AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictionless table ( x-y plane) and is hinged to it at the point A, so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity $ω$ .
At time T, when the side B C is parallel to the x-axis, a force F is applied on B along B C (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T.

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

(1) $\sqrt{6} {mlω}^{2}$

(2) $\sqrt{9} {mlω}^{2}$

(3) $\sqrt{5} {mlω}^{2}$

(4) $\sqrt{3} {mlω}^{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Angular acceleration of system about point A is

$α = \frac{τ_{A}}{I_{A}} = \frac{(F) (\frac{\sqrt{3}}{2} l)}{2 {ml}^{2}} = \frac{\sqrt{3} F}{4 ml}$

Now, acceleration of COM along x-axis is

$a_{x} = rα = (\frac{1}{\sqrt{3}}) (\frac{\sqrt{3} F}{4 ml}) or a_{x} = \frac{F}{4 m}$

Now, let $F_{x}$ be the force applied by the hinge along x-axis.

Then,

$F_{x} + F = (3 m) a_{x} or F_{x} + F = (3 m) (\frac{F}{4 m}) or F_{x} + F = \frac{3}{4} F or F_{x} = - \frac{F}{4}$

Further if $F_{y}$ be the force applied by the hinge along yaxis. Then,

$F_{y} = centripetal force F_{y} = \sqrt{3} {mlω}^{2}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!