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A rod of length 'L' is hung from its one end and a mass 'm' is attached to its free
end. What tangential velocity must be imparted to 'm', so that it reaches the top of
the vertical circle? (g = acceleration due to gravity)
Options:
end. What tangential velocity must be imparted to 'm', so that it reaches the top of
the vertical circle? (g = acceleration due to gravity)
Solution:
1871 Upvotes
Verified Answer
The correct answer is:
$2 \sqrt{\mathrm{gL}}$
$2 \sqrt{g L}$
$h=2 L$
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} 2 \mathrm{L}$
$v_{2}=4 g L$
$v=2 \sqrt{g L}$
$h=2 L$
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} 2 \mathrm{L}$
$v_{2}=4 g L$
$v=2 \sqrt{g L}$
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