A uniform rod of mass M and length $l$ is hinged at point O and is free to rotate on a horizontal smooth surface. Point O is at a distance of $\frac{ℓ}{4}$ from one end of the rod. A sharp impulse p₀ $2 \sqrt{130} kgm / s$ applied along the surface at one end of the rod as shown in figure $[\tan θ = \frac{9}{7}]$ .If $ω b e t h e a n g u l a r v e l o c i t y o f t h e r o d i m m e d i a t e l y a f t e r t h e i m p a c t a n d P b e t h e i m p u l s e$
on the rod due to the hinge , then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$ω = \frac{24 P_{0} \sin θ}{5 m l}$

$ω = \frac{36 P_{0} \cos θ}{7 m l}$

$P = P_{0} \sqrt{{(2 \sin θ)}^{2} + {(3 \cos θ)}^{2}}$

P= $P_{0} \sqrt{\sin^{2} θ + {(\frac{2}{7} \cos θ)}^{2}}$

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Angular impulse = change in angular momentum
$\Rightarrow \frac{3 ℓ}{4} P_{0} \cos θ = [\frac{{mℓ}^{2}}{12} + \frac{{mℓ}^{2}}{16}] ω \Rightarrow ω = \frac{36 P_{0} \cos θ}{7 mℓ}$

Velocity of centre of mass after hit is
$V_{CM} = ω \frac{ℓ}{4} = \frac{9 P_{0} \cos θ}{7 m}$
Let the rectangular components of impulse by the hinge be $P_{x} and P_{y}$
$P_{x} = P_{0} \sin θ and P_{0} \cos θ + P_{y} = {mV}_{cm} \to P_{y} = \frac{2 P_{0} \cos θ}{7}$
$∴$ Impulse by the hinge has a magnitude