figure. If the maximum height attained by the object is $\frac{7v_0^2}{10g}$ ($g =$ acceleration due to gravity), the object is a . . . . . . .

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