Question

A rod and a block are of same mass. Initially rod is in horizontal position. What will be acceleration of tip of the rod just after the system is released from this position shown in figure?

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Solution

The correct option is **C** 3g8

Let us consider that just after releasing the angular acceleration of rod is α & linear acceleration of block is a, then from constraint theory a=lα, where l is the length of rod.

For rod, T×l−mg×12=ml23α

For block, mg - T = ma

Solving above equations we get,

g2=4a3

⇒a=3g8 which is also the acceleration of free end of the rod.

Let us consider that just after releasing the angular acceleration of rod is α & linear acceleration of block is a, then from constraint theory a=lα, where l is the length of rod.

For rod, T×l−mg×12=ml23α

For block, mg - T = ma

Solving above equations we get,

g2=4a3

⇒a=3g8 which is also the acceleration of free end of the rod.

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