A light rod of length L is hinged to a plank of mass m. The plank is lying on the edge of a horizontal table such that the rod can swing freely in the vertical plane without any hindrance from the table. A particle of mass m is attached to the end of the rod and system is released from θ=0∘position (see figure) Assume that friction between the plank and the table is large enough to prevent it from slipping and the smallest normal force applied by the plank on the table is N. Again if that friction is absent everywhere , the speed of the plank when the rod makes θ=180∘ is u .Then

(a) Let the speed of the particle be u at position θEnergy conservation gives–12mu2=mgL(1−cos⁡θ)And the equation for centripetal force i T+mgcos⁡θ=mu2LT=2mg(1−cos⁡θ)−mgcos⁡θ T=mg(2−3cos⁡θ) For the plank normal force will be minimum when Tcos⁡θ is maximum That is when cos⁡θ(2−3cos⁡θ)is maximum⇒−sin⁡θ(2−3cos⁡θ)+cos⁡θ(3sin⁡θ)=0 ⇒−sin⁡θ(2−3cos⁡θ)+cos⁡θ(3sin⁡θ)=0 ⇒2sin⁡θ=6sin⁡θcos⁡θ ⇒cos⁡θ=13 ∴Nmin=mg−mgcos⁡θ(2−3cos⁡θ) =mg−mg⋅132−3⋅13=2mg3 Momentum conservation tells us that the velocity of the particle and the plank must be equal and opposite in horizontal direction when the rod gets vertical. ∴12mu2+12mu2=2mgL⇒u=4gL

A light rod of length L is hinged to a plank of mass m. The plank is lying on the edge of a horizontal table such that the rod can swing freely in the vertical plane without any hindrance from the table. A particle of mass m is attached to the end of the rod and system is released from θ=0∘position (see figure) Assume that friction between the plank and the table is large enough to prevent it from slipping and the smallest normal force applied by the plank on the table is N. Again if that friction is absent everywhere , the speed of the plank when the rod makes θ=180∘ is u .Then

(a) Let the speed of the particle be u at position θEnergy conservation gives–12mu2=mgL(1−cos⁡θ)And the equation for centripetal force i T+mgcos⁡θ=mu2LT=2mg(1−cos⁡θ)−mgcos⁡θ T=mg(2−3cos⁡θ) For the plank normal force will be minimum when Tcos⁡θ is maximum That is when cos⁡θ(2−3cos⁡θ)is maximum⇒−sin⁡θ(2−3cos⁡θ)+cos⁡θ(3sin⁡θ)=0 ⇒−sin⁡θ(2−3cos⁡θ)+cos⁡θ(3sin⁡θ)=0 ⇒2sin⁡θ=6sin⁡θcos⁡θ ⇒cos⁡θ=13 ∴Nmin=mg−mgcos⁡θ(2−3cos⁡θ) =mg−mg⋅132−3⋅13=2mg3 Momentum conservation tells us that the velocity of the particle and the plank must be equal and opposite in horizontal direction when the rod gets vertical. ∴12mu2+12mu2=2mgL⇒u=4gL

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A light rod of length L is hinged to a plank of mass m. The plank is lying on the edge of a horizontal table such that the rod can swing freely in the vertical plane without any hindrance from the table. A particle of mass m is attached to the end of the rod and system is released from $θ = 0^{\circ}$ position (see figure)

Assume that friction between the plank and the table is large enough to prevent it from slipping and the smallest normal force applied by the plank on the table is N. Again if that friction is absent everywhere , the speed of the plank when the rod makes $θ = 180^{\circ}$ is u .Then

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N = $\frac{3 m g}{4}$

$u = \sqrt{4 g l}$

$u = \sqrt{6 g l}$

N = $\frac{2 m g}{3}$

answer is B, C.

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Detailed Solution

(a) Let the speed of the particle be u at position $θ$ Energy conservation gives–
$\frac{1}{2} {mu}^{2} = mgL (1 - \cos θ)$
And the equation for centripetal force i $T + mgcos θ = \frac{{mu}^{2}}{L}$
$T = 2 mg (1 - \cos θ) - mgcos θ T = mg (2 - 3 \cos θ)$

For the plank normal force will be minimum when $Tcos θ$ is maximum

That is when $\cos θ (2 - 3 \cos θ)$ is maximum
$\Rightarrow - \sin θ (2 - 3 \cos θ) + \cos θ (3 \sin θ) = 0 \Rightarrow - \sin θ (2 - 3 \cos θ) + \cos θ (3 \sin θ) = 0 \Rightarrow 2 \sin θ = 6 \sin θcos θ \Rightarrow \cos θ = \frac{1}{3} ∴ N_{min} = mg - mgcos θ (2 - 3 \cos θ) = mg - mg \cdot \frac{1}{3} (2 - 3 \cdot \frac{1}{3}) = \frac{2 mg}{3}$

Momentum conservation tells us that the velocity of the particle and the plank must be equal and opposite in horizontal direction when the rod gets vertical. $∴ \frac{1}{2} {mu}^{2} + \frac{1}{2} {mu}^{2} = 2 mgL \Rightarrow u = \sqrt{4 gL}$