First slide

Circular motion in horizontal plane

Question

A string of length l = 1 m is fixed at one end and carries a mass of 100gm at other end. The string makes $\sqrt{5} / π$ revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical ? Take g = 10 m/s²

Moderate

A

$30^{\circ}$
B

$45^{\circ}$
C

$60^{\circ}$
D

$75^{\circ}$

Solution

The different forces are shown in fig. (10). From figure
$Tsin θ = \frac{{mv}^{2}}{r} = {mω}^{2} r = {mω}^{2} lsin θ . . . . (1)$
and $Tcos θ = mg$
From eq. (1),
$T = {mω}^{2} l = {mω}^{2} (∵ l = 1 m)$
From eq. (2) ${mω}^{2} \cos θ = mg$
$∴ \cos θ = \frac{g}{ω^{2}} = \frac{g}{4 π^{2} n^{2}} = \frac{10}{4 π^{2} (5 / π^{2})}$
or $\cos θ = 0 .5 i.e., θ = 60^{\circ}$

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