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  • A block of mass 1kg slides with velocity on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about

A block of mass 1kg slides with velocity v=6m/s on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle \theta before momentarily coming to rod. If the rod has mass M=2kg and length 1m. The value of \theta is approx ( take g=10m/s^2).
Option: 1 63^{o}
 
Option: 2 55^{o}
Option: 3 69^{o}  
Option: 4 49^{o}

Answers (1)

best_answer

\\ \text{Angular momentum conservation}\\ mvr=\left[\frac{Ml^2}{3}+ml^2\right] \omega \\ \Rightarrow 1 \times 6 \times 1=\left[\frac{2}{3}+1 \right ] \omega\\ \Rightarrow \omega=\frac{18}{5} rad/s\\ \text{Applying energy conservation}\\ \frac{1}{2}I \left[\frac{18}{5} \right ]^2+ Mg\frac{l}{2}=\frac{1}{2}I(0)+Mg\left[\frac{l}{2}+\frac{l}{2}(1-cos \theta) \right ]+mgl(1-cos \theta)\\ \Rightarrow I=\frac{Ml^2}{3}+ml^2=\frac{5}{3}\\ \Rightarrow (1-cos \theta) 20=\frac{1}{2} \times \frac{5}{3} \times \left[\frac{18}{5} \right]^2\\ \Rightarrow \theta \approx 63^o

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avinash.dongre

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