Problem 72 - University Physics with Modern Physics

Question

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45\(^\circ\) with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? What is the tension in the string (b) when it makes an angle of 45\(^\circ\) with the vertical, (c) as it passes through the vertical?

Step-by-Step Solution

Answer

(a) Speed is 1.81 m/s. (b) Tension at 45° is 2.32 N. (c) Tension at vertical is 1.67 N.

Step 1: Determine the initial potential energy at the highest point

At the highest point, all the energy of the pendulum is potential. The height \( h \) of the pendulum above the lowest point can be calculated using trigonometry: \( h = L(1 - \cos\theta) \), where \( L = 0.80 \) m is the length and \( \theta = 45^\circ \). Thus, \( h = 0.80(1 - \cos 45^\circ) = 0.80(1 - \frac{\sqrt{2}}{2}) \approx 0.165 \) m. The initial potential energy \( PE \) is \( PE = mgh \), where \( g = 9.81 \) m/s². Substitute the known values to get the potential energy: \( PE = 0.12 \times 9.81 \times 0.165 \approx 0.194 \) J.

Step 2: Calculate the speed at the vertical position using conservation of energy

As the pendulum swings down to the vertical position, the potential energy is converted to kinetic energy (\( KE \)). Using conservation of energy: \( \text{Initial PE} = \text{Final KE} \). So, \( \frac{1}{2}mv^2 = 0.194 \). Solve for \( v \): \( v = \sqrt{\frac{2 \times 0.194}{0.12}} \approx 1.81 \) m/s.

Step 3: Calculate the tension at 45 degrees from the vertical

At an angle of 45°, the forces are the tension \( T \) and the gravitational force component along the string \( mg\cos\theta \) and perpendicular to the motion \( mg\sin\theta \). The centripetal force at this position is provided by the difference \( T - mg\cos\theta \). Using \( T - mg\cos 45^\circ = \frac{mv^2}{r} \) where \( v^2 = 2gL(1 - \cos\theta) \) from previous steps at 45°. Calculate to find \( T \approx 1.48 + 0.84 = 2.32 \) N.

Step 4: Compute the tension at the vertical position

When the pendulum is vertical, the tension is larger due to the need to provide centripetal force to maintain motion. Thus, \( T = mg + \frac{mv^2}{L} \). Using \( v = 1.81 \) m/s from step 2, we have \( \frac{0.12 \times 1.81^2}{0.80} = 0.49 \) N and therefore \( T = 0.12 \times 9.81 + 0.49 \approx 1.67 \) N.

Key Concepts

Conservation of Energy

The principle of Conservation of Energy is a fundamental concept that states energy in a closed system is conserved and cannot be created or destroyed. In physics problems, such as pendulums, this principle is essential for understanding how energy transitions between different forms.

For a simple pendulum, energy alternates between potential energy (\(PE\)) and kinetic energy (\(KE\)). At the highest point of the swing, the pendulum's energy is purely potential due to its elevated position. Here, potential energy is calculated as \(PE = mgh\), where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is height above the lowest point.

• When the pendulum falls to its lowest point, this potential energy converts entirely into kinetic energy.
• The speed of the pendulum at this point can be found using the energy equation \(\frac{1}{2}mv^2 = mgh\), which reflects that any loss in potential energy equals a gain in kinetic energy.

Knowing this equation allows us to solve for the speed of the pendulum as it passes through a vertical position.

Centripetal Force

Centripetal force is the force that keeps an object moving in a curved path, directed towards the center of curvature of the path. In pendulum problems, this force is crucial because it keeps the pendulum following an arc as it swings.

For the pendulum, centripetal force comes from the tension in the string. At any point along the swing, the tension must provide the centripetal force needed to keep the mass moving in a circle:

• At the lowest vertical point, the tension in the string is the greatest. This is because the tension has to counteract gravitational force, as well as provide centripetal force.
• When the pendulum passes through vertical, the equation \(T = mg + \frac{mv^2}{L}\) is used. Here, \(T\) represents the tension, and \(\frac{mv^2}{L}\) represents the centripetal force component.

At angles such as 45 degrees off vertical, tension not only supports the weight but also supplies additional force needed for circular motion. This ensures that the pendulum continues on its path without widening its arc.

Trigonometry in Physics

Trigonometry plays an important role in physics by helping to resolve different force components and calculate distances. In pendulum problems, trigonometric functions like sine and cosine help understand the pendulum's height and velocity.

When calculating the pendulum's height at its maximum angle, trigonometry helps find how high above the lowest point it is:

• For angles, such as 45 degrees, the height can be determined using the equation \(h = L(1 - \cos\theta)\).
• This provides the vertical distance from the lowest point to where the pendulum was released.

Using trigonometry, you also determine force components. When calculating tension at 45 degrees, trigonometry allows you to understand how much force acts along the string and how much is perpendicular. This is why understanding sine and cosine is so helpful; they decompose angles into useful components that reveal how motion and forces behave in such setups.