Problem 91 - University Physics with Modern Physics

Question

A block with mass m is revolving with linear speed \(v_1\) in a circle of radius \(r_1\) on a frictionless horizontal surface (see Fig. E10.40). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_2\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_1\) and the radius \(r_1\). (b) Use \(W = \int_{r1}^{r2} \overrightarrow{T} (r) \cdot d\overrightarrow{r}\) to calculate the work done by \(\overrightarrow{T}\) when \(r\) changes from \(r_1\) to \(r_2\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

Step-by-Step Solution

Answer

(a) Tension: \(T = \frac{m v_1^2 r_1^2}{r^3}\); (b) Work done: \(\frac{m v_1^2 r_1^2}{2} \left( \frac{1}{r_2^2} - \frac{1}{r_1^2} \right)\); (c) Work done equals change in kinetic energy.

Step 1: Understand the relationship among quantities

The block is in uniform circular motion. For such a motion, the centripetal force \(F_c\), necessary to keep the block moving in a circle, is provided by the tension in the string. The formula for centripetal force is \(F_c = \frac{mv^2}{r}\). Initially, the speed is \(v_1\) and radius \(r_1\). When the string is pulled, the radius changes to \(r_2\).

Step 2: Determine expression for tension T

Using conservation of angular momentum, we have \(m v_1 r_1 = m v r\), which gives \(v = \frac{v_1 r_1}{r}\). Substitute \(v\) in the centripetal force formula: \(T = \frac{m v^2}{r} = \frac{m (\frac{v_1 r_1}{r})^2}{r} = \frac{m v_1^2 r_1^2}{r^3}\).

Step 3: Calculate work done using tension as a function of r

The work done \(W\) by tension is an integral from \(r_1\) to \(r_2\): \[W = \int_{r1}^{r2} T(r) \, dr = \int_{r1}^{r2} \frac{m v_1^2 r_1^2}{r^3} \, dr\].Calculate the integral:\[W = \frac{m v_1^2 r_1^2}{-2} \left[\frac{1}{r^2}\right]_{r1}^{r2} = \frac{m v_1^2 r_1^2}{2} \left( \frac{1}{r_2^2} - \frac{1}{r_1^2} \right)\].

Step 4: Find change in kinetic energy

The initial kinetic energy \(K_1 = \frac{1}{2} m v_1^2\), and the final kinetic energy \(K_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m \left(\frac{v_1 r_1}{r_2}\right)^2\). Thus, the change in kinetic energy is:\[\Delta K = K_2 - K_1 = \frac{1}{2} m \frac{v_1^2 r_1^2}{r_2^2} - \frac{1}{2} m v_1^2\].

Step 5: Verify if work done equals change in kinetic energy

From Step 3, work done is \(\frac{m v_1^2 r_1^2}{2} \left( \frac{1}{r_2^2} - \frac{1}{r_1^2} \right)\). The change in kinetic energy found in Step 4 is \(\frac{1}{2} m v_1^2 \left(\frac{r_1^2}{r_2^2} - 1\right)\). These equations match, confirming that the work done by tension equals the change in kinetic energy.

Key Concepts

Centripetal Force

Centripetal force is essential for keeping an object moving in a circular path. It acts towards the center of the circle and ensures the object does not deviate from its path. In our scenario, the centripetal force required to keep the block moving in its circular path is provided by the tension in the string. You can calculate this force using the equation:

• \(F_c = \frac{mv^2}{r} \)

where \(m\) is the mass of the object, \(v\) is the velocity, and \(r\) is the radius of the circle.
As the string is pulled and the radius of the circle changes, the tension in the string must adjust to ensure the block continues its circular motion. Understanding how tension changes with radius is a key part of solving such problems. It's important to note that as the radius decreases, the velocity of the block increases to conserve angular momentum. This relationship plays a crucial role in determining the ever-important centripetal force.

Kinetic Energy

Kinetic energy is the energy of motion. It helps us understand how energy is transferred and transformed in dynamic systems. For our block in circular motion, the kinetic energy is given by the familiar formula:

• \(K = \frac{1}{2}mv^2\)

where \(m\) is the mass and \(v\) is the velocity.
Initially, the block has a certain kinetic energy based on its speed \(v_1\) when the radius is \(r_1\). However, when the radius reduces to \(r_2\), the block's speed changes to keep its angular momentum constant, which in turn alters its kinetic energy.
The change in kinetic energy, when the block moves from one circular path to another, signifies the work that's done on the system. In this exercise, when the radius of the path decreases, the speed increases, reflecting a change in kinetic energy. Recognizing how to calculate and interpret these changes is vital for understanding the energy dynamics of the block.

Work-Energy Theorem

The work-energy theorem is a fundamental principle that ties the concept of work done on an object to the change in its kinetic energy. Simply put, the work done by all forces acting on an object will result in a change in its kinetic energy. In our example, we calculate the work done by the tension in the string as it pulls the block from a radius \(r_1\) to \(r_2\). The work done is given by the integral:

• \(W = \int_{r1}^{r2} T(r) \cdot dr\)

where \(T(r)\) is the tension as a function of radius \(r\).
The work-energy theorem states that this work equals the change in kinetic energy of the block. This is confirmed when calculated values of work done and the change in kinetic energy match. This theorem acts as a valuable tool for solving problems where forces cause changes in motion. By understanding this principle, you can better analyze energy transformations in dynamic systems.