Question 193: Work, Energy and Power & Circular Motion

Question

A body of mass 100 gram is tied to a spring of spring constant $8 \mathrm{~N} / \mathrm{m}$, while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angular speed $8 \mathrm{rad} / \mathrm{s}$ then the ratio of extension in the spring to its natured length will be

Accepted Answer

Mass: $m = 100\,	ext{g} = 0.1\,	ext{kg}$
Spring constant: $k = 8\,	ext{N/m}$
Angular speed: $\omega = 8\,	ext{rad/s}$
Motion: Mass is moving in circle on smooth surface, spring attached to center.
We are asked: ratio of
extension
($x$) to
natural length
($l$).
Step 1: Force Balance
Centripetal force is provided by spring force:
$ kx = m \omega^2 r $
Here $r$ is radius of circular motion = natural length $l$ + extension $x$:
$ r = l + x $
So:
$ kx = m \omega^2 (l + x) $
Step 2: Plug Known Values
Substitute $m=0.1$, $\omega = 8$, $k=8$:
$ 8x = (0.1)(64)(l+x) $
$ 8x = 6.4(l+x) $
Step 3: Rearrange
$ 8x - 6.4x = 6.4 l $
$ 1.6 x = 6.4 l $
$ \frac{x}{l} = \frac{6.4}{1.6} = 4 $
Final Answer:
$ \frac{	ext{extension}}{	ext{natural length}} = 4:1 $
Ã¢ÂÂ Correct Option:
D (4:1)

Answer

$1: 1$

Answer

$8: 1$

Answer

$2: 1$

Answer

$4: 1$

Work, Energy and Power & Circular Motion

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