A uniform rod has mass $m = 2 k g$ and length $l = 13 m$ . One end of the rod is pulled with a constant velocity of $v_{0} = 34 m / s$ along a frictionless horizontal floor in the negative x direction. The other end is moving along a parabolic fixed curve. The equation of the parabola is $x^{2} = 20 y$ . Find the angular velocity of the rod (in rad/s) when the end point ‘B’ is at (10, 5)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Question Image

$\tan ϕ = \frac{d y}{d x} = \frac{2 x}{20} = \frac{2 \times 10}{20} = 1 ϕ = 45 ° {\vec{v}}_{B} = {\vec{v}}_{A} + \vec{ω} \times \vec{A B} \Rightarrow \frac{v_{B}}{\sqrt{2}} (- \overset{\land}{i} - \overset{\land}{j}) = 34 (- \overset{\land}{i}) + ω \times 13 {\sin θ \overset{\land}{i} + \cos θ (- \overset{\land}{j})} \Rightarrow (\frac{- v_{B}}{\sqrt{2}}) \overset{\land}{i} + (\frac{- v_{B}}{\sqrt{2}}) \overset{\land}{j} = \overset{\land}{i} {- 34 + 13 ω \times \frac{5}{13}} + (- 13 ω \times \frac{12}{13}) \overset{\land}{j} \Rightarrow \frac{v_{B}}{\sqrt{2}} = 34 - 5 ω & \frac{v_{B}}{\sqrt{2}} = 12 ω \Rightarrow 34 - 5 ω = 12 ω \Rightarrow ω = 2 r a d / s$

Ans. 2

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET