Problem 2 - College Physics

Question

A stone with a mass of 0.80 \(\mathrm{kg}\) is attached to one end of a string 0.90 \(\mathrm{m}\) long. The string will break if its tension exceeds 60.0 \(\mathrm{N} .\) The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Make a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without breaking the string.

Step-by-Step Solution

Answer

The stone's maximum speed without breaking the string is approximately 8.22 m/s.

Step 1: Draw the Free-Body Diagram

Consider the forces acting on the stone. There are two main forces: the tension in the string, which acts as the centripetal force and is directed toward the center of the circle, and the gravitational force acting downward. Since the motion is horizontal and on a frictionless table, only the tension needs to be considered for circular motion.

Step 2: Identify Forces and Equations

The key force acting here is the tension, which equals the centripetal force required to keep the stone moving in a circle. The equation for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( F_c \) is the centripetal force (tension here), \( m \) is the mass of the stone, \( v \) is the velocity, and \( r \) is the radius of the circle.

Step 3: Set up the Equation

Insert the known values into the centripetal force equation. Here, \( F_c = T \), where \( T \) is the tension (maximum 60.0 N), \( m = 0.80 \text{ kg} \), and \( r = 0.90 \text{ m} \). So, the equation becomes \( 60.0 = \frac{0.80v^2}{0.90} \).

Step 4: Solve for Maximum Speed

Rearrange the equation to solve for the velocity \( v \). First, multiply both sides by \( 0.90 \), giving \( 60.0 \cdot 0.90 = 0.80v^2 \). Simplifying gives \( 54.0 = 0.80v^2 \). Divide both sides by 0.80 to isolate \( v^2 \): \( v^2 = \frac{54.0}{0.80} = 67.5 \). Finally, take the square root of both sides to find \( v \): \( v = \sqrt{67.5} \approx 8.22 \text{ m/s} \).

Key Concepts

Tension in a String

Tension in a string is a crucial element when talking about forces, especially in scenarios involving rotational movement like in this problem. When a stone is attached to a string and whirled in a circle, the string experiences a force that stretches it, known as tension. This tension works towards the center of the circle to keep the stone moving in its path. The tension isn't constant; it changes depending on how fast the stone is moving and the radius of the circle. Importantly, if the stone moves too quickly, the tension can reach a limit and cause the string to break. In this exercise, the maximum tension the string can handle is given as 60.0 N. Staying below this limit ensures the stone continues its circular path without snapping the string. Understanding and calculating this tension helps determine safe speed thresholds during the stone's motion.

Free-Body Diagram

A free-body diagram is a visual representation used to show the relative magnitude and direction of all forces acting on an object. For the stone in this problem, let's consider the forces. You will typically only draw these forces acting on the object, isolating it from everything else. In vertical or horizontal circle problems like this one, two forces come into play:

• The gravitational force, which pulls the stone downward.
• The tension force, which acts as the centripetal force, pulling the stone inward toward the center of the circle.

While gravity is present, it doesn't affect the horizontal motion if the surface is frictionless. Therefore, the diagram will have an arrow pointing inward, indicating the tension as the central force ensuring circular motion. This visual greatly aids in understanding the forces involved and assists in setting up accurate equations for calculations.

Circular Motion

Circular motion occurs when an object moves along the circumference of a circle under the influence of a centripetal force. This is exactly what happens with the stone whirling in a circle. In such motion:

• Velocities are constantly changing direction, making it a case of accelerated motion even if speed remains constant.
• The force required to maintain this circular motion is called the centripetal force, which, in this problem, is provided by the tension in the string.

Understanding circular motion isn't only about the path but also understanding how forces apply to maintain that path. Here, we see how tension works to ensure the stone continues moving in a circle instead of flying out of its path. The key is to keep the tension below the maximum allowed so the string does not break. Understanding these principles helps in predicting the motion characteristics and in solving related physics problems.

Maximum Speed Calculation

Calculating the maximum speed involves understanding and manipulating the relationship given by centripetal force. The formula for centripetal force is \[ F_c = \frac{mv^2}{r} \]where:

• \(F_c\) is the centripetal force, which equals the tension here (a maximum of 60.0 N)
• \(m\) is the mass of the stone (0.80 kg)
• \(v\) is the speed of the stone
• \(r\) is the radius of the circle (0.90 m)

By inserting the given values into the formula, we isolate and solve for \(v\), the velocity. This involves simple algebraic manipulation:1. Plug the numbers: \(60.0 = \frac{0.80v^2}{0.90}\)2. Rearrange to find \(v^2\) by multiplying both sides by 0.90: \(54.0 = 0.80v^2\)3. Divide by 0.80: \(v^2 = 67.5\)4. Take the square root to find \(v\): \(v \approx 8.22 \text{ m/s}\)This calculated speed of approximately 8.22 m/s is the maximum speed the stone can reach before the tension exceeds the maximum limit, potentially breaking the string.