Problem 31 - College Physics

Question

A wheel of radius \(1.5 \mathrm{~m}\) rotates at a uniform speed. If a point on the rim of the wheel has a centripetal acceleration of \(1.2 \mathrm{~m} / \mathrm{s}^{2},\) what is the point's tangential speed?

Step-by-Step Solution

Answer

The point's tangential speed is approximately \(1.34\,\text{m/s}\).

Step 1: Understanding the Problem

We are asked to find the tangential speed of a point on the rim of a wheel given its centripetal acceleration. We are given two pieces of information: the radius of the wheel, which is \( r = 1.5 \, \text{m} \), and the centripetal acceleration \( a_c = 1.2 \, \text{m/s}^2 \).

Step 2: Formula for Centripetal Acceleration

The formula for centripetal acceleration is \( a_c = \frac{v^2}{r} \), where \( v \) is the tangential speed and \( r \) is the radius of the path. We will use this formula to solve for \( v \).

Step 3: Rearranging the Formula

To find \( v \), rearrange the formula \( a_c = \frac{v^2}{r} \) to solve for \( v \). This gives us \( v^2 = a_c \cdot r \).

Step 4: Substituting the Values

Substitute the given values into the rearranged formula: \( v^2 = 1.2 \, \text{m/s}^2 \times 1.5 \, \text{m} \).

Step 5: Calculating the Expression

Calculate \( v^2 = 1.2 \, \text{m/s}^2 \times 1.5 \, \text{m} = 1.8 \, \text{m}^2/\text{s}^2 \).

Step 6: Finding the Tangential Speed

Take the square root to solve for \( v \): \( v = \sqrt{1.8 \, \text{m}^2/\text{s}^2} \).

Step 7: Final Calculation

Calculate the square root: \( v \approx 1.34 \, \text{m/s} \). This is the tangential speed of the point on the rim of the wheel.

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a key concept in understanding how objects move in a circular path. This acceleration points towards the center of the circle in which the object is traveling. It's important to know that centripetal acceleration is what keeps the object moving in a circular path rather than flying off in a straight line.

In mathematical terms, centripetal acceleration (\(a_c\)) is expressed as:

• \(a_c = \frac{v^2}{r}\)

where:

• \(v\) is the tangential speed of the object
• \(r\) is the radius of the circular path

This formula tells us that the faster an object moves (\(v\) is larger) or the tighter the circle (\(r\) is smaller), the greater the centripetal acceleration needed to keep it moving in that circle. By understanding this relationship, you can calculate any of the three variables if you have the other two. This way, if you are ever given a scenario with a certain centripetal acceleration and radius, you can find out how fast the object is moving.

Uniform Circular Motion

Uniform circular motion refers to the motion of an object traveling at a constant speed along a circular path. In this type of motion, although the speed remains constant, the velocity is continuously changing direction since velocity is a vector quantity that depends on direction as well as speed.

Due to this constant change in direction, an object in uniform circular motion experiences centripetal acceleration, even though its speed isn't increasing or decreasing. This means every point on the circular path has the same centripetal acceleration magnitude, which directly affects the tangential speed of the object.

• Even if the speed is constant, the specific direction changes at every moment, causing acceleration.
• This is why uniform circular motion is often considered an example of accelerated motion.

Understanding uniform circular motion is fundamental in fields like physics and engineering, especially when analyzing systems involving rotation, such as wheels, engines, or planetary orbits.

Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause these motions. It focuses on concepts such as velocity, acceleration, displacement, and time.

In the context of circular motion, kinematics becomes a useful tool to analyze how objects behave as they move around a circular path. In this specific scenario with the wheel's rim, we relate the centripetal acceleration and the radius to find the tangential speed of a point on the rim.

• This involves understanding how different forms of motion - linear and angular - relate to each other.
• Kinematics can help distinguish between different types of speed and acceleration, like linear vs. angular or tangential vs. centripetal.

The benefit of kinematics lies in its ability to break down complex movements into simple, understandable parts, and solve them systematically, like using the formula for centripetal acceleration to determine speeds. Grasping these fundamental principles allows you to approach and solve real-world problems more effectively.