Problem 59 - University Physics with Modern Physics

Question

A \(0.100-\mathrm{kg}\) potato is tied to a string with length \(2.50 \mathrm{m},\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Step-by-Step Solution

Answer

Speed: 7.00 m/s; Tension: 2.94 N

Step 1: Identify the Energy Conservation Principle

To find the speed of the potato at the lowest point, we first apply the principle of conservation of mechanical energy. At the top, all the potato's energy is potential energy, and at the lowest point, it is converted entirely into kinetic energy. The potential energy (PE) at the top will equal the kinetic energy (KE) at the bottom.Potential Energy at the top is given by: \[ PE = mgh \]where \( m = 0.100 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( h = 2.50 \text{ m} \) as the height is equal to the length of the string. Kinetic Energy at the bottom is given by: \[ KE = \frac{1}{2}mv^2 \]Setting PE equal to KE gives:\[ mgh = \frac{1}{2}mv^2 \]

Step 2: Solve for Speed at Lowest Point

Using conservation of energy, solve for the velocity \( v \) at the lowest point:\[ mgh = \frac{1}{2}mv^2 \]\[ 9.81 \times 2.50 = \frac{1}{2}v^2 \]Cancel \( m \) from both sides and solve the equation:\[ 2gh = v^2 \]\[ v = \sqrt{2gh} \]\[ v = \sqrt{2 \times 9.81 \times 2.50} \]\[ v \approx 7.00 \text{ m/s} \]

Step 3: Calculate Tension in the String at Lowest Point

At the lowest point, the forces acting on the potato are tension in the string and gravitational force. The tension has to provide the centripetal force needed to keep the potato in circular motion.The relation is given by:\[ T - mg = \frac{mv^2}{r} \]where \( T \) is the tension, \( r \) is the radius (equal to the string's length, 2.50 m), and \( v \) (velocity) is found in Step 2.Solving for \( T \):\[ T = mg + \frac{mv^2}{r} \]Substitute the values:\[ T = 0.100 \times 9.81 + \frac{0.100 \times (7.00)^2}{2.50} \]\[ T = 0.981 + 1.96 \]\[ T \approx 2.94 \text{ N} \]

Step 4: Final Step: Summary

The speed of the potato at the lowest point is approximately 7.00 m/s, and the tension in the string at that point is approximately 2.94 N.

Key Concepts

Centripetal Force

Centripetal force is the invisible hand that keeps an object moving in a circle, pulling it toward the center of motion. Imagine swinging a ball on a string; the string acts as the centripetal force, keeping the ball from flying off in a straight line.​
For an object moving in a circle, the centripetal force can be calculated using the formula:

• \[ F_c = \frac{mv^2}{r} \]

Where:

• \( F_c \) is the centripetal force,
• \( m \) is the mass,
• \( v \) is the velocity,
• \( r \) is the radius of the circle.

In the case of our potato, as it swings downwards, the string must exert enough force to keep pulling it along the circular path, maintaining the potato's motion.​
This force is crucial in problems dealing with circular motion, such as roller coasters, planets orbiting stars, or even potatoes swinging from strings!
It's important to understand that while the centripetal force is essential for circular motion, it is not an actual new force, but rather a result of other forces, like gravity and tension.

Potential Energy

Potential energy is stored energy that an object has due to its position or shape. For example, a potato held high above the ground possesses gravitational potential energy because of its height. As it is released and falls, this stored energy transforms into kinetic energy of motion.​
Gravitational potential energy can be calculated using:

• \[ PE = mgh \]

Where:

• \( m \) is the mass,
• \( g \) is the acceleration due to gravity (9.81 m/s²),
• \( h \) is the height above the ground.

In our exercise, when the potato is held straight out horizontally, it is at its highest point, maximizing its potential energy.​
When the potato is released, this energy converts into kinetic energy as it reaches the lowest point of its swing, demonstrating the principle of conservation of energy. This conversion helps us calculate the speed of the potato at any point of its motion by using energy balance.

Tension in a String

Tension in a string is the force exerted by the stretched string when it pulls back against the force that is stretching it. If you imagine our potato swinging from a string, the tension in the string is one of the key forces acting on it as it swings in a circular arc.

• Tension must be present to provide the centripetal force keeping the potato in its circular path.
• It also must counteract the force of gravity pulling the potato downward.

To find the tension, you can use the formula:

• \[ T = mg + \frac{mv^2}{r} \]

Where:

• \( T \) is the tension,
• \( m \) is the mass,
• \( g \) is the acceleration due to gravity,
• \( v \) is the velocity at the lowest point,
• \( r \) is the radius of the circular motion (length of the string).

At the lowest point of the swing, tension is greatest because it equals the centripetal force required to keep it moving in a circle, plus the force to balance out gravity. This means that if you were holding the string, you'd feel it pull more strongly at the lowest point of the swing. Understanding tension will allow you to predict how strong a string needs to be before it might break, crucial for anything involving pendulums or swings.