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(B) The velocity at the bottommost point is $v_0 = 10 \, m/s$. The radius is $R = 2.5 \, m$. The critical velocity required to complete the vertical c

Vedclass · Accepted Answer

$N_1 > 0$ for motion in $CDA$,$N_2 > 0$ for motion in $ABC$

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(B) The velocity at the bottommost point is $v_0 = 10 \, m/s$. The radius is $R = 2.5 \, m$. The critical velocity required to complete the vertical circle is $v_c = \sqrt{5gR} = \sqrt{5 	imes 10 	imes 2.5} = \sqrt{125} \approx 11.18 \, m/s$. Since $v_0 At any angle $	heta$ from the bottom,the radial equation is $N - mg \cos 	heta = \frac{mv^2}{R}$,where $N$ is the normal force. By conservation of energy,$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgR(1 - \cos 	heta)$,so $v^2 = v_0^2 - 2gR(1 - \cos 	heta) = 100 - 50(1 - \cos 	heta) = 50 + 50 \cos 	heta$.Substituting $v^2$ into the radial equation: $N = mg \cos 	heta + \frac{m}{R}(50 + 50 \cos 	heta) = mg \cos 	heta + 20mg + 20mg \cos 	heta = mg(20 + 21 \cos 	heta)$.For the bottom half (where $\cos 	heta > 0$),$N > 0$,meaning the ball presses against the outer wall $(N_2)$.For the top half (where $\cos 	heta Thus,$N_2 > 0$ for the lower part $(ABC)$ and $N_1 > 0$ for the upper part $(CDA)$.

List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

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