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Byju's Answer

Standard X

Physics

Faraday's Experiment

A thin unifor...

Question

A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel force of lever arm $I$ . One force is of magnitude $F$ and acts at one extreme end. The length of the rod is:

\frac{2 (F + m a) I}{m a}

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I (1 + \frac{F}{m a})

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\frac{(F + m a) I}{2 m a}

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\frac{m a I}{m a + F}

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Solution

The correct option is A $\frac{2 (F + m a) I}{m a}$
Let L be the length of the rod and F₁ the magnitude of other force C is the center of the rod.

$F_{1} - F = m a$

$F_{1} = F + m a . . . . (i)$

Now, net torque about center of the rod should be zero

$F (\frac{L}{2}) = F_{1} (\frac{L}{2} - I)$
now, put the value of $F_{1}$ from equation (i)

$F (\frac{L}{2}) = F + m a (\frac{L}{2} - I)$

$F (\frac{L}{2}) = F + m a (\frac{L}{2}) - (F + m a) I$

$F (\frac{L}{2}) = F (\frac{L}{2}) + m a (\frac{L}{2}) - (F + m a) I$

$L = \frac{2 (F + m a) I}{m a}$

Hence, the length of the rod is $\frac{2 (F + m a) I}{m a}$

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A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel force of lever arm I. One force is of magnitude F and acts at one extreme end. The length of the rod is:

A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel force of lever arm $I$ . One force is of magnitude $F$ and acts at one extreme end. The length of the rod is: