Problem 39 - College Physics

Question

A light string of length of \(56.0 \mathrm{~cm}\) connects two small square blocks, each with a mass of \(1.50 \mathrm{~kg} .\) The system is placed on a slippery (frictionless) sheet of horizontal ice and spun so that the two blocks rotate uniformly about their common center of mass, which itself does not move. They are supposed to rotate with a period of \(0.750 \mathrm{~s}\). If the string can exert a force of only \(100 \mathrm{~N}\) before it breaks, determine whether this string will work.

Step-by-Step Solution

Answer

The string will work as the tension is 59.6 N, which is less than the breaking tension of 100 N.

Step 1: Determine relevant forces

To decide whether the string will break, we need to find out the tension in the string due to the centripetal force required to keep the blocks rotating in a circle. The centripetal force is provided by the tension in the string that keeps each block in circular motion.

Step 2: Calculate the radius of the circle

Both blocks together form the rotating system. As they are connected by a string and rotate about the center of mass, each block is at a distance of half the length of the string from the center. Therefore, the radius (r) is proportional to half the string length, i.e., \( r = \frac{56.0 \text{ cm}}{2} = 28.0 \text{ cm} = 0.28 \text{ m} \).

Step 3: Convert period to angular velocity

The period \( T \) of rotation is given as \( 0.750 \text{ s} \). First, we find the angular velocity \( \omega \) using the formula \( \omega = \frac{2\pi}{T} \). Substitute \( T = 0.750 \): \[ \omega = \frac{2\pi}{0.750} \approx 8.38 \text{ rad/s} \]

Step 4: Calculate centripetal force per block

Each block requires a centripetal force \( F_c \), given by:\[ F_c = m \cdot r \cdot \omega^2 \] where \( m = 1.50 \text{ kg} \), \( r = 0.28 \text{ m} \), and \( \omega \approx 8.38 \text{ rad/s} \):\[ F_c = 1.50 \cdot 0.28 \cdot (8.38)^2 \approx 29.8 \text{ N} \]

Step 5: Calculate total tension in the string

Since there are two blocks, the total centripetal force (which is also the total tension in the string) is twice the centripetal force needed for one block. Hence:\[ F_{total} = 2 \cdot 29.8 \approx 59.6 \text{ N} \]

Step 6: Determine if the string can withstand the force

The tension in the string when both blocks rotate is \( 59.6 \text{ N} \). The maximum tension the string can withstand without breaking is \( 100 \text{ N} \). Since \( 59.6 \text{ N} < 100 \text{ N} \), the string will not break.

Key Concepts

Uniform Circular Motion

Uniform circular motion occurs when an object moves in a circle at a constant speed. While speed is constant, velocity is not, because direction changes continuously. This change in direction is due to the presence of a net force acting towards the center of the circular path.

This inward force is called the centripetal force. In the context of a rotating system like the one described in the exercise, each block moves uniformly in a circle, maintained by this force. The net force (here, the tension in the string) is what makes uniform circular motion possible. Without this tension, the blocks would move off in a straight line due to inertia.

So, even if the objects seem to move at a constant pace along their path, they are under the action of continuous acceleration towards the center. This is why understanding the force is crucial when assessing whether the string will support the motion given the tension limit of 100 N.

Angular Velocity

Angular velocity (\(\omega\)) represents how quickly an object rotates or revolves around a center point, and it is measured in radians per second. It differs from linear velocity, which measures how fast an object moves along a path.

In the given problem, you are asked to determine whether the string's tension will exceed its breaking point when two blocks spin in a circle. Angular velocity is crucial here because it helps calculate the centripetal force required for such motion. First, you convert the time for one full rotation (period) to angular velocity using the formula \(\omega = \frac{2\pi}{T}\). Plugging in the period (0.750 s), you find \(\omega \approx 8.38 \text{ rad/s}\).

This angular velocity is used to compute the force necessary to maintain circular motion, highlighting its integral role in understanding rotational dynamics.

Tension in String

Tension in a string or rope results from the opposing forces trying to pull it apart. In the context of circular motion, tension provides the centripetal force that enables an object to follow a curved path.

To understand how much tension the string is under, we can calculate the force required to keep each block traveling in a circle. Using the formula \(F_c = m \cdot r \cdot \omega^2\), we find each block demands around 29.8 N of force. For two blocks, this translates to a total tension of about 59.6 N.

It's essential to recognize the role of tension in circular motion problems. Without the right amount of tension, the objects won't stay on their path. In this case, 59.6 N is far below the maximum the string can handle (100 N), ensuring it won't break. This assessment is crucial for engineering applications where safety and material limits matter.