91Ó°ÊÓ

• 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.

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

Draw the free-body diagram of the block.

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

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

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:

$$\begin{gathered} \mathrm{Ó\neg }=\frac{v}{r} \\ =\frac{3.534m/s}{0.75m}\left(\frac{1rev}{2\mathrm{Ï€°ù²¹»å}}\right)\left(\frac{60s}{1min}\right) \\ =45rev/min \end{gathered}$$

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:

$$\begin{gathered} \mathrm{Ó\neg }=\frac{v}{r} \\ =\frac{2.349m/s}{0.75m}\left(\frac{1rev}{2\mathrm{Ï€°ù²¹»å}}\right)\left(\frac{60s}{1min}\right) \\ =29.9rev/min \end{gathered}$$

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$T_1\sin \mathrm{θ}<mg$ . It is required that$\mathrm{θ}\text{'}>\mathrm{θ}$ and let$T_1\sin \mathrm{θ}\text{'}=mg$ . So, the block will still rotate around the axis, but the rotational angle$\left(90°-\mathrm{θ}\text{'}\right)$ 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.