JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 3)
A conical pendulum of length 1 m makes an angle $$\theta $$ = 45o w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms−2 )
_9th_April_Morning_Slot_en_3_1.png)
0.4 m/s
4 m/s
0.2 m/s
2 m/s
Explanation
FBD of pendulum is :
_9th_April_Morning_Slot_en_3_2.png)
$$\therefore\,\,\,$$ T sin $$\theta $$ = $${{m{v^2}} \over r}$$
T cos $$\theta $$ = mg
$$\therefore\,\,\,$$ tan $$\theta $$ = $${{{v^2}} \over {rg}}$$
$$ \Rightarrow $$$$\,\,\,$$ tan45o = $${{{v^2}} \over {rg}}$$
$$ \Rightarrow $$$$\,\,\,$$ v2 = rg
$$ \Rightarrow $$$$\,\,\,$$ v = $$\sqrt {0.4 \times 10} $$ = 2 m/s
$$\therefore\,\,\,$$ T sin $$\theta $$ = $${{m{v^2}} \over r}$$
T cos $$\theta $$ = mg
$$\therefore\,\,\,$$ tan $$\theta $$ = $${{{v^2}} \over {rg}}$$
$$ \Rightarrow $$$$\,\,\,$$ tan45o = $${{{v^2}} \over {rg}}$$
$$ \Rightarrow $$$$\,\,\,$$ v2 = rg
$$ \Rightarrow $$$$\,\,\,$$ v = $$\sqrt {0.4 \times 10} $$ = 2 m/s
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