A 4.00-kg block is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in Fig. P5.104 and the tension in the upper string is 80.0 N.

(a) What is the tension in the lower cord?

(b) How many revolutions per minute does the system make?

(c) Find the number of revolutions per minute at which the lower cord just goes slack.

(d) Explain what happens if the number of revolutions per minute is less than that in part (c).

• The mass of the block,$m=4kg$ .
• The tension in the upper string,$T_1=80N$ .

According to Newton’s second law, the force is the product of the mass and acceleration of the object. The force that acts on a string or wire when pulled by forces acting in opposite directions is known as tension force.

The given situation is as follows.

From the above figure, the distance r is calculated as follows.

$$\begin{gathered} r^2+{\left(1m\right)}^2={\left(1.25m\right)}^2 \\ r=0.75m \end{gathered}$$

The angle is calculated as follows.

$$\begin{gathered} \theta ={\sin }^{-1}\left(\frac{1}{1.25}\right) \\ =53.13° \end{gathered}$$

Draw the free-body diagram of the block.

Using Newton’s second law, the net force acting along the y-axis is given by,

$$\begin{gathered} \sum _{}{F}_y=0 \\ {T}_1\sin \left(53.13°\right)-mg-T_2\sin \left(53.13°\right)=0 \\ {T}_2=\frac{T_1\sin \left(53.13°\right)-mg}{\sin \left(53.13°\right)}..........\left(1\right) \end{gathered}$$

Here,$T_2$is tension in the lower string, and g is the acceleration due to gravity.

Substitute$9.8m/s^2$for g, 4 kg for m , and 80 N for$T_1$in the equation (1), and we get,

$$\begin{gathered} T_2=\frac{\left(80N\right)\sin \left(53.13°\right)-\left(4kg\right)\left(9.8m/s^2\right)}{\sin \left(53.13°\right)} \\ =31.0N \end{gathered}$$

Therefore, the tension in the lower cord is 31.0 N .

Let the rotational speed of the block be v .

The centripetal force of the block is given by,

$\frac{m{v}^2}{r}=T_1\cos \left(53.13°\right)+T_2\cos \left(53.13°\right).........\left(2\right)$

Substitute 31.0 N for $T_2$, 4 kg for m , 0.750 m for r , and 80 N for$T_1$ in the equation (2), and we get,

$$\begin{gathered} \frac{\left(4kg\right)v^2}{0.750m}=\left(80N\right)\cos \left(53.13°\right)+\left(31.0N\right)\cos \left(53.13°\right) \\ {v}^2=12.4875 \\ v=\sqrt{12.4875} \\ =3.534m/s \end{gathered}$$

Calculate the angular velocity as follows:

$$\begin{gathered} \omega =\frac{v}{r} \\ =\frac{3.534m/s}{0.75m}\left(\frac{1rev}{2\mathrm{\pi rad}}\right)\left(\frac{60s}{1min}\right) \\ =45rev/min \end{gathered}$$

Therefore, the number of revolutions is 45 rev/min .

The lower cord goes slack means$T_2=0$ then,

$$\begin{gathered} T_1\sin \left(53.13°\right)=mg \\ {T}_1=\frac{\left(4kg\right)\left(9.81m/s^2\right)}{\sin \left(53.13°\right)} \\ =49.05N \end{gathered}$$

Calculate the rotational speed of the lock as follows:

$$\begin{gathered} \frac{\left(4kg\right)v^2}{0.750m}=\left(49.05N\right)\cos \left(53.13°\right) \\ v=2.349m/s \end{gathered}$$

Calculate the angular velocity as follows:

$$\begin{gathered} \omega =\frac{v}{r} \\ =\frac{2.349m/s}{0.75m}\left(\frac{1rev}{2\mathrm{\pi rad}}\right)\left(\frac{60s}{1min}\right) \\ =29.9rev/min \end{gathered}$$

Therefore, the number of revolutions is 29.19 rev/min .

Suppose the number of revolutions per minute is less than 29.9 rev/min , then$T_1\sin \theta <mg$ . It is required that$\theta \text{'}>\theta$ and let$T_1\sin \theta \text{'}=mg$ . So, the block will still rotate around the axis, but the rotational angle$\left(90°-\theta \text{'}\right)$ will be smaller.

Therefore, if the revolution is reduced, then the angle must be changed so that the vertical component of$T_1$ will increase to achieve Newton’s law.