Question 14: Laws of Motion

Question

A uniform rod AB of length L = 1 m is sliding along two mutually perpendicular surfaces OP and OQ as shown in the figure.

When the rod subtends an angle $$	heta$$ = 30$$^\circ$$ with OQ, the end B has a velocity $$\sqrt3$$ m/s. The velocity of end A at that time is

Accepted Answer

According to the question,
Let OB = x, OA = y and x = L cos$$	heta$$
So, by Pythagoras theorem,
$$\Rightarrow$$ x
2
+ y
2
= L
2
Differentiating w.r.t. 't', we get
$$2x{{dx} \over {dt}} + 2y{{dy} \over {dt}} = 0$$
$$2x{v_B} + 2y{v_A} = 0$$ [$$\because$$ $${{dx} \over {dt}} = {v_B}$$ and $${{dy} \over {dt}} = {v_A}$$]
$$ \Rightarrow {v_A} =  - {x \over y}{v_B}$$
$$ \Rightarrow \left| {{v_A}} ight| = {x \over y}{v_B} = {v_B}\cot 	heta $$ [$$\because$$ $${x \over y} = \cot 	heta $$]
$$ = \sqrt 3  	imes \cot 30^\circ $$
$$ = \sqrt 3  	imes {1 \over {\sqrt 3 }}$$ = 1 m/s

Answer

1 m/s

Answer

0.5 m/s

Answer

$$\sqrt3$$ m/s

Answer

$${1 \over {\sqrt 3 }}$$ m/s

Laws of Motion

A uniform rod AB of length L = 1 m is sliding along two mutually perpendicular surfaces OP and OQ as shown in the figure. When the rod subtends an angle $\theta = 30^\circ with OQ, the end B has a velocity \sqrt3$ m/s. The velocity of end A at that time is

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