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Q.

A rod of mass M and length L is placed in a horizontal plane with one end hinged about the vertical axis. A vertical force F=Mg2 is applied at a distance 5L6 from the hinged end. The angular acceleration of the rod will be

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a

3g4L

b

4g5L

c

5g4L

d

4g3L

answer is B.

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Detailed Solution

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We know that τ=Iα… (i) 

I=ML23 (moment of inertia of rod wrt axis passing from one end and perpendicular to its length)

Now τ=(force) (perpendicular distance)

τ=Mg25L6

from eq. (i)

Mg25L6=ML23(α)

solving this

⇒α=5g4L

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