Problem 26 - College Physics

Question

\(\bullet\) A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2} .\) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start, (b) after it has turned through \(60.0^{\circ},\) and \((\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)

Step-by-Step Solution

Answer

(a) Start: Tangential = 0.180 m/s², Radial = 0, Resultant = 0.180 m/s² (b) 60°: Tangential = 0.180 m/s², Radial = 0.377 m/s², Resultant ≈ 0.418 m/s² (c) 120°: Tangential = 0.180 m/s², Radial = 0.753 m/s², Resultant ≈ 0.775 m/s²

Step 1: Compute Tangential Acceleration

The tangential acceleration (\(a_t\)) can be calculated using the formula \(a_t = \alpha r\), where \(\alpha = 0.600\, \mathrm{rad/s^2}\) is the angular acceleration and \(r = 0.300\, \mathrm{m}\) is the radius of the flywheel. Thus, \(a_t = 0.600 \times 0.300 = 0.180\, \mathrm{m/s^2}\).

Step 2: Compute the Radial Acceleration at the Start (θ = 0 radians)

At the start, the radial acceleration \(a_r = \omega^2 r\). The angular velocity \(\omega = \alpha t\). Initially, \(t = 0\), hence \(\omega = 0\), leading to \(a_r = 0 \, \mathrm{m/s^2}\).

Step 3: Compute the Tangential Acceleration After 60° Turn

The tangential acceleration remains constant throughout, so \(a_t = 0.180 \, \mathrm{m/s^2}\).

Step 4: Compute the Radial Acceleration After 60° Turn

Convert 60° to radians: \(\theta = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \, \mathrm{radians}\). Using \(\omega^2 = 2\alpha\theta\), we find \(\omega^2 = 2 \times 0.600 \times \frac{\pi}{3} = 1.200 \cdot \frac{\pi}{3}\). Thus, \(a_r = \omega^2 r = 1.200 \cdot \frac{\pi}{3} \times 0.300 = 0.377 \,\mathrm{m/s^2}\).

Step 5: Compute the Tangential Acceleration After 120° Turn

The tangential acceleration remains constant, \(a_t = 0.180 \, \mathrm{m/s^2}\).

Step 6: Compute the Radial Acceleration After 120° Turn

Convert 120° to radians: \(\theta = 120 \times \frac{\pi}{180} = \frac{2\pi}{3} \, \mathrm{radians}\). Using \(\omega^2 = 2\alpha\theta\), we get \(\omega^2 = 2 \times 0.600 \times \frac{2\pi}{3}\). Thus, \(a_r = \omega^2 r = 1.200 \cdot \frac{2\pi}{3} \times 0.300 = 0.753 \, \mathrm{m/s^2}\).

Step 7: Compute Resultant Acceleration at Each Turning Point

Resultant acceleration \(a = \sqrt{a_t^2 + a_r^2}\). Initially, \(a = \sqrt{0.180^2 + 0^2} = 0.180\, \mathrm{m/s^2}\). After 60°, \(a = \sqrt{0.180^2 + 0.377^2} \approx 0.418\, \mathrm{m/s^2}\). After 120°, \(a = \sqrt{0.180^2 + 0.753^2} \approx 0.775\, \mathrm{m/s^2}\).

Key Concepts

Tangential Acceleration

Tangential acceleration is a measure of how quickly a point on the edge of a rotating object, like a flywheel, speeds up as it moves along a circular path. It's directly tied to the angular acceleration, which tells us how fast the rotation itself is speeding up or slowing down. In the given exercise, the flywheel has an angular acceleration of \(0.600 \, \text{rad/s}^2\). Because the flywheel is rotating, each point on its rim has a tangential acceleration that can be computed using the formula:

• \(a_t = \alpha r\)

Here, \(\alpha\) is the angular acceleration, and \(r\) is the radius of the flywheel, which is \(0.300 \, \text{m}\). Hence, substituting the given values, the tangential acceleration \(a_t\) is calculated as \(0.180 \, \text{m/s}^2\). This constant acceleration means the speed at which a point on the rim moves along its path is increasing at a steady rate.

Radial Acceleration

Radial acceleration, also known commonly as centripetal acceleration, ensures that a point on a rotating body continues its circular path. Unlike tangential acceleration, which changes the speed of the point, radial acceleration changes the direction. In other words, it keeps the point going in a circle rather than moving off in a straight line. Radial acceleration depends on the angular velocity (\(\omega\)) of the wheel and the radius of the circle. It's given by:

• \(a_r = \omega^2 r\)

At the very start of the motion, the angular velocity is zero because the flywheel starts from rest, making the radial acceleration also zero. As the wheel turns, such as after it has rotated through \(60^\circ\) or \(120^\circ\), the angular velocity increases, and so does the radial acceleration. This is because more speed requires more force to keep the point moving in a circle, due to the sharp change in direction at higher speeds.

Resultant Acceleration

Resultant acceleration is the total acceleration a point feels as it experiences both tangential and radial accelerations. This is important because it represents the actual path and speed changes of the point on the rim. The resultant acceleration is computed using the Pythagoras theorem because tangential and radial accelerations act at right angles to each other:

• \(a = \sqrt{a_t^2 + a_r^2}\)

Initially, when the flywheel just starts its motion, the resultant acceleration equals the tangential acceleration since the radial component is zero (the wheel starts from rest). As the wheel rotates, for instance, after \(60^\circ\) or \(120^\circ\), the radial component increases due to the rising angular velocity, making the resultant acceleration more significant. Understanding this helps in comprehending the influence of both acceleration types on an object's circular motion.