Problem 9 - University Physics with Modern Physics

Question

\(\mathrm{A} 0.800\) -kg ball is tied to the end of a string 1.60 \(\mathrm{m}\) long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

Step-by-Step Solution

Answer

Complete circle: both zero. Semicircle: tension zero, gravity -25.1 J.

Step 1: Understanding the Problem

We need to determine the total work done on the ball by the tension in the string and by gravity for one complete circle and for a semicircle from the lowest to the highest point. Let's consider these separately for parts (a) and (b).

Step 1: Work Done by Tension (Complete Circle)

In circular motion, the tension in the string does not perform work on the ball because the tension force is always perpendicular to the direction of displacement at every point in the circle. Therefore, for one complete circle, the work done by tension is:\[W_T = 0 \, \text{Joules}\]

Step 2: Work Done by Gravity (Complete Circle)

The gravitational force is conservative, and the displacement is closed (starting and ending at the same position). The net work done by gravity over one complete circle is zero:\[W_G = 0 \, \text{Joules}\]

Step 3: Work Done by Tension (Semicircle)

Similar to the full circle case, the work done by tension in moving the ball from the lowest to the highest point (semicircle) is still zero because the tension is always perpendicular to the displacement:\[W_T = 0 \, \text{Joules}\]

Step 4: Work Done by Gravity (Semicircle)

When moving from the lowest point to the highest point, the work done by gravity can be calculated from the change in gravitational potential energy, as the force of gravity is conservative. The height difference is equal to the diameter of the circle (2 times the radius, r = 1.60 m), so:\[\Delta h = 2 \times 1.60 = 3.20 \, \text{m}\]The work done by gravity:\[W_G = -mgh = -(0.800 \, \text{kg})(9.81 \, \text{m/s}^2)(3.20 \, \text{m}) = -25.1 \, \text{Joules}\]

Step 6: Conclusion

For the complete circle, work done by tension is zero and also zero by gravity. For the semicircle from the lowest to the highest point, work done by tension is zero, and by gravity it is -25.1 Joules.

Key Concepts

Work Done by Tension

In a vertical circular motion, the tension in the string is a central force that constantly changes direction to keep the object moving in a circular path. While the tension is critical for maintaining the path, it does not contribute to any work being done on the object. The reason lies in the definition of work itself. Work done by a force is calculated using the formula: - Work = Force × Displacement × cos(θ) Here, θ is the angle between the force and the displacement. For tension, this angle is always 90 degrees since the tension in the string acts towards the center, while the displacement is along the circumference of the path. Since the cosine of 90 degrees is zero: - The work done by tension = Force × Displacement × cos(90) = 0 This holds true for all parts of the circle, whether it's a complete circle or just a semicircle. Thus, for the scenario presented: - The work done by tension is always zero.

Work Done by Gravity

Gravity plays a crucial role in circular motion, especially when the path is vertical. It is considered a conservative force, which means the work done by gravity is related only to the change in height, irrespective of the path taken. For a complete circle: - The object returns to its initial height. Thus, the net change in height is zero. Consequently, the work done by gravity across a full circle is zero because there is no overall displacement in the vertical direction. In the specific case of a semicircle: - The object moves from the lowest point to the highest point. - This involves a definitive change in height, which equals twice the radius of the circular path (since the lowest to the highest point is the diameter). The work done by gravity is given by the expression: - Work done by gravity = -mgh Where: - **m** is the mass of the object, - **g** is the acceleration due to gravity (9.81 m/s²), and - **h** is the change in height, which is the diameter here (2 × 1.60 m = 3.20 m). Thus, when moving from the lowest to the highest point: - The work done by gravity = -0.800 kg × 9.81 m/s² × 3.20 m = -25.1 Joules. This negative work signifies that gravity acts in the direction opposite to the displacement, opposing the motion.

Gravitational Potential Energy

Gravitational potential energy (GPE) is energy stored due to an object's position in a gravitational field. In circular motion, especially when dealing with vertical circles, changes in height affect an object's GPE. The formula for gravitational potential energy is: - GPE = mgh Where: - **m** is the mass of the object, - **g** is the gravitational acceleration (9.81 m/s²), and - **h** is the height relative to a reference point. In a vertical circular path: - At the lowest point, the GPE is at its minimum because the height is the least relative to the highest point. - At the highest point, the height is maximum, hence the GPE is at its maximum. The change in gravitational potential energy as an object moves from the lowest to the highest point reflects the work done against gravity. This energy change is what results in the negative work calculated earlier. It's important to understand that, during this motion: - Energy is conserved but transforms from kinetic to potential and vice versa. Gaining a clear grasp of gravitational potential energy is vital as it connects the concepts of force, work, and energy in motion.