Question

A string is tied to a rock, and the rock is twirled around in a circle at a constant speed. If gravity is ignored and the period of the circular motion is doubled, the tension in the string is a) reduced to \(\frac{1}{4}\) of its original value. b) reduced to \(\frac{1}{2}\) of its original value. c) unchanged d) increased to 2 times its original value. e) increased to 4 times its original value.

Short answer

Question: When the period of the circular motion is doubled, the tension in the string becomes: a) 1/4 of its original value b) 1/2 of its original value c) Twice its original value d) Four times its original value

Step-by-step solution

Understand the concept of centripetal force

Centripetal force is the force required to keep an object moving in a circular path. For an object of mass \(m\) moving in a circle of radius \(r\) with an angular velocity \(\omega\), the centripetal force \(F_c\) is given by the formula: \(F_c = m\cdot r\cdot \omega^2\).

Express the period in terms of angular velocity

We have the period of the circular motion, denoted by \(T\). Now let's get the relationship between period (T) and angular velocity (\(\omega\)). Angular velocity can also be expressed in terms of the period of motion using the formula: \(\omega = \frac{2\pi}{T}\).

Substitute the period and angular velocity into the centripetal force formula

Replace the \(\omega\) in the centripetal force formula with the period-angular velocity relationship: \(F_c = m\cdot r\cdot \left(\frac{2\pi}{T}\right)^2\).

Calculate the original tension

Let's denote the tension in the string as \(F_T\). In the initial situation, the centripetal force is equal to the tension: \(F_{T1} = m\cdot r\cdot \left(\frac{2\pi}{T}\right)^2\).

Calculate the new tension when the period is doubled

When the period is doubled, the new period (\(T_2\)) is equal to \(2\cdot T\). The new tension in the string (\(F_{T2}\)) becomes: \(F_{T2} = m\cdot r\cdot \left(\frac{2\pi}{2\cdot T}\right)^2\).

Compare the original tension with the new tension

Now we will find the ratio of the new tension \(F_{T2}\) to the original tension \(F_{T1}\). \(\frac{F_{T2}}{F_{T1}} = \frac{m\cdot r\cdot \left(\frac{2\pi}{2\cdot T}\right)^2}{m\cdot r\cdot \left(\frac{2\pi}{T}\right)^2}\) The mass \(m\) and radius \(r\) gets canceled: \(\frac{F_{T2}}{F_{T1}} = \frac{\left(\frac{2\pi}{2\cdot T}\right)^2}{\left(\frac{2\pi}{T}\right)^2} = \left(\frac{T}{2\cdot T}\right)^2 =\frac{1}{4}\) Hence, the tension in the string is reduced to \(\frac{1}{4}\) of its original value when the period of the circular motion is doubled. The correct answer is (a).

Key concepts

Circular Motion

Circular motion is a physical phenomenon that occurs when an object moves in a path that forms a circle. At any given point, the object is travelling tangential to the circle, and there's a constant change in the direction of the velocity, which means the object is accelerating, even if it is maintaining a constant speed.

In the context of the exercise provided, a rock tied to a string and twirled around showcases a perfect example of uniform circular motion. Why uniform? Because the speed is constant. However, centripetal force—the force directed towards the center of the circle—is necessary to maintain this motion. This force keeps pulling the rock inward, preventing it from moving in a straight line, which it would do due to its inertia.

Improving understanding of circular motion involves grasping the invisible 'tug-of-war' between inertia wanting to propel the rock outward and centripetal force keeping it on the circular path. This tug-of-war is evident in the tension of the string, representing centripetal force.

Period of Motion

The period of motion, denoted as T, is the time it takes for the rock—on our string from the exercise—to complete one full rotation. A complete rotation means the rock has traveled the circumference of the circle it's describing.

As part of the solutions to the exercise, an important concept highlighted was how the period relates to the centripetal force. The direct relationship between the period and the tension in the string could be a tad confusing. When you double the period of motion (increasing T), the rock completes the circle at a slower pace. As a result, the centripetal force required to keep the rock moving in its circle decreases.

Incorporating this understanding, it becomes more intuitive why, when T increases, the tension—which is equivalent to the centripetal force—decreases. To drive the point home, using the formulae provided helps in solidifying this inversely proportional relationship between the period of motion and the centripetal force required.

Angular Velocity

Angular velocity, symbolized as \(\omega\), represents how fast the rock moves around the circle. It is the angle the rock sweeps per unit time, usually measured in radians per second. An interesting point in our exercise was to interpret how the angular velocity influences the centripetal force.

As shown in the solution steps, angular velocity and period share a special connection: \(\omega = \frac{2\pi}{T}\). This equation tells you that if the period doubles, angular velocity is halved. The decrease in angular velocity signifies that the rock is making fewer rotations per second, which, in turn, means a reduction in centripetal force. This understanding garners an appreciation for how angular velocity directly impacts the tension in the string, offering a clearer visualization of the scenario presented in the problem.

When a student wonders why doubling the period reduces the tension by a quarter, highlighting the squared relationship in the centripetal force formula (\(F_c = m\cdot r\cdot \omega^2\)) clarifies the concept significantly. A halved angular velocity, when squared, results in a quarter of its original value, hence, a fourfold decrease in tension.