In the arrangement shown, the rod is freely pivoted at point O and is in contact with the equilateral triangular block (whose front view is shown in figure) which can move on the horizontal frictionless ground. As the block is given a speed v forward, the rod rotates about point O . Find the angular velocity of rod in $rad / s$ at the instant when $θ = 30^{\circ}$ , $v = 20 m/s, a = 1$ meter.

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

answer is 5.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$AB = vdt sin θ$ … (i) (along the circular arc)
   $\begin{array}{r} OA = \frac{AN}{sin θ} \\ ds = rd θ \end{array}$
Question Image
vdt $sin θ = OAd θ$
( There is no relative motion to surface of rod between point on rod and the point on triangular block).
   $\begin{array}{r} vdt sin θ = \frac{AN}{sin θ} \cdot d θ \\ vdtsin θ = \frac{a}{sin θ} d θ \\ \frac{d θ}{dt} = \frac{v}{a} {sin}^{2} θ \\ \frac{d θ}{dt} = 5 rad / s \end{array}$
Or    $ω = \frac{v \sin θ}{O A} = \frac{v \sin^{2} θ}{a} .$