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Question: .A rod of length L is sliding such that one of its ends is always in contact with a vertical wall an...

.A rod of length L is sliding such that one of its ends is always in contact with a vertical wall and its other end is always in contact with horizontal surface. Just after the rod is released from rest, the magnitude of acceleration of rod at this instant will be -

A

a+b\frac { a + b } { \ell }

B

a2b2\frac { \sqrt { \left| \mathrm { a } ^ { 2 } - \mathrm { b } ^ { 2 } \right| } } { \ell }

C

a2+b2\frac { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } { \ell }

D

None

Answer

a2+b2\frac { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } { \ell }

Explanation

Solution

At the initial moment, angular velocity of rod is zero. Acceleration of end B of rod with respect to end A is shown in figure. Centripetal acceleration of point B with respect to A is zero (Q w2l = 0)

So at the initial moment, acceleration of end B with respect to end A is perpendicular to the rod which is equal to a2+b2\sqrt { a ^ { 2 } + b ^ { 2 } }

arel = l a

a2+b2\frac { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } } } { \ell } = a where a is angular acceleration