A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant, if we want the tension in the string to be doubled, then the new speed should nearly be

Question

A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant, if we want the tension in the string to be doubled, then the new speed should nearly be (A) 2.25 rpm (B) 7 rpm (C) 10 rpm (D) 14 rpm

Tardigrade Admin · Accepted Answer

F = m ω2 r = mr (2 n π)2 thus F × n 2 n = resolution per second. thus (2 F/F)=( n prime 2/ n ) or n prime=√2 n =5 √2 rpm =7 rpm

Q. A mass of $2 k g$ is whirled in a horizontal circle by means of a string at an initial speed of $5$ revolutions per minute. Keeping the radius constant, if we want the tension in the string to be doubled, then the new speed should nearly be

$2.25 r p m$

$7 r p m$

$10 r p m$

$14 r p m$

$F = m ω^{2} r$
$= m r (2 nπ)^{2}$
thus $F \times n^{2} n =$ resolution per second.
thus $\frac{2 F}{F} = \frac{n ^{'2}}{n}$
or $n^{'} = 2 n$
$= 52 r p m$
$= 7 r p m$

Q. A mass of 2kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant, if we want the tension in the string to be doubled, then the new speed should nearly be

2.25rpm

7rpm

10rpm

14rpm