The problem in the image describes a conical pendulum, where a small object (bob) is attached to a string and moves in a horizontal circle, sweeping out a cone. We are asked to find:

1. (a) The tension in the string
2. (b) The period of the motion

Given Data:

• Mass of the bob, $m = 0.0160 \, \text{kg}$
• Length of the string, $L = 0.953 \, \text{m}$
• Circumference of the circular path, $C = 1.25 \, \text{m}$

Solution Steps:

Part (a) Tension in the String:

We can begin by calculating the radius $r$ of the circular path. The relationship between the circumference and radius is: $C = 2\pi r$ Solving for $r$: $r = \frac{C}{2\pi} = \frac{1.25}{2\pi} \approx 0.199 \, \text{m}$

Now, let's analyze the forces on the bob. There are two main forces acting on it:

1. Tension in the string, $T$, acting along the string.
2. The gravitational force, $mg$, acting vertically downward.

The bob moves in a horizontal circle, so the tension has both a vertical and a horizontal component:

• The vertical component of tension balances the weight: $T \cos(\theta) = mg$
• The horizontal component provides the centripetal force necessary for circular motion: $T \sin(\theta) = \frac{m v^2}{r}$

To solve for $T$, we need $\theta$, the angle the string makes with the vertical. From the geometry of the conical pendulum: $\sin(\theta) = \frac{r}{L} = \frac{0.199}{0.953} \approx 0.209$ This gives $\theta \approx 12.08^\circ$.

Now using $\cos(\theta)$: $\cos(\theta) \approx \cos(12.08^\circ) \approx 0.978$

Substituting into the vertical force balance equation: $T \cdot 0.978 = mg \quad \Rightarrow \quad T = \frac{mg}{0.978} = \frac{(0.0160 \, \text{kg})(9.81 \, \text{m/s}^2)}{0.978} \approx 0.160 \, \text{N}$

Thus, the tension in the string is approximately 0.160 N.

Part (b) Period of the Motion:

The period $T_{\text{period}}$ is the time it takes for one complete revolution, and is related to the velocity $v$ and the radius $r$ by: $v = \frac{C}{T_{\text{period}}} = \frac{1.25}{T_{\text{period}}}$

To find $v$, we use the horizontal force balance equation $T \sin(\theta) = \frac{m v^2}{r}$, so: $0.160 \cdot \sin(12.08^\circ) = \frac{(0.0160) v^2}{0.199}$ Solving for $v$: $v^2 = \frac{0.160 \cdot 0.209 \cdot 0.199}{0.0160} \quad \Rightarrow \quad v \approx 0.456 \, \text{m/s}$

Now, using the formula for the period: $T_{\text{period}} = \frac{C}{v} = \frac{1.25}{0.456} \approx 2.74 \, \text{seconds}$

Thus, the period of the motion is approximately 2.74 seconds.

Do you have any questions, or would you like further details on any part of the solution?

Related Questions:

1. How does the length of the string affect the tension in a conical pendulum?
2. What would happen to the period if the mass of the bob was doubled?
3. How is the period of motion related to the speed of the bob in a conical pendulum?
4. What is the effect of gravity on the tension in the string?
5. How does the angle $\theta$ influence the speed of the bob in circular motion?

Tip:

For conical pendulum problems, always break the tension force into vertical and horizontal components and use the geometry of the system to relate angle, radius, and string length.