Problem 100 - Physics for Scientists and Engineers with Modern Physics

Question

A ball of mass \(m=1.0 \mathrm{~kg}\) at the end of a thin cord of length \(r=0.80 \mathrm{~m}\) revolves in a vertical circle about point \(\mathrm{O},\) as shown in Fig. \(5-56 .\) During the time we observe it, the only forces acting on the ball are gravity and the tension in the cord. The motion is circular but not uniform because of the force of gravity. The ball increases in speed as it descends and decelerates as it rises on the other side of the circle. At the moment the cord makes an angle \(\theta=30^{\circ}\) below the horizontal, the ball's speed is \(6.0 \mathrm{~m} / \mathrm{s}\). At this point, determine the tangential acceleration, the radial acceleration, and the tension in the cord, \(F_{\mathrm{T}}\). Take \(\theta\) increasing downward as shown.

Step-by-Step Solution

Answer

Tangential acceleration is 4.9 m/s², radial acceleration is 45 m/s², and tension is 53.48 N.

Step 1: Identify Forces Acting on the Ball

At any point on the circle, two forces act on the ball: gravity (downwards, with magnitude \(mg\)) and the tension in the string \(F_T\). These forces contribute to the ball's acceleration both radially and tangentially due to the circular motion.

Step 2: Calculate Tangential Acceleration

The component of gravitational force contributing to the tangential acceleration \(a_t\) is \(mg \sin \theta\). Use the formula \[ a_t = g \sin \theta \]where \(g = 9.8 \mathrm{~m/s^2}\) and \(\theta = 30^{\circ}\).Substitute the values:\[ a_t = 9.8 \sin 30^{\circ} = 4.9 \mathrm{~m/s^2} \]

Step 3: Calculate Radial Acceleration

The radial acceleration \(a_r\) can be calculated using the centripetal acceleration formula, \[ a_r = \frac{v^2}{r} \]where \(v = 6.0 \mathrm{~m/s}\) and \(r = 0.8 \mathrm{~m}\).Substitute the values:\[ a_r = \frac{6.0^2}{0.8} = 45 \mathrm{~m/s^2} \]

Step 4: Calculate Tension in the Cord

The tension in the cord is the result of both centripetal force required for circular motion and the gravitational force's radial component. Using the formula:\[ F_T = m \cdot a_r + mg \cos \theta \]where \(m = 1.0 \mathrm{~kg}\), \(g = 9.8 \mathrm{~m/s^2}\), and \(\theta = 30^{\circ}\), substitute the values:\[ F_T = 1.0 \cdot 45 + (1.0)(9.8)(\cos 30^{\circ}) \]Calculate:\[ F_T = 45 + 9.8 \cdot 0.866 = 53.48 \mathrm{~N} \]

Key Concepts

Tangential Acceleration

In circular motion, tangential acceleration refers to the rate of change of the object's speed along the circular path. It is the acceleration that affects how fast or slow an object moves along its circular path.
For our ball on a string, the force of gravity plays a crucial role. It causes the ball to speed up as it moves downward. This change in speed is what we call tangential acceleration.

When the ball is at an angle of \( \theta = 30^{\circ} \) below the horizontal, the component of gravitational force affecting this speed change is \( mg \sin \theta \). Using the formula:

• \( a_t = g \sin \theta \)
• Where \( g = 9.8 \mathrm{~m/s^2} \)
• \( \theta = 30^{\circ} \)

Substituting the given values into the formula results in \( a_t = 4.9 \mathrm{~m/s^2} \).
This value means that the ball is accelerating at a rate of \( 4.9 \mathrm{~m/s^2} \) in the direction tangential to the circular path.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, is the acceleration that keeps an object moving in a circular path. It acts toward the center of the circle, essentially "pulling" the object inward to maintain its circular motion.
In our scenario, it's responsible for keeping the ball stable along its path as it travels around the circle. This acceleration rises because of the constant change in direction of the ball's velocity.

To find the radial acceleration \( a_r \), use the formula:

• \( a_r = \frac{v^2}{r} \)
• Where \( v = 6.0 \mathrm{~m/s} \)
• \( r = 0.8 \mathrm{~m} \)

Plug in these values to get \( a_r = 45 \mathrm{~m/s^2} \).
This number means that the ball experiences a force strong enough to keep it moving in its circular path, even as it wants to fly outward due to its inertia.

Tension in a String

Tension in the string is a pivotal force during circular motion, responsible for sustaining the ball's path around the circle. It acts along the length of the string and is formed by two main components: the need for centripetal force (to maintain the circle) and the gravitational force's radial component.

To calculate the tension \( F_T \) in the string, consider:

• The centripetal force component: \( m \cdot a_r \)
• The gravitational force component aligned radially: \( mg \cos \theta \)
• Combine these in the tension formula: \( F_T = m \cdot a_r + mg \cos \theta \)

With \( m = 1.0 \mathrm{~kg} \), \( g = 9.8 \mathrm{~m/s^2} \), and \( \theta = 30^{\circ} \), fill in the known values:
\( F_T = 1.0 \cdot 45 + 9.8 \cdot 0.866 = 53.48 \mathrm{~N} \).
This value signifies the string needs to exert a considerable force, 53.48 Newtons, to balance both the centripetal needs of the motion and the pull of gravity acting on the ball.