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5.Work, Energy, Power and Collision
normal
The mass of the bob of a simple pendulum of length $L$ is $m$. If the bob is left from its horizontal position then the speed of the bob and the tension in the thread in the lowest position of the bob will be respectively
A
$\sqrt {2gL}$ and $3\,mg$
B
$3\,mg$ and $\sqrt {2gL}$
C
$2\,mg$ and $\sqrt {2gL}$
D
$3\,gL$ and $3\,mg$
Solution
$\mathrm{O}+\mathrm{mgL}=\frac{1}{2} \mathrm{mv}_{\mathrm{L}}^{2}+\mathrm{O}$
$\Rightarrow \quad V_{L}=\sqrt{2 g L}$
at lowest point $:-$
$\Rightarrow \mathrm{T}_{\mathrm{L}}-\mathrm{mg}=\frac{\mathrm{mv}_{\mathrm{L}}^{2}}{\mathrm{L}}$
$\mathrm{T}_{\mathrm{L}}=3 \mathrm{mg}$
Standard 11
Physics
