Problem 69 - University Physics with Modern Physics

Question

A classic 1957 Chevrolet Corvette of mass 1240 \(\mathrm{kg}\) starts from rest and speeds up with a constant tangential acceleration of 3.00 \(\mathrm{m} / \mathrm{s}^{2}\) on a circular test track of radius 60.0 \(\mathrm{m}\) . Treat the car as a particle. (a) What is its angular acceleration? (b) What is its angular speed 6.00 s after it starts? (c) What is its radial acceleration at this time? (d) Sketch a view from above showing the circular track, the car, the velocity vector, and the acceleration component vectors 6.00 s after the car starts. (e) What are the magnitudes of the total acceleration and net force for the car at this time? (f) What angle do the total acceleration and net force make with the car's velocity at this time?

Step-by-Step Solution

Answer

(a) 0.050 rad/s e) 6.20 m/s², 7688 N f) 62° with velocity

Step 1: Understand Circular Motion Basics

When an object moves in a circular path, it possesses both tangential and radial components of acceleration. For an object moving with constant tangential acceleration, we can link linear and angular quantities through the formulas: \(a_t = r \alpha\), where \(a_t\) is tangential acceleration, \(\alpha\) is angular acceleration, and \(r\) is the radius of the circular path.

Step 2: Calculate Angular Acceleration

Given the tangential acceleration \(a_t = 3.00 \, \mathrm{m/s^2}\) and the radius \(r = 60.0 \, \mathrm{m}\), the angular acceleration \(\alpha\) can be calculated using the equation \(\alpha = \frac{a_t}{r}\).\[\alpha = \frac{3.00 \, \mathrm{m/s^2}}{60.0 \, \mathrm{m}} = 0.0500 \, \mathrm{rad/s^2}\]

Step 3: Calculate Angular Speed After 6 Seconds

Use the formula for angular speed: \(\omega = \omega_0 + \alpha t\), where \(\omega_0\) is the initial angular speed (0, since the car starts from rest), \(\alpha\) is angular acceleration, and \(t = 6.00 \, \mathrm{s}\).\[\omega = 0 + (0.050 \, \mathrm{rad/s^2})(6.00 \, \mathrm{s}) = 0.300 \, \mathrm{rad/s}\]

Step 4: Calculate Radial Acceleration at 6 Seconds

The radial acceleration of an object moving in a circle is given by \(a_r = \omega^2 r\). Using \(\omega = 0.300 \, \mathrm{rad/s}\) and \(r = 60.0 \, \mathrm{m}\):\[a_r = (0.300 \, \mathrm{rad/s})^2 \times 60.0 \, \mathrm{m} = 5.40 \, \mathrm{m/s^2}\]

Step 5: Sketch Velocity and Acceleration Vectors

Imagine the car moving counterclockwise on the circular track. The velocity vector is tangent to the path. The radial acceleration vector points towards the center of the circle, while the tangential acceleration vector aligns with the car's velocity direction.

Step 6: Calculate Total Acceleration and Net Force

The total acceleration \(a_t\) of the car is the vector sum of the tangential and radial accelerations. Use Pythagorean theorem:\[|a_t| = \sqrt{a_{t}^2 + a_{r}^2} = \sqrt{(3.00)^2 + (5.40)^2} \, \mathrm{m/s^2} = 6.20 \, \mathrm{m/s^2}\]The net force \(F\) is given by \(F = ma_t\), where \(m\) is the mass of the car.\[F = (1240 \, \mathrm{kg})(6.20 \, \mathrm{m/s^2}) = 7688 \, \mathrm{N}\]

Step 7: Calculate the Angle of Total Acceleration and Force

The angle \(\theta\) between the total acceleration and the velocity is found using the relation \(\tan(\theta) = \frac{a_r}{a_t}\):\[\theta = \tan^{-1}\left(\frac{5.40}{3.00}\right) = 61.9^\circ\]

Key Concepts

Angular Acceleration

Angular acceleration is all about how fast the rotational speed of an object changes over time. If we think of our Chevrolet moving around a circular track, angular acceleration (\(\alpha\)) tells us how quickly the car speeds up as it rotates around the track.
For our car, angular acceleration can be calculated using the formula:

• \( \alpha = \frac{a_t}{r} \)

where \(a_t = 3.00 \, \mathrm{m/s^2}\) is the tangential acceleration and \(r = 60.0 \, \mathrm{m}\) is the radius of the track.
Plugging in the values, we find:

• \( \alpha = \frac{3.00 \, \mathrm{m/s^2}}{60.0 \, \mathrm{m}} = 0.0500 \, \mathrm{rad/s^2} \)

This means that the car's rate of change of its angular velocity is 0.0500 radians per second squared.
Understanding angular acceleration is crucial because a higher angular acceleration means a quicker increase in speed as the car travels around the track.

Radial Acceleration

Radial acceleration is key to understanding how an object maintains its circular path. It's the component of acceleration that points towards the center of the circle. Imagine something pulling the car inward to keep it in its circular motion.
For our Chevrolet, the radial acceleration (\(a_r\)) after 6 seconds can be calculated with:

• \( a_r = \omega^2 r \)

where \(\omega\) is the angular speed at 6 seconds, calculated as \(0.300 \, \mathrm{rad/s}\).
So, using \(r = 60.0 \, \mathrm{m}\), we get:

• \( a_r = (0.300 \, \mathrm{rad/s})^2 \times 60.0 \, \mathrm{m} = 5.40 \, \mathrm{m/s^2} \)

This component is crucial, as without this inward pull, the car would simply fly off in a straight line! Think of radial acceleration as the car's link to the track's center, helping it stay on course.

Tangential Acceleration

Tangential acceleration refers to how quickly the car is speeding up along the track -- think about pressing the gas pedal and feeling the car zoom faster.
In this scenario, our Chevrolet starts from rest with a given tangential acceleration of 3.00 \, \mathrm{m/s^2}. This means it's consistently speeding up at this rate along the edge of the circular path. Tangential acceleration affects the car's velocity, increasing it over time.
So, after 6 seconds, the car wouldn't just be moving at the initial velocity. Instead:

• \( \text{final speed} = \text{initial speed} + a_t \times t \)
• Since it starts from rest, the initial speed is 0, so it just becomes \( a_t \times t \)

In this particular situation, the tangential acceleration not only changes the linear speed but also influences the circular motion, as it directly relates to angular acceleration through the formula: \(a_t = r \alpha\).
Understanding tangential acceleration helps us see how linear speed influences circular motion and provides insights into how quickly our car gains speed on the track.