Problem 62 - Physics for Scientists and Engineers with Modern Physics

Question

(II) An object moves in a circle of radius \(22 \mathrm{~m}\) with its speed given by \(v=3.6+1.5 t^{2},\) with \(v\) in meters per second and \(t\) in seconds. At \(t=3.0 \mathrm{~s},\) find \((a)\) the tangential acceleration and \((b)\) the radial acceleration.

Step-by-Step Solution

Answer

Tangential acceleration: 9.0 m/s^2; Radial acceleration: 13.29 m/s^2.

Step 1: Understand the Problem

We are given the speed of an object moving in a circle as a function of time: \( v = 3.6 + 1.5t^2 \). We need to find the tangential and radial accelerations at \( t = 3.0 \) seconds. The radius of the circle is \( 22 \) meters.

Step 2: Find the Tangential Acceleration

Tangential acceleration \( a_t \) is the derivative of speed with respect to time. Compute the derivative of \( v = 3.6 + 1.5t^2 \): \[ a_t = \frac{dv}{dt} = \frac{d}{dt}(3.6 + 1.5t^2) = 3.0t \]Substitute \( t = 3.0 \) into the equation:\[ a_t = 3.0 \times 3.0 = 9.0 \, \text{m/s}^2 \].

Step 3: Calculate the Speed at t = 3.0 seconds

We need the speed at \( t = 3.0 \) seconds to find the radial acceleration. Substitute \( t = 3.0 \) into the speed equation:\[ v = 3.6 + 1.5(3.0)^2 = 3.6 + 1.5 \times 9 = 3.6 + 13.5 = 17.1 \text{ m/s} \].

Step 4: Compute the Radial Acceleration

Radial (centripetal) acceleration \( a_r \) is given by:\[ a_r = \frac{v^2}{r} \], where \( v \) is the speed and \( r \) is the radius. Substitute \( v = 17.1 \text{ m/s} \) and \( r = 22 \text{ m} \):\[ a_r = \frac{(17.1)^2}{22} = \frac{292.41}{22} \approx 13.29 \, \text{m/s}^2 \].

Step 5: Summarize the Results

The tangential acceleration at \( t = 3.0 \) seconds is \( 9.0 \text{ m/s}^2 \), and the radial acceleration is approximately \( 13.29 \text{ m/s}^2 \).

Key Concepts

Tangential Acceleration

Tangential acceleration is all about how quickly an object's speed changes as it travels along a circular path. Think of it like when you're stepping on the gas pedal in a car, causing it to speed up. Here, the speed of the object is given by the formula \( v = 3.6 + 1.5t^2 \). To find the tangential acceleration, we need to determine how this speed changes over time. This is done by calculating the derivative of the speed with respect to time, which defines the tangential acceleration.
Using calculus, we derive \( v \) with respect to \( t \) to get \( a_t = 3.0t \). Once we know the formula for \( a_t \), finding the tangential acceleration at any given time \( t \) is straightforward. Simply plug the \( t \) value into this expression.
For our specific problem, at \( t = 3.0 \) seconds, the tangential acceleration is \( a_t = 3.0 \times 3.0 = 9.0 \text{ m/s}^2 \). This means at 3 seconds, the object's speed is increasing at a rate of 9 meters per second squared.

Radial Acceleration

Radial acceleration, also known as centripetal acceleration, points toward the center of the circular path. It keeps the object moving in its circular course rather than flying off in a straight line. Whenever an object travels in a circle, interpreting its radial acceleration is key, as it maintains the object's circular motion.
For radial acceleration, we use the formula \( a_r = \frac{v^2}{r} \). Here, \( v \) stands for speed, and \( r \) is the radius of the circle. In our example, we've already calculated that at \( t = 3.0 \) seconds, the speed is \( v = 17.1 \text{ m/s} \). With a circle radius of 22 meters, we substitute these into the formula to get \( a_r = \frac{(17.1)^2}{22} = \frac{292.41}{22} \approx 13.29 \text{ m/s}^2 \).
This result indicates that at 3 seconds, an acceleration of approximately 13.29 meters per second squared is needed to keep the object moving in its circular path.

Motion in a Circle

Motion in a circle involves both these accelerations working together. When an object travels in a circle, it's influenced by both the tangential and radial accelerations.

• **Tangential Acceleration (\( a_t \))**: changes the speed of the object along the circular path.
• **Radial Acceleration (\( a_r \))**: changes the direction of the object's motion, pulling it toward the center of the circle.

This combination keeps objects moving in a circular motion and allows us to predict how they will behave over time. If there's an increase in tangential acceleration, the object speeds up along the path. On the other hand, strong radial acceleration ensures it stays on the path, preventing any deviation into a straight line. Together, these concepts form the basis of understanding circular motion and how forces interact to produce paths like the orbit of planets or the movement of cars on circular tracks.