Q. A uniform rod of length `l` hinged at the lower end is free to rotate in the vertical plane. If the rod is held vertically in the beginning and then released the angular acceleration of the rod...

Q. A uniform rod of length `l` hinged at the lower end is free to rotate in the vertical plane. If the rod is held vertically in the beginning and then released the angular acceleration of the rod when it makes an angle of 45° with the horizontal (`I= (ml^2)/3` ) :-

A)  `(3g)/(2sqrt2l)`

B) `(6g)/(sqrt2l)`

C) `(sqrt2g)/l`

D) `(2g)/l`

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mvcdc | Student, Graduate | (Level 2) Associate Educator

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We know that the rod has length l and is free to rotate. Also, it makes an angle of `45^o` with the horizontal.

First, we note that `tau = I times omega` , where `tau` is the torque, `I` the moment of inertia and `w` the angular acceleration.` `

We are given the moment of inertia: `I = (ml^2)/3` and we want to look for the angular acceleration. To do so, we first need to solve for the torque acting on the rod. Torque is given by the following:` `

`tau = (mgl)/2 cos(theta)` , which we can express, by substituting the angle, as:

`tau = (mgl)/2 cos(45^o) = (mgl)/2 * (sqrt(2)/2) = (sqrt(2)mgl)/4`

Now that we have an expression for the torque, we solve for the angular acceleration:

`tau = I times w rArr w = tau / I`

`omega = [(sqrt(2)mgl)/4]/[(ml^2)/3]`

`omega = (3sqrt(2)g)/(4l)`

Note that this is equivalent to:

`omega = (3g)/(2sqrt(2)l)`

[To get the transformation, simply multiply either expression by `sqrt(2)/sqrt(2)` to get the other.]` `

Hence, answer is A

Sources:

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