A person of mass M is sitting on a swing of l | vedclass

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(A) At the lowest point,the velocity of the person is $V_0 = \omega A = \sqrt{\frac{g}{L}} (	heta_0 L) = 	heta_0 \sqrt{gL}$.When the person stands u

Vedclass · Accepted Answer

$Mgl(1 + 	heta_0^2)$

Vedclass · Answer

(A) At the lowest point,the velocity of the person is $V_0 = \omega A = \sqrt{\frac{g}{L}} (	heta_0 L) = 	heta_0 \sqrt{gL}$.When the person stands up,the distance of the center of mass from the pivot changes from $L$ to $L-l$. By the conservation of angular momentum about the pivot point:$M V_0 L = M V_1 (L-l) \implies V_1 = V_0 \frac{L}{L-l} = V_0 (1 - \frac{l}{L})^{-1} \approx V_0 (1 + \frac{l}{L})$.The work-energy theorem states: $W_g + W_p = \Delta KE$.Here,$W_g = -Mgl$ (work done against gravity as the center of mass rises).$\Delta KE = \frac{1}{2} M (V_1^2 - V_0^2) = \frac{1}{2} M [V_0^2 (1 + \frac{l}{L})^2 - V_0^2] \approx \frac{1}{2} M V_0^2 (1 + \frac{2l}{L} - 1) = M V_0^2 \frac{l}{L}$.Substituting $V_0^2 = 	heta_0^2 gL$:$\Delta KE = M (	heta_0^2 gL) \frac{l}{L} = Mgl 	heta_0^2$.Thus,$W_p = Mgl + \Delta KE = Mgl + Mgl 	heta_0^2 = Mgl(1 + 	heta_0^2)$.

$A$ person of mass $M$ is sitting on a swing of length $L$ and swinging with an angular amplitude $\theta_0$. If the person stands up when the swing passes through its lowest point,the work done by him,assuming that his centre of mass moves by a distance $l$ $(l << L)$,is close to

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