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**Problem 6.77:** We have a mass \( m \) attached to a string of length \( L \), which is wound around a nail as shown in the figure. The mass is released from a horizontal position and allowed to swing downward. We need to solve for:

1. The tension in the string when it makes an angle of \( 23^\circ \) with the vertical.
2. The total work done to bring the string to this position.
3. The work done by the tension in the string.

### Solution Approach:

Let's break down the problem step-by-step:

#### (a) Tension in the string at \( 23^\circ \) with the vertical

1. **Gravitational Potential Energy Change:** Initially, the mass \( m \) is at the same height as the pivot point when it is horizontal. As it swings to an angle of \( 23^\circ \), it drops a certain vertical distance, causing a change in gravitational potential energy.
   
   The vertical height \( h \) at angle \( \theta = 23^\circ \) can be calculated using:
   \[
   h = L(1 - \cos \theta)
   \]
   
2. **Conservation of Energy:** The initial potential energy will be converted into kinetic energy and the work done by tension. At angle \( \theta = 23^\circ \), the kinetic energy is given by:
   \[
   mgh = \frac{1}{2}mv^2
   \]
   Solving for \( v \), we find the speed of the mass at that point.

3. **Centripetal Force and Tension:** The tension \( T \) at that point provides the necessary centripetal force along with balancing the gravitational component along the string. Using Newton's second law in the radial direction:
   \[
   T - mg\cos \theta = \frac{mv^2}{L}
   \]

4. Substitute values to find \( T \).

#### (b) Total Work Done

The total work done in bringing the string to the \( 23^\circ \) position is equivalent to the change in potential energy from the starting point to this position.

\[
\text{Work Done} = mgh = mgL(1 - \cos \theta)
\]

#### (c) Work Done by Tension

The tension force is always perpendicular to the displacement of the mass along its circular path. Hence, the work done by the tension force is zero, as there is no component of tension in the direction of the motion.

Would you like detailed calculations for each part, or have any specific questions about the solution?

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Here are some related questions for further exploration:

1. What would the tension be at different angles as the mass swings down?
2. How would the tension change if the angle was larger, such as 45 degrees?
3. Can we find the maximum tension in the string throughout the swing?
4. How would the solution change if the mass was attached to a string of different length \( L \)?
5. What happens to the work done by gravity if the mass swings up after reaching the bottom?

**Tip:** When working with energy conservation in pendulum problems, always identify the height changes relative to the starting point to simplify potential energy calculations.

The string in Figure 6-44 is wound around a nail as shown. The end of the string is attached to a mass m, that is released from the horizontal position, starting from rest. Derive an expression for (a) the tension in the string when it makes an angle of 23° with the vertical, (b) the total work done to bring the string to the position in part (a), (c) the work done by the tension.

Solve for the tension, work done by gravity, and work done by tension in a pendulum motion where a mass m swings from the horizontal to an angle of 23° with the vertical. This problem uses principles of energy conservation and trigonometry, making it ideal for physics students.

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