First slide

Rigid body doing transational plus rotational motion

Question

Lower end B of a thin rod AB of length 1 m is made to move with a constant velocity of 1 m/s as shown in figure. Initial angle of inclination the rod with horizontal is $90^{0} .$ Then magnitude of angular acceleration of the rod at t = 0.5 second is

Difficult

A

$\frac{3}{2} rad / s^{2}$
B

$\frac{1}{2} rad / s^{2}$
C

$\frac{2}{3} rad / s^{2}$
D

$\frac{1}{3} rad / s^{2}$

Solution

At time t, Let $θ$ be the angle of inclination of the rod with horizontal and at that instant of time ‘e’ is the instantaneous centre of velocity of the rod. Therefore instantaneous angular velocity of the rod is given by $ω = \frac{V_{B}}{BC} = \frac{V}{l \sin θ}$
$\Rightarrow α = \frac{dω}{dt} = - \frac{V}{l} cosec θ \cot θ$
Now, $\cos θ = \frac{vt}{l} and at t = 0 .5 \sec, \cos θ = \frac{1 \times 0 .5}{1} = \frac{1}{2}$
$\Rightarrow θ = 60^{0} .$
$∴ At t = 0 .5 s, α = - \frac{1}{V} cosec 60^{0} . \cot 60^{0} rad / \sec^{2}$
$\Rightarrow α = - \frac{2}{\sqrt{3}} . \frac{1}{\sqrt{3}} rad / \sec^{2} = - \frac{2}{3} rad / \sec^{2} .$

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