Question

Two strings are tied $6\text{m}$ apart on a straight vertical rod and on the other end, a mass of $100\text{g}$ is tied as shown in figure.


Each string is $5\text{m}$ long. Tensions $T_1$ and $T_2$ are experienced by the string when the rod is rotated with $60\text{rpm}$.
Take $g=10{\text{m/s}}^2$ and ${\pi }^2\simeq 10$ for calculation.

• A. Value of $T_1$ is $10.83\text{N}$
• B. Value of $T_2$ is $9.17\text{N}$
• C. Angular velocity of rod is $2\pi \text{rad/s}$
• D. Angular velocity of rod is $\pi \text{rad/s}$

Answer 1

Answer: (A), (B), (C)

Angular velocity of rod is $2\pi \text{rad/s}$

Given, $m=100\text{g}=0.1\text{kg}$
The angular speed of the rod is,
$\omega =\frac{2\pi N}{60}=\frac{2\pi \times 60}{60}=2\pi \text{rad/s}$
Drawing the FBD of block from the rotating frame by applying a centrifugal force $\left(m{\omega }^{2}r\right)$ in the radially outward direction.

FBD of block:


Applying the condition of vertical equilibrium,
$T_1\sin \theta =T_2\sin \theta +mg...\left(i\right)$

and applying the condition of horizontal equilibrium,
$T_1\cos \theta +T_2\cos \theta =mr{\omega }^2$
$\Rightarrow (T_1+T_2)\cos \theta =mr{\omega }^2....\left(ii\right)$
From the geometry of rod and string, we get the following triangle.


Here, radius of the circular path is the base of triangle,
$r=\sqrt{5^2-3^2}=4\text{m}$
$\Rightarrow \sin \theta =\frac{3}{5}\&\cos \theta =\frac{4}{5}$
On substituting the values in Eq.$\left(ii\right)$,
$\Rightarrow (T_1+T_2)\times \frac{4}{5}=0.1\times 4\times (2\pi {)}^2$
Using ${\pi }^2=10$, we get
$T_1+T_2=20....\left(iii\right)$
Substituting the values in Eq.$\left(i\right)$, we get
$\frac{3}{5}{T}_1-\frac{3}{5}{T}_2=1$
$\Rightarrow T_1-T_2=\frac{5}{3}.....\left(iv\right)$ .

From Eq. $\left(iii\right)\&\left(iv\right)$,
$2T_1=\frac{65}{3}$
$\Rightarrow T_1=10.83\text{N}$
and $T_2=20-10.83$
$\therefore T_2\simeq 9.17\text{N}$
Thus, options $\left(a\right),\left(b\right),\left(c\right)$ are correct.