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4-1.Newton's Laws of Motion
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A body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1\, cm$. If the angular velocity is doubled, the elongation in the spring is $5\, cm$. The original length of the spring is ......... $cm$
A$15$
B$12$
C$16$
D$10$
(JEE MAIN-2013)
Solution
Let the length of the spring is $l$. When the system is whirled round in a horizontal circle the centripetal force is given by
$F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=mr{{\omega }^{2}}$
then $r=l+$ elongation Given: elongation $=1 \;cm$ (in the first case)
or angular velocity $\omega$ the force required is
${{F}_{1}}=m(l+1){{\omega }^{2}}=kx=k\times 1=k$
$k=m(l+1){{\omega }^{2}}$ …….(i)
${{F}_{2}}=m(l+5){{(2\omega )}^{2}}=kx=k\times 5=5k$
$5k=4m(l+5){{\omega }^{2}}$…….(ii)
Now, dividing Eq. $(i)$ by Eq.$(ii)$, we get
$\frac{k}{5k}=\frac{m(l+1){{\omega }^{2}}}{4m(l+5){{\omega }^{2}}}$
$5(l+1)=4(l+5)$
$l=20-5=15\,cm$
$F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=mr{{\omega }^{2}}$
then $r=l+$ elongation Given: elongation $=1 \;cm$ (in the first case)
or angular velocity $\omega$ the force required is
${{F}_{1}}=m(l+1){{\omega }^{2}}=kx=k\times 1=k$
$k=m(l+1){{\omega }^{2}}$ …….(i)
${{F}_{2}}=m(l+5){{(2\omega )}^{2}}=kx=k\times 5=5k$
$5k=4m(l+5){{\omega }^{2}}$…….(ii)
Now, dividing Eq. $(i)$ by Eq.$(ii)$, we get
$\frac{k}{5k}=\frac{m(l+1){{\omega }^{2}}}{4m(l+5){{\omega }^{2}}}$
$5(l+1)=4(l+5)$
$l=20-5=15\,cm$
Standard 11
Physics
