Rotational Motion

JEE Main level, single correct, medium

219 of 280

Question 219 of 280

A block of mass m=1kgm=1kg slides with velocity v=6ms−1v=6m{s}^{-1} on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about OO and swings as a result of the collision making angle θ\theta before momentarily coming to rest. if the rod has mass M=2kg,M=2kg, and length ℓ=1m,\ell =1m, the value of θ\theta is approximately(takeg=10ms−2)\left.(takeg=10m{s}^{-2}\right)

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Answer: (A) 63∘63^{\circ}

Angular momentum
mvl=(ml2+2ml23)ωmvl=\left(m{l}^{2}+\dfrac{2m{l}^{2}}{3}\right)\omega
mvl=53ml2ωmvl=\dfrac{5}{3}m{l}^{2}\omega
ω=3v5l\omega =\dfrac{3v}{5l}
12Iω2=2mgl2(1−cos⁡θ)+mgl(1−cos⁡θ)\dfrac{1}{2}{I\omega }^{2}=2mg\dfrac{l}{2}\left(1-\cos \theta \right)+mgl\left(1-\cos \theta \right)
12(53ml2)9v225l2=2mgl(1−cos⁡θ)\dfrac{1}{2}\left(\dfrac{5}{3}m{l}^{2}\right)\dfrac{9{v}^{2}}{25{l}^{2}}=2mgl\left(1-\cos \theta \right)
35×2mv2=2mgl(1−cosθ)\dfrac{3}{5\times 2}{mv}^{2}=2mgl\left(1-cos\theta \right)
310×362×10=1−cos⁡θ\dfrac{3}{10}\times \dfrac{36}{2\times 10}=1-\cos \theta
1−2750=cos⁡θ1-\dfrac{27}{50}=\cos \theta
cos⁡θ=2350=0.46\cos \theta =\dfrac{23}{50}=0.46
θ=cos⁡−1(0.46)=63∘\theta ={\cos }^{-1}\left(0.46\right)=63^{\circ}

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