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Chapter 11: Problem 78

A wheel of radius \(0.250 \mathrm{~m},\) moving initially at \(43.0 \mathrm{~m} / \mathrm{s},\) rolls to a stop in \(225 \mathrm{~m}\). Calculate the magnitudes of its (a) linear acceleration and (b) angular acceleration. (c) Its rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis. Find the magnitude of the torque about the central axis due to friction on the wheel.

Short Answer

Expert verified
(a) Linear acceleration: 4.11 m/s虏, (b) Angular acceleration: 16.44 rad/s虏, (c) Torque: 2.5482 Nm.

Step by step solution

01

Determine the Linear Acceleration

To find the linear acceleration, use the formula for motion with constant acceleration: \[v^2 = u^2 + 2as\]where \(v = 0\) m/s (final velocity), \(u = 43.0\) m/s (initial velocity), and \(s = 225\) m (distance). Solving for \(a\):\[0 = (43.0)^2 + 2a(225)\]\[a = \frac{-(43.0)^2}{2 \times 225}\]\[a = -4.11 \mathrm{~m/s^2}\]Thus, the linear acceleration is \(4.11 \mathrm{~m/s^2}\).
02

Calculate the Angular Acceleration

The relationship between linear acceleration \(a\) and angular acceleration \(\alpha\) is given by:\[a = r\alpha\]where \(r = 0.250 \mathrm{~m}\) is the radius of the wheel. Solving for \(\alpha\):\[\alpha = \frac{a}{r} = \frac{4.11}{0.250}\]\[\alpha = 16.44 \mathrm{~rad/s^2}\]The angular acceleration is \(16.44 \mathrm{~rad/s^2}\).
03

Find the Torque Due to Friction

Torque \(\tau\) can be found using the relation from rotational dynamics:\[\tau = I \alpha\]where \(I = 0.155 \mathrm{~kg \cdot m^2}\) is the rotational inertia and \(\alpha = 16.44\) rad/s虏. Substituting the values:\[\tau = 0.155 \times 16.44\]\[\tau = 2.5482 \mathrm{~Nm}\]The torque due to friction is \(2.5482 \mathrm{~Nm}\).

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Key Concepts

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

Linear Acceleration
Linear acceleration is a concept that describes how the velocity of an object changes with time along a straight path. In simpler terms, it tells us how quickly an object is speeding up or slowing down in a straight line. To find linear acceleration, we use the formula \[v^2 = u^2 + 2as\], where:
  • \(v\) is the final velocity. In our problem, the wheel comes to a stop, so \(v=0\) m/s.
  • \(u\) is the initial velocity. For our wheel, \(u = 43.0\) m/s.
  • \(s\) is the distance over which the wheel travels, which is 225 m in this case.
The equation is solved for \(a\), the linear acceleration, which gives us \(a = -4.11\) m/s虏. The negative sign indicates that the wheel is slowing down, which is also called deceleration. This helps us understand how fast the wheel's speed is decreasing until it stops.
Angular Acceleration
Angular acceleration is the rate at which the angular velocity of an object changes with time. For objects moving in a circle, this is an important concept because it helps us understand how quickly the object is rotating faster or slower. In this example, the angular acceleration \(\alpha\) is connected to the linear acceleration by the formula \(a = r\alpha\), where:
  • \(a\) is the linear acceleration, already found at 4.11 m/s虏.
  • \(r\) is the radius of the wheel, given as 0.250 m.
By rearranging the formula to solve for \(\alpha\), we have \(\alpha = \frac{a}{r}\), resulting in \(\alpha = 16.44\) rad/s虏. This tells us that for each meter per second squared of linear deceleration, the wheel's angular speed decreases more significantly due to its relatively small radius.
Torque
Torque is a measure of the rotational force applied to an object, causing it to rotate around an axis. It is similar to how force causes linear acceleration, but in this case, it brings about angular acceleration. In the context of this problem, torque due to friction is what slows the wheel down. The formula to find torque is \(\tau = I \alpha\), where:
  • \(\tau\) is the torque.
  • \(I\) is the rotational inertia of the object, which depends on how mass is distributed around the axis. In our problem, \(I = 0.155\) kg路m虏.
  • \(\alpha\) is the angular acceleration, previously determined to be 16.44 rad/s虏.
Using these values, we find the torque \(\tau = 2.5482\) Nm. This result shows the amount of rotational force needed to bring the wheel to a stop due to friction. Torque is a crucial concept in understanding how rotational motion is influenced by forces.

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Most popular questions from this chapter

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