Problem 86 - University Physics with Modern Physics

Question

A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.40). 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 m/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 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 0.351 m.

Step 1: Understand the Problem

We need to find the radius at which the tension in the string, due to the block's circular motion, reaches 30.0 N and causes the string to break.

Step 2: Use Conservation of Angular Momentum

Since there is no external torque, the angular momentum is conserved. Initially, angular momentum is given by \( L_i = m r_i v_i \), and when the radius changes, it becomes \( L_f = m r_f v_f \). Equate the two: \( m r_i v_i = m r_f v_f \).

Step 3: Relate Tangential Speed to Radius

From the conservation equation \( r_i v_i = r_f v_f \), express \( v_f \) in terms of \( r_f \): \( v_f = \frac{r_i v_i}{r_f} \).

Step 4: Connect Tension and Centripetal Force

The centripetal force required to keep the block moving in a circle is provided by the tension in the string: \( F_c = \frac{m v_f^2}{r_f} \). Set \( F_c = 30.0 \) N, as it's the breaking tension.

Step 5: Solve for the Radius When the String Breaks

Substitute \( v_f \) from Step 3 into the expression for centripetal force: \[ 30.0 = \frac{m \left(\frac{r_i v_i}{r_f}\right)^2}{r_f} \] Simplify and solve for \( r_f \).

Step 6: Perform Calculations

Using \( m = 0.130 \) kg, \( r_i = 0.800 \) m, \( v_i = 4.00 \) m/s, plug the values into the equation: \[ 30.0 = \frac{0.130 \times \left( \frac{0.800 \times 4.00}{r_f} \right)^2}{r_f} \] Simplify and solve this equation for \( r_f \). After some algebra, we find \( r_f \approx 0.351 \) m.

Key Concepts

Centripetal Force

Centripetal force is essential in understanding circular motion. It is the force that keeps an object moving in a circle, directed towards the center of the circle.
In this exercise, the centripetal force is provided by the tension in the string.
This force must be strong enough to keep the block moving in its circular path. But if the force exceeds a certain threshold, in this case, 30.0 N, the string will break. Understanding how centripetal force relates to tension helps us analyze what happens when the string reaches its breaking point.

• The centripetal force formula is: \( F_c = \frac{m v^2}{r} \), where
  
  • \( F_c \) is the centripetal force.
  • \( m \) is the mass of the block.
  • \( v \) is the tangential speed.
  • \( r \) is the radius of circular motion.

When calculating the breaking tension of the string, we equate the centripetal force to the tension force, since they are essentially the same force for this setup. Recognizing this relationship allows us to find the point, or the radius, at which the string breaks due to tension.

Tangential Speed

Tangential speed refers to the speed of an object moving along the circular path. It is the linear speed of the object at any point on the circle.
In this case, as the radius of the circular motion decreases, the tangential speed increases to conserve angular momentum.
That means as the string is pulled, shortening the radius, the block moves faster along its circular path.This happens due to conservation of angular momentum, which states that unless acted upon by an external torque, the angular momentum of the system remains constant. This concept can be expressed as:

• Initial angular momentum: \( L_i = m r_i v_i \)
• Final angular momentum: \( L_f = m r_f v_f \)
• The conservation equation: \( m r_i v_i = m r_f v_f \)

Using this relationship, you can solve for the final tangential speed \( v_f \) when the radius changes. This rearranges to:\( v_f = \frac{r_i v_i}{r_f} \)Ensuring this principle helps to understand why the tangential speed changes as the radius is adjusted.

Radii of Circular Motion

The radius of circular motion is the distance from the center of the path to the object moving along the circle.
In this exercise, understanding how the radius affects the motion and forces in the system is crucial.
Starting with a radius of 0.800 m, the radius decreases as the string is pulled. One key point is how changing the radius affects other aspects of the motion, such as tangential speed and centripetal force.

• A decrease in radius leads to an increase in tangential speed due to conservation of angular momentum.
• Forces balance out differently as radius changes; tension must account for stronger centripetal force at a shorter radius.

The interaction between radius, speed, and force is an excellent example of how interconnected these ideas are in circular motion. The exercise involves calculating the exact radius at which the string breaks when tension reaches 30.0 N. This process involves using the centripetal force formula along with the conservation of angular momentum. Understanding these relations helps solve not only physics problems but also grasp real-world situations where forces in circular motion are at play.