Question

A bowling ball weighing 71.2 N 116.0 lb2 is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 m/s. At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

Short answer

Acceleration: 4.64 m/s² upward; Tension: 104.94 N.

Step-by-step solution

Identify Forces and Setup Coordinate System

In a pendulum, the forces involved are tension (T) in the rope and gravitational force (mg). Since the ball is swinging, we'll use a radial coordinate system with centripetal force pointing towards the center (upward) and gravitational force acting downward.

Calculate Radial (Centripetal) Acceleration

The radial (centripetal) acceleration can be calculated using the formula: \[ a_r = \frac{v^2}{r} \]where \( v = 4.20 \, \text{m/s} \) is the speed and \( r = 3.80 \, \text{m} \) is the length of the rope. Substituting the values:\[ a_r = \frac{4.20^2}{3.80} = \frac{17.64}{3.80} = 4.64 \, \text{m/s}^2 \]

Determine the Direction of Acceleration

Since the bowling ball is at the lowest point in its swing, the direction of acceleration is upward (towards the center of the circular path), aligning with the tension in the rope.

Analyze Forces on Bowling Ball

At the lowest point, two forces act: the upward tension \( T \) and the downward gravitational force \( mg \). The centripetal force is provided by the net force directed towards the center, thus:\[ T - mg = ma_r \]

Solve for Tension (T) in the Rope

Rearrange the equation from Step 4:\[ T = ma_r + mg \]Given \( mg = 71.2 \, \text{N} \), use \( a_r = 4.64 \, \text{m/s}^2 \) and the mass \( m \) can be derived as \( \frac{71.2}{9.8} \approx 7.27 \, \text{kg} \). Substitute into the equation:\[ T = 7.27 \times 4.64 + 71.2 = 33.74 + 71.2 = 104.94 \, \text{N} \]

Conclusion

To summarize, the acceleration of the bowling ball is \(4.64 \, \text{m/s}^2\), directed upward. The tension in the rope when the ball is at its lowest point is \(104.94 \, \text{N}\).

Key concepts

Centripetal Acceleration

Centripetal acceleration occurs when an object moves along a circular path, constantly changing direction. Although the speed may remain constant, the direction change means the velocity is varying, creating acceleration. This acceleration always points towards the center of the circular path.
A simple formula to calculate centripetal acceleration is:

• \( a_r = \frac{v^2}{r} \)

where \( v \) is the velocity of the object and \( r \) is the radius of the circle. For example, in the case of our bowling ball pendulum, the velocity is \( 4.20 \, \text{m/s} \) and the radius is the length of the rope, \( 3.80 \, \text{m} \). Substituting into the equation gives the result of \( 4.64 \, \text{m/s}^2 \) for the centripetal acceleration.
Remember, the direction of this acceleration is always inward, towards the center of the pendulum's oscillatory path, ensuring the pendulum continues its circular trajectory.

Tension Force

Tension force is crucial in pendulum motion as it acts along the length of the rope, maintaining the connection between the pendulum bob and its pivot point. It is a pulling force, opposing gravitational pull to keep the object in circular motion.
When the pendulum swings, tension isn't constant. Its value depends on the pendulum's position and velocity. At the lowest point in the swing, where our bowling ball has the highest speed, tension force is greatest. This is because tension must counteract both the gravitational force and provide the needed centripetal force for acceleration.
To find the tension at this lowest point, we use the equation:

• \( T = ma_r + mg \)

where \( m \) is the mass, \( a_r \) is the centripetal acceleration, and \( g \) is the gravitational acceleration (\( 9.8 \, \text{m/s}^2 \)). This tension supports the gravitational force and contributes to the centripetal force needed to keep the ball moving in a circle.

Gravitational Force

Gravitational force acts vertically downward on the pendulum bob and is a constant force given by \( mg \), where \( m \) is mass and \( g \) is gravitational acceleration. This force pulls the pendulum towards Earth's center.
In our scenario, gravitational force contributes to the downward pull on the bowling ball, creating a maximum at points swung furthest from the vertical. However, when at the bottom, this force contributes to the centripetal force requirement for circular motion. Here, the gravitational force equals the weight of the bowling ball, which has been calculated as \( 71.2 \, \text{N} \).
It's important to understand that while gravitational force remains constant, it doesn't often act alone. In pendulum motion analysis, it's combined with tension to either aid or counteract forces experienced, particularly when analyzing forces like in this exercise. Keeping gravitational influence in check allows us to analyze movements and tensions effectively in pendular swings.