A uniform circular disc of radius ‘R’ is rolling without slipping on a rough horizontal surface with a constant acceleration ‘a’. Then the radius of curvature of trajectory of point ‘A’ of the disc relative to the ground at the given instant as shown in the figure is found to be $a \sqrt{b} R$ . Then $a + b =$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Velocity of point ‘A’ $V_{A} = \sqrt{V^{2} + ω^{2} R^{2}} = v \sqrt{2}$ normal acceleration of point A,

$a_{A} (n) = ω^{2} R \cos 45^{0} + α R \cos 45^{0} - a \cos 45^{0}, a_{A (n)} = \frac{ω^{2} R}{\sqrt{2}} = \frac{V^{2}}{\sqrt{2} R}$
$∴$ Radius of curvature of trajectory of point ‘A’ relative to the ground is
$r = \frac{{(V_{A})}^{2}}{^{a} A (n)} = \frac{{(V \sqrt{2})}^{2}}{\frac{V^{2}}{\sqrt{2} R}} = 2 \sqrt{2} R$

Watch 3-min video & get full concept clarity

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!

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