Question

Figure 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.040 kg, the string has length L=0.90 mand negligible mass, and the bob follows a circular path of circumference 0.94 m. What are

(a) the tension in the string and

(b) the period of the motion?

Short answer

$$\begin{gathered} a)T=0.4N \\ b)t=1.9s \end{gathered}$$

Step-by-step solution

Given data

• The circumference of the circular path “r” is 0.94 m .
• The length of the cord “L” is 0.9 m .
• The mass of bob “m” is 0.04 kg .

To understand the concept

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.

Calculate the angle made by the cord.

Use Newton’s second law of motion to calculate the tension in the cord. After calculating the tension, calculate the velocity of the bob and the time required to complete one revolution.

(a) Calculate the tension in the string

The radius of the circular path R,

$R=\frac{r}{2\mathrm{\pi }}=\frac{0.94\mathrm{m}}{2\mathrm{\pi }}=0.15\mathrm{m}$

The angle cord makes with horizontal:

$\theta ={\cos }^{-1}\left(R/L\right)={\cos }^{-1}\left(0.15\mathrm{m}/0.94\mathrm{m}\right)=80°$

To calculate tension, apply Newton’s second law of motion in a vertical direction:

$T\sin \theta =mg$

Substitute the values in the above expression, and we get,

$$\begin{gathered} T=\frac{\left(0.04\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\sin \left(80°\right)} \\ =0.4\mathrm{N} \end{gathered}$$

Thus, the tension in the string is 0.4 N.

(b) Calculate the period of the motion

To calculate the velocity of bob, apply Newton’s second law of motion horizontal direction:

$$\begin{gathered} T\cos \theta =\frac{m{v}^2}{R} \\ v=\sqrt{\frac{R\times T\cos \theta }{m}} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v=\sqrt{\frac{\left(0.15\mathrm{m}\right)\times \left(0.4\mathrm{N}\right)\cos \left(80°\right)}{0.04\mathrm{kg}}} \\ =0.49\mathrm{m}/\mathrm{s} \end{gathered}$$

To calculate the time to complete one revolution:

$$\begin{gathered} t=\frac{0.94\mathrm{m}}{0.49\mathrm{m}/\mathrm{s}} \\ =1.9\mathrm{s} \end{gathered}$$

Thus, the period of motion is 1.9 s.