Problem 52 - College Physics

Question

A simple pendulum of length \(2.00 \mathrm{~m}\) is released from a horizontal position. When it makes an angle of \(30^{\circ}\) from the vertical, determine (a) its angular acceleration, \((\mathrm{b})\) its centripetal acceleration, and \((\mathrm{c})\) the tension in the string. Assume the bob's mass is \(1.50 \mathrm{~kg}\).

Step-by-Step Solution

Answer

(a) Angular acceleration: \(2.45 \mathrm{~rad/s^2}\); (b) Centripetal acceleration: \(2.62 \mathrm{~m/s^2}\); (c) Tension in string: \(16.67 \mathrm{~N}\).

Step 1: Understand the Problem

We have a simple pendulum with a string length of \(2.00 \mathrm{~m}\) and a mass of \(1.50 \mathrm{~kg}\). We need to find the angular acceleration, centripetal acceleration, and tension in the string when the pendulum makes a \(30^{\circ}\) angle with the vertical.

Step 2: Calculate Angular Acceleration

The angular acceleration \( \alpha \) of a simple pendulum is given by the formula \( \alpha = \frac{g}{L} \sin(\theta) \), where \( g = 9.81 \mathrm{~m/s^2} \) is the acceleration due to gravity, \( L = 2.00 \mathrm{~m} \) is the length of the pendulum, and \( \theta = 30^{\circ} \) is the angle from the vertical. Plugging in the values:\[\alpha = \frac{9.81}{2.00} \sin(30^{\circ}) = \frac{9.81}{2.00} \times 0.5 = 2.4525 \mathrm{~rad/s^2}\]

Step 3: Calculate Centripetal Acceleration

The centripetal acceleration \( a_c \) of a pendulum bob is \( a_c = \frac{v^2}{L} \), where \( L = 2.00 \mathrm{~m} \). At an angle \( \theta \), the speed \( v \) is acquired from energy conservation principles, \( v = \sqrt{2gL(1 - \cos(\theta))} \). First, find \( v \):\[v = \sqrt{2 \times 9.81 \times 2.00 \times (1 - \cos(30^{\circ}))} = \sqrt{2 \times 9.81 \times 2.00 \times (1 - 0.866)}\]\[v = \sqrt{2 \times 9.81 \times 2.00 \times 0.134} = \sqrt{5.256} \approx 2.29 \mathrm{~m/s}\]Then, calculate \( a_c \):\[a_c = \frac{v^2}{L} = \frac{(2.29)^2}{2.00} \approx 2.62 \mathrm{~m/s^2}\]

Step 4: Calculate Tension in the String

The tension \( T \) in the string can be calculated using Newton's second law in the direction along the string: \( T - mg\cos(\theta) = m\cdot a_c \), where \( m = 1.50 \mathrm{~kg} \). Solve for \( T \):\[T = m \cdot a_c + mg\cos(\theta) = 1.50 \times 2.62 + 1.50 \times 9.81 \times 0.866\]\[T = 3.93 + 12.74 \approx 16.67 \mathrm{~N}\]

Step 5: Final Calculation Summary

- Angular acceleration: \(2.45 \mathrm{~rad/s^2}\)- Centripetal acceleration: \(2.62 \mathrm{~m/s^2}\)- Tension in the string: \(16.67 \mathrm{~N}\)

Key Concepts

Angular Acceleration

Angular acceleration refers to how quickly the angular velocity of an object changes with time. For a pendulum, this happens due to gravity pulling on it, striving to bring it back to its equilibrium position. The formula for finding angular acceleration \( \alpha \) in a pendulum is given by:

• \( \alpha = \frac{g}{L} \sin(\theta) \)

Here, \( g \) is the acceleration due to gravity (approximately \( 9.81 \mathrm{~m/s^2} \)), \( L \) is the length of the pendulum, and \( \theta \) is the angle with respect to the vertical.
The sine component, \( \sin(\theta) \), tells us how much of the gravitational pull contributes to the pendulum's motion perpendicular to the string's length. Since \( \sin(30^\circ) \) is \( 0.5 \), this reduces the effect to half its potential at this angle.
For the pendulum in our problem with a length of \( 2.00 \mathrm{~m} \), the angular acceleration works out to be about \( 2.45 \mathrm{~rad/s^2} \). Understanding this helps in predicting how the pendulum will speed up as it swings downward.

Centripetal Acceleration

Centripetal acceleration is crucial in understanding pendular motion. It refers to the net acceleration acting toward the center of the circular path that the pendulum bob follows. This ensures it stays on its circular arc as it swings back and forth.
To find the centripetal acceleration \( a_c \), we use the relationship:

• \( a_c = \frac{v^2}{L} \)

where \( v \) is the tangential speed of the pendulum bob and \( L \) is the length of the pendulum. The speed \( v \) can be determined by the energy conservation principle, finding the kinetic energy at its current position as potential energy converts.
In the given exercise, the calculated speed \( v \) of the pendulum bob is approximately \( 2.29 \mathrm{~m/s} \). Plugging this value into the centripetal acceleration formula gives \( a_c \approx 2.62 \mathrm{~m/s^2} \). This highlights how quickly and tightly the pendulum follows its circular path, significantly influenced by its velocity and the string's length.

Tension in String

Tension in the string of a pendulum is the force exerted by the string on the pendulum bob to not only support its weight but also provide the centripetal force required as it swings.
This tension can be calculated by applying Newton's second law in the direction along the string:

• \( T = m \cdot a_c + mg\cos(\theta) \)

Here, \( T \) is the tension, \( m \) is the mass of the pendulum bob, \( a_c \) is the centripetal acceleration, and \( \theta \) is the angle from the vertical.
The term \( mg\cos(\theta) \) accounts for the component of gravitational force along the direction of the string, which combines with the centripetal component \( m \cdot a_c \). In our example, these forces sum up to give a tension of approximately \( 16.67 \mathrm{~N} \), which balances the total forces acting on the pendulum bob.