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3-2.Motion in Plane
medium
If the radius of curvature of the path of two particles of same masses are in the ratio $1 : 2$, then in order to have constant centripetal force, their velocity, should be in the ratio of
A$1:4$
B$4:1$
C$\sqrt 2:1$
D$1\,\,:\,\,\sqrt 2 $
Solution
$(d)$ The centripetal force, $F = \frac{{m{v^2}}}{r}$
$⇒$ $r = \frac{{m{v^2}}}{F}$
$⇒$ $r \propto {v^2}$ or $v \propto \sqrt r $ (If $m$ and $F$ are constant),
$⇒$ $\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{r_1}}}{{{r_2}}}} = \sqrt {\frac{1}{2}} $
$⇒$ $r = \frac{{m{v^2}}}{F}$
$⇒$ $r \propto {v^2}$ or $v \propto \sqrt r $ (If $m$ and $F$ are constant),
$⇒$ $\frac{{{v_1}}}{{{v_2}}} = \sqrt {\frac{{{r_1}}}{{{r_2}}}} = \sqrt {\frac{1}{2}} $
Standard 11
Physics
