Question 1: A bob is whirled in a horizontal plane by means of a string with an initial speed of \omega \mathrm{rpm}. The tension in the string is T . If speed becomes 2 \omega while keeping the same radius, the tension in the string becomes :
(1) T
(2) 4 T
(3) \frac{\mathrm{T}}{4}
(4) \sqrt{2} \mathrm{~T}
Answer: Option (2)
Explanation:
When a bob is whirled in a horizontal circle, the required centripetal force is provided by the tension in the string.
The centripetal force is given by F=\frac{m v^{2}}{r},
where m is the mass of the bob,
v is its linear speed,
and r is the radius of the circular path.
Thus, the tension in the string is T=\frac{m v^{2}}{r}.
If the angular speed is \omega,
then the linear speed is v=r\omega.
Substituting, we get T=m r \omega^{2}.
This shows that tension is directly proportional to the square of angular speed,
i.e., T \propto \omega^{2}.
When the speed becomes 2\omega,
the new tension T' is
T' \propto (2\omega)^{2}=4\omega^{2}.
Hence, T'=4T.
Therefore, the tension in the string becomes four times the original value.