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Chapter 10: Problem 32

A car starts from rest and moves around a circular track of radius \(30.0 \mathrm{~m}\). Its speed increases at the constant rate of \(0.500 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the magnitude of its net linear acceleration \(15.0\) s later? (b) What angle does this net acceleration vector make with the car's velocity at this time?

Short Answer

Expert verified
(a) The magnitude of net acceleration is approximately 1.940 m/s虏. (b) The angle with the car's velocity is approximately 75.96掳.

Step by step solution

01

Calculate tangential acceleration

The tangential acceleration (\(a_t\)) is given as the rate at which speed increases, which is constant at \(0.500 \, \mathrm{m/s^2}\). Therefore, the tangential acceleration is \(a_t = 0.500 \, \mathrm{m/s^2}\).
02

Calculate final speed after 15 seconds

Since the car starts from rest, the speed (\(v\)) after time \(t\) seconds can be calculated using the formula:\[ v = a_t \times t \]Substitute the given values:\[ v = 0.500 \, \mathrm{m/s^2} \times 15.0 \, \mathrm{s} = 7.5 \, \mathrm{m/s} \]
03

Calculate centripetal acceleration

The centripetal acceleration (\(a_c\)) can be calculated using the formula:\[ a_c = \frac{v^2}{r} \]where \(v = 7.5 \, \mathrm{m/s}\) and \(r = 30.0 \, \mathrm{m}\). Thus:\[ a_c = \frac{(7.5 \, \mathrm{m/s})^2}{30.0 \, \mathrm{m}} = \frac{56.25}{30.0} = 1.875 \, \mathrm{m/s^2} \]
04

Calculate the magnitude of the net acceleration

The magnitude of the net acceleration (\(a_{net}\)) is found by combining the tangential and centripetal accelerations:\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]Substitute the known values:\[ a_{net} = \sqrt{(0.500 \, \mathrm{m/s^2})^2 + (1.875 \, \mathrm{m/s^2})^2} = \sqrt{0.25 + 3.515625} = \sqrt{3.765625} \approx 1.940 \, \mathrm{m/s^2} \]
05

Calculate the angle of net acceleration with velocity

The angle (\(\theta\)) of the net acceleration vector with respect to the car's velocity can be found using the formula:\[ \theta = \arctan\left(\frac{a_c}{a_t}\right) \]Substitute the calculated accelerations:\[ \theta = \arctan\left(\frac{1.875}{0.500}\right) = \arctan(3.75) \\theta \approx 75.96^\circ \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangential Acceleration
Tangential acceleration is the component of acceleration that is parallel to the direction of motion in circular motion. It essentially represents the rate of change of the object's speed along its path. In the context of the given exercise, we consider how the car's speed increases over time due to this constant acceleration.
In this scenario, the tangential acceleration is provided as a constant value of \(0.500 \, \mathrm{m/s^2}\). This means the speed of the car increases steadily, adding \(0.500 \, \mathrm{m/s}\) to its velocity every second.
  • To calculate the speed of the car after a certain time, the formula used is \(v = a_t \times t\).
  • This is a straightforward calculation since the car starts from rest.
In particular, after \(15\) seconds, the car reaches a speed of \(7.5 \, \mathrm{m/s}\) due to the constant tangential acceleration.
Centripetal Acceleration
Centripetal acceleration is key in circular motion and acts perpendicular to the direction of the car's velocity, always pointing towards the center of the circular path. This acceleration doesn't change the speed of the car but ensures it stays on the circular track.
To calculate centripetal acceleration, the formula used is \(a_c = \frac{v^2}{r}\), where:
  • \(v\) is the car's linear speed, here \(7.5 \, \mathrm{m/s}\).
  • \(r\) is the radius of the circle, given as \(30.0 \, \mathrm{m}\).
The car, consequently, experiences a centripetal acceleration of \(1.875 \, \mathrm{m/s^2}\). This acceleration is crucial in maintaining the circular path as it prevents the car from moving off the track.
Net Linear Acceleration
The net linear acceleration in circular motion is the vector sum of two distinct accelerations: tangential and centripetal acceleration. It gives the overall acceleration of the object in motion and determines both the change in speed and direction.
This can be determined using the Pythagorean theorem since the accelerations are perpendicular:\[ a_{net} = \sqrt{a_t^2 + a_c^2} \]Where:
  • \(a_t = 0.500 \, \mathrm{m/s^2}\) (tangential acceleration)
  • \(a_c = 1.875 \, \mathrm{m/s^2}\) (centripetal acceleration)
Plugging in these values provides a net linear acceleration of \(1.940 \, \mathrm{m/s^2}\).
This net acceleration vector forms an angle with the velocity vector of the car, highlighting the combined effect of both changing speed and the direction of motion. Calculating this angle requires:\[ \theta = \arctan\left(\frac{a_c}{a_t}\right) \approx 75.96^\circ \]This angle shows how these accelerations interplay to influence the car's motion on a circular path.

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Most popular questions from this chapter

A bicyclist of mass \(70 \mathrm{~kg}\) puts all his mass on each downwardmoving pedal as he pedals up a steep road. Take the diameter of the circle in which the pedals rotate to be \(0.40 \mathrm{~m}\), and determine the magnitude of the maximum torque he exerts about the rotation axis of the pedals.

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