A bead of mass m slides without friction on t | vedclass

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(D) Let the top of the hoop be the reference level for gravitational potential energy $(U_g = 0)$.At the top,the spring is stretched by $x_i = 2R - R

Vedclass · Accepted Answer

$\sqrt{3 R g+\frac{k R^2}{m}}$

Vedclass · Answer

(D) Let the top of the hoop be the reference level for gravitational potential energy $(U_g = 0)$.At the top,the spring is stretched by $x_i = 2R - R = R$. The initial energy is $E_i = U_{g,i} + U_{s,i} + K_i = 0 + \frac{1}{2}kR^2 + 0 = \frac{1}{2}kR^2$.When the spring length is $R$,the bead is at an angle $	heta = 60^\circ$ from the vertical,as the distance from the bottom is $R$ and the radius is $R$,forming an equilateral triangle.The height of the bead from the top is $h = R + R \cos 60^\circ = R + \frac{R}{2} = \frac{3R}{2}$.The gravitational potential energy at this point is $U_{g,f} = -mg(\frac{3R}{2})$.The spring is at its natural length,so $U_{s,f} = 0$.The final kinetic energy is $K_f = \frac{1}{2}mv^2$.By the conservation of mechanical energy: $E_i = E_f \implies \frac{1}{2}kR^2 = -\frac{3mgR}{2} + \frac{1}{2}mv^2$.Multiplying by $2/m$: $\frac{kR^2}{m} = -3gR + v^2$.Thus,$v = \sqrt{3gR + \frac{kR^2}{m}}$.

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