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Chapter 5: Problem 52

A disk rotates with constant angular acceleration. The initial angular speed of the disk is \(2 \pi\) rad/s. After the disk rotates through \(10 \pi\) radians, the angular speed is \(7 \pi \mathrm{rad} / \mathrm{s} .\) (a) What is the magnitude of the angular acceleration? (b) How much time did it take for the disk to rotate through \(10 \pi\) radians? (c) What is the tangential acceleration of a point located at a distance of \(5.0 \mathrm{cm}\) from the center of the disk?

Short Answer

Expert verified
a) The magnitude of the angular acceleration is \(\frac{9}{4}π \,\text{rad/s}^2\). b) The time it took for the disk to rotate through \(10π\) radians is \(\frac{20}{9}\,\text{s}\). c) The tangential acceleration of a point located 5.0 cm from the center of the disk is \(0.1125π \,\text{m/s}^2\).

Step by step solution

01

Write down the given information

Write down the initial angular speed (\(ω_0\)), the final angular speed (\(ω\)), and the change in angle (\(θ-θ_0\)): \(ω_0 = 2π \,\text{rad/s}\) \(ω = 7π \,\text{rad/s}\) \(θ - θ_0 = 10π \,\text{rad}\)
02

Determine the angular acceleration (a)

Use equation 3 to solve for the angular acceleration (\(α\)): \(ω^2 = ω_0^2 + 2α(θ - θ_0)\) Solve for \(α\): \(α = \frac{ω^2 - ω_0^2}{2(θ - θ_0)}\) Plug in the given values: \(α = \frac{(7π)^2 - (2π)^2}{2(10π)}\) \(α = \frac{45π^2}{20π}\) \(α = \frac{9}{4}π \,\text{rad/s}^2\) The magnitude of the angular acceleration is \(\frac{9}{4}π \,\text{rad/s}^2\).
03

Determine the time it took (b)

Use equation 2 to solve for the time it took for the disk to rotate through \(10π\) radians (\(t\)): \(ω = ω_0 + αt\) Solve for \(t\): \(t = \frac{ω - ω_0}{α}\) Plug in the given values: \(t = \frac{(7π) - (2π)}{\frac{9}{4}π}\) \(t = \frac{5π}{\frac{9}{4}π}\) \(t = \frac{20}{9}\,\text{s}\) The time it took for the disk to rotate through \(10π\) radians is \(\frac{20}{9}\,\text{s}\).
04

Determine the tangential acceleration (c)

The tangential acceleration (\(a_t\)) of a point located at a distance \(r\) from the center of the disk is related to the angular acceleration by the following equation: \(a_t = αr\) We are given that the distance \(r\) is 5.0 cm, which we must convert to meters: \(r = 5.0 \,\text{cm} = 0.050 \,\text{m}\) Plug in the values for \(α\) and \(r\): \(a_t = \frac{9}{4}π \,\text{rad/s}^2 \cdot 0.050 \,\text{m}\) \(a_t = \frac{9}{4} \cdot(0.050) \cdot π \,\text{m/s}^2\) \(a_t = 0.1125π \,\text{m/s}^2\) The tangential acceleration of a point located 5.0 cm from the center of the disk is \(0.1125π \,\text{m/s}^2\).

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