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Figure 8-23 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0\). As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?

Step by step solution

01

Understand the Initial Setup

We are given a rod of length \(L = 2.00\, \text{m}\) attached to a ball of mass \(m = 5.00\, \text{kg}\). It's initially released from an angle \(\theta_0 = 30.0^\circ\) with an initial velocity of zero.

02

Calculate Work Done by Gravitational Force (a)

The gravitational work done can be calculated using the change in height. The height at release is \(h_0 = L \cdot (1 - \cos \theta_0)\). At the lowest point, height, \(h = 0\). So, \(h_0 = 2(1 - \cos 30^\circ) = 2(1 - \frac{\sqrt{3}}{2}) = 1 - \frac{\sqrt{3}}{4}\). Use \(W = mgh_0\), where \(g = 9.81\, \text{m/s}^2\). Substitute values to get \(W = 5 \times 9.81 \times (1 - \frac{\sqrt{3}}{4})\).

03

Calculate the Change in Gravitational Potential Energy (b)

The change in gravitational potential energy \(\Delta U = U_f - U_i\), where \(U_f = 0\) at the lowest point and \(U_i = mgh_0\) at the initial release height. Thus, \(\Delta U = - mgh_0 = -W\) calculated in the previous step.

04

Determine Gravitational Potential Energy at Release (c)

If the potential energy is zero at the lowest point, then at release, it is \(mgh_0\). Hence, the potential energy is \(U = 5 \cdot 9.81 \cdot (2 - 2 \cdot \frac{\sqrt{3}}{2})\) also using height calculated as before.

05

Effect of Increasing the Release Angle (d)

Increasing \(\theta_0\) increases the initial height \(h_0\), thus increasing both the work done by gravity and the gravitational potential energy at release, making both values larger.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy

Gravitational potential energy is a type of potential energy stored by an object due to its height above the ground and the force of gravity. In this exercise, we deal with a ball attached to a rod, pivoting in vertical motion. Gravitational potential energy is given by the formula \( U = mgh \), where:

• \( m \) is the mass in kilograms,
• \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth),
• \( h \) is the height in meters above a specified reference point.

The height \( h \) is crucial since gravitational potential energy directly depends on it. At the highest point of its swing, the ball has maximum potential energy. When defining potential energy as zero at the rod's lowest point, any other configuration's potential energy is calculated relative to that point.

Work Done by Gravitational Force

The concept of work done involves moving a force through a distance. When gravitational force acts on an object, the work done is calculated by the change in the object's gravitational potential energy. Specifically:

• Work done by gravity can be described with \( W = mgh \), where \( h \) is the change in height.

In the setup provided, as the ball moves from its initial height to the lowest point, gravity does work on it. Here, the ball starts at a height calculated from the potential energy equation and descends to the lowest point where potential energy is zero. The work done by gravity causes this energy change and is equivalent to the work required to lift the ball back to its initial height against gravity.

Mechanics

Mechanics focuses on the motion of objects and the forces acting upon them. It separates into branches such as kinematics and dynamics. In this case, we are particularly dealing with a branch concerned with motion and forces like gravity. The problem involves understanding how gravitational forces influence a swinging ball attached to a rod. Key principles applied include:

• Newton's laws of motion, which describe the relationship between the motion of an object and the forces acting on it.
• Conservation of energy, where gravitational potential energy and kinetic energy can interchange.

When the ball is released, it's acted upon by the force of gravity, pulling it through the circular path designated by the rod length. The problem explores how forces and energy interchange as the ball moves under gravitational influence.

Rotational Motion

Rotational motion involves objects moving in circular paths around a point (pivot, in this case). With the rod pivoting in a vertical circle, we observe rotational dynamics, including:

• Torque, which is the force causing an object to rotate and depends on the lever arm's length.
• Centripetal force, required to keep an object moving in a circle, directed towards the rotation's center.

The ball's pivot about one end means the entire system experiences rotational motion. The angle of release matters here— a larger initial angle increases the height and thus the potential energy, affecting how energetic the system is as it rotates downward. This rotation analysis applies not just to simple pendulums like this setup but also to more complex systems such as rotating wheels or orbiting planets.