Problem 85 - Physics for Scientists and Engineers with Modern Physics

Question

A ball is attached to a horizontal cord of length \(\ell\) whose other end is fixed, Fig. \(8-42 .(a)\) If the ball is released, what will be its speed at the lowest point of its path? \((b)\) A peg is located a distance \(h\) directly below the point of attachment of the cord. If \(h=0.80 \ell,\) what will be the speed of the ball when it reaches the top of its circular path about the peg?

Step-by-Step Solution

Answer

(a) Speed at lowest point: \( \sqrt{2g\ell} \). (b) Speed at top around peg: \( \sqrt{0.8g\ell} \).

Step 1: Understanding the Problem

A ball on a string is released from rest and swings down to its lowest point. We need to find its speed at this lowest point, considering it swings in a vertical circle.

Step 2: Analyze the Energy Conservation in Part (a)

At the starting point, the ball has gravitational potential energy and no kinetic energy, since it's released from rest. The potential energy is converted into kinetic energy at the lowest point. Using energy conservation: \[ mgh = \frac{1}{2}mv^2 \] where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h = \ell \) is the height the ball falls.

Step 3: Solve for Speed at Lowest Point

From the energy conservation equation in Step 2, we get: \[ mgh = \frac{1}{2}mv^2 \Rightarrow \ell g = \frac{1}{2}v^2 \] Solving for \( v \), we have: \[ v = \sqrt{2g\ell} \] This is the speed of the ball at the lowest point of its path.

Step 4: Understanding Part (b)

A peg is located a distance \( h = 0.80\ell \) directly below the cord's fixed point. The ball will start moving in a smaller circle around the peg. We need to calculate the ball's speed at the top of this new circular path.

Step 5: Analyze Energy Conservation for Part (b)

Before reaching the peg, the ball's total mechanical energy at the bottom equals its maximum kinetic energy \( K_1 \). At the top of its path around the peg, the ball has potential energy \( mgh_2 \) and kinetic energy \( \frac{1}{2}mv^2 \): \[ K_1 = \frac{1}{2}m v_{bottom}^2 = mgh_2 + \frac{1}{2}mv_{top}^2 \]\(h_2 = 2(0.8\ell) - \ell = 0.6\ell \)We substitute \(v_{bottom} = \sqrt{2g\ell} \) from Step 3.

Step 6: Solve for Speed at Top Around Peg

We substitute and simplify:\[ \frac{1}{2}m(\sqrt{2g\ell})^2 = mg(0.6\ell) + \frac{1}{2}mv_{top}^2 \]\[ mg\ell = mg(0.6\ell) + \frac{1}{2}mv_{top}^2 \]\[ \ell g = 0.6\ell g + \frac{1}{2} v_{top}^2 \]Solve for \( v_{top} \):\[ 0.4\ell g = \frac{1}{2} v_{top}^2 \]\[ v_{top} = \sqrt{0.8g\ell} \]

Step 7: Conclusion

The speed of the ball at the lowest point of its path is \( \sqrt{2g\ell} \) and the speed of the ball at the top of its path around the peg is \( \sqrt{0.8g\ell} \).

Key Concepts

Energy Conservation

Energy conservation is a fundamental principle in physics stating that the total energy within an isolated system remains constant over time. In this exercise, energy conservation plays a key role in solving for the speed of the ball at different points along its path.

Initially, the ball is held at a certain height, and its energy is purely gravitational potential energy:

• Potential energy at the start is given by \(E_p = mgh\), where \(h\) represents the height.
• Once released, this potential energy gradually converts into kinetic energy as the ball gains speed.

At the lowest point of the trajectory, all the initial potential energy has turned into kinetic energy:

• Kinetic energy is given by \(E_k = \frac{1}{2}mv^2\).
• The equation \(mgh = \frac{1}{2}mv^2\) captures this conversion.
• Simplifying, we find the speed at the lowest point: \(v = \sqrt{2g\ell}\).

Energy conservation also helps solve part (b) of the exercise. As the ball swings to the top of the smaller circular path around the peg:

• The total mechanical energy must remain constant.
• The increase in potential energy must equal the decrease in kinetic energy.

By writing out and simplifying the energy conservation equation again:

• We find the speed of the ball at the top of the new circular path: \(v_{top} = \sqrt{0.8g\ell}\).

Through these steps, energy conservation allows us to predict the ball's behavior without knowing its mass or other details about the system.

Kinematics

Kinematics is the study of motion without considering the forces that cause it. It involves analyzing speed, velocity, acceleration, and displacement to understand how an object moves.

In this exercise, kinematics helps us understand the ball's motion as it swings downward and upwards:

• The ball starts from rest, so its initial velocity at the top is zero.
• As the ball falls, its velocity increases due to the acceleration stemming from gravity.
• At the lowest point, it attains maximum speed.
• As the ball moves in a circular path around the peg, its velocity changes again, especially when reaching the top of this path.

Using kinematics, we know that the velocity at any point in the ball's motion can be described using the equation:

• For velocity at the lowest point: velocity \(v = \sqrt{2g\ell}\).
• At the highest point of the smaller circle, the speed decreases, resulting in \(v_{top} = \sqrt{0.8g\ell}\).

These calculations show how kinematic descriptions of motion align with our energy conservation analysis. By examining the changes in speed and using mathematical formulas, we anticipate the motion characteristics of the ball throughout its path.

Circular Motion

Circular motion describes objects moving along a circular path. This path can vary in size and rate of motion. A distinctive feature of circular motion is the centripetal force acting towards the center of the circle, keeping the object in motion.

In this problem, the ball undergoes circular motion twice:

• Initially, around the fixed point of the cord. After release, the ball swings in a circular arc until reaching the lowest point.
• Secondly, around the peg, forming a smaller circle once it moves over the peg.

During these motions:

• The ball's speed is greatest at the lowest point due to gravitational energy converting to kinetic energy.
• Centripetal force is vital to maintain the motion along the circular paths. This force depends on the velocity and the radius of the path.
• In part (b), when the ball begins to circle around the peg, the path radius becomes smaller, which affects the centripetal force and speed.

Understanding circular motion helps explain how speed and direction change as the ball moves from one circular path to another. It elucidates why the ball needs different energies and speeds depending on its position within the circular motion.