A particle of mass m is attached to a rod of length ‘ $l$ ’ and it rotates in a circle with angular velocity $ω$ . An observer P is rigidly fixed on the rod at a distance $L / 2$ from the centre. The acceleration of ‘m’ and the pseudo force on ‘m’ from the reference frame of P must be respectively

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Zero, zero

zero, $\frac{m ω^{2} L}{2}$

$ω^{2} L, \frac{m ω^{2} L}{2}$

zero, $m ω^{2} L$

answer is D.

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Detailed Solution

Velocity of mass ‘m’ as seen by ‘P’ is
$\begin{array}{l} \vec{V_{m}} - \vec{V_{p}} + \vec{W} \times (\vec{r_{p}} - \vec{r_{m}}) \\ = W L \hat{j} - W \frac{L}{2} \hat{j} + w (\hat{k} + (\frac{- L}{2} \hat{i})) \\ = \frac{W L}{2} \hat{j} - \frac{W L}{2} \hat{j} = 0 \end{array}$
Acceleration of mass ‘m’ as seen by ‘P’
$\begin{array}{l} = - w^{2} L \hat{i} - (- w^{2} \frac{L}{2} \hat{i}) + 0 + w^{2} (\hat{k}) \times [\hat{k} \times (\frac{- w L}{2} \hat{i})] + 0 \\ = - w^{2} L \hat{i} + w^{2} \frac{L}{2} \hat{i} + w^{2} \frac{L}{2} \hat{i} \\ = 0 \end{array}$
Isndo force of ‘m’ as seen by ‘P’
$\begin{array}{l} = m [- \vec{a_{p}} + \vec{α} \times (\vec{r_{P}} - \vec{r_{m}}) + \vec{w} \times (\vec{w} \times (\vec{r_{P}} - \vec{r_{m}})) + 2 \vec{w} \times \vec{v_{r e l}}] \\ = m [- \vec{a_{m}} + \vec{a} o f m . w . r . t p] \\ = m [- (- w^{2} L \hat{i}) + 0] \\ = m w^{2} L \hat{i} \end{array}$