Problem 96 - University Physics with Modern Physics

Question

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.42). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N} .\) What is the radius of the circle when the string breaks?

Step-by-Step Solution

Answer

The radius of the circle when the string breaks is approximately 0.44 m.

Step 1: Understanding Centripetal Force

The centripetal force required to keep the block moving in a circle is given by the formula: \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the tangential speed, and \( r \) is the radius. We know the initial conditions: \( m = 0.250 \) kg, \( v = 4.00 \) m/s, and \( r = 0.800 \) m.

Step 2: Calculating Initial Centripetal Force

Using the initial conditions, calculate the initial centripetal force. \( F_c = \frac{(0.250 \text{ kg})(4.00 \text{ m/s})^2}{0.800 \text{ m}} = \frac{4.00}{0.800} = 5.00 \text{ N} \). The initial centripetal force is 5.00 N.

Step 3: Determining New Tangential Speed

When the radius changes, the tangential speed has to be recalculated using the conservation of angular momentum, which states \( mvr = \text{constant} \). Initially, \( (0.250)(4.00)(0.800) = 0.800 \text{ kg} \cdot \text{m}^2/\text{s} \). If the new radius is \( r \), then the new speed \( v' \) satisfies \( (0.250)v'r = 0.800 \) yielding \( v' = \frac{0.800}{0.250r} = \frac{3.20}{r} \).

Step 4: Substituting Break Point Conditions

The string breaks at 30.0 N, which is the new centripetal force \( F_c' = \frac{m(v')^2}{r} \). Set up the equation: \( 30.0 = \frac{0.250(\frac{3.20}{r})^2}{r} \) which simplifies to \( 30.0 = \frac{0.250(10.24)}{r^3} \).

Step 5: Solving for the Break Radius

From the equation obtained, \( 30.0 = \frac{2.56}{r^3} \). Solving for \( r^3 \) gives \( r^3 = \frac{2.56}{30.0} = 0.08533 \). Taking the cube root, \( r = (0.08533)^{1/3} \approx 0.44 \text{ m} \).

Key Concepts

Conservation of Angular Momentum

The concept of conservation of angular momentum is crucial in understanding circular motion, especially when the radius of the motion changes. Angular momentum, denoted as \( L \), is a vector quantity that represents the rotational analogue of linear momentum. It is given by the equation \( L = mvr \), where \( m \) is the mass, \( v \) is the tangential speed, and \( r \) is the radius of the circle. When the string in the exercise is pulled, the radius \( r \) decreases, but the angular momentum \( L \) of the system must remain constant in the absence of external torques (according to the conservation law). This means the product \( mvr \) remains unchanged even when \( r \) changes. As a result, any decrease in \( r \) must be compensated by an increase in \( v \) (tangential speed), ensuring \( mvr = \text{constant} \).In practical terms, this principle helps us determine the new speed \( v' \) of the mass as the string shortens. For students underlining conservation principles, remember: no external torque equals constant angular momentum, influencing all components involved.

Tangential Speed

In circular motion, tangential speed \( v \) refers to how fast an object is moving along the edge of the circle at any point. It's a linear measure of the object's speed as it "glides" around the circle, but it's different from rotational speed, which considers the angle turned per unit time.When the problem mentions finding a new tangential speed \( v' \) after changing the radius \( r \), it requires careful consideration of how linear speed adjusts with a changing radius. Using the principle of conservation of angular momentum, we see that as the radius decreases, the tangential speed has to increase to preserve the product \( mvr \). This relation is common in systems like ice skaters pulling in their arms to spin faster or planets orbiting closer to the sun, moving at a higher speed. Understanding how tangential speed dynamically changes with radius adjustments helps clarify many physical scenarios, not just in theory but in real-life applications.

Newton's Laws of Motion

Newton's laws of motion lay the foundational framework for understanding how forces affect motion, including circular motion where centripetal forces are involved. Specifically, Newton's second law plays a crucial role here, describing how the force (\( F \)) is related to mass (\( m \)) and acceleration (\( a \)) through \( F = ma \).In circular motion, the acceleration involved is the centripetal acceleration, pointing toward the center of the circular path. This is where the formula \( F_c = \frac{mv^2}{r} \) comes into play, representing the centripetal force needed to keep an object moving in a circle of radius \( r \) at speed \( v \). The centripetal force, being the net force acting toward the center, indicates how much force is necessary to redirect the object's path continuously. Newton's laws also help explain why the string, mentioned in the exercise, ultimately breaks. As the radius becomes smaller and the mass's speed increases, the necessary centripetal force eventually exceeds the breaking strength of the string, causing it to snap. Understanding how these forces interact and lead to such outcomes is essential to mastering physics concepts on circular motion.