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Chapter 8: Problem 28

The blades of a windmill start from rest and rotate with an angular acceleration of \(0.236 \mathrm{rad} / \mathrm{s}^{2} .\) How much time elapses before a point on a blade experiences the same value for the magnitudes of the centripetal acceleration and tangential acceleration?

Short Answer

Expert verified
The time elapses before a point on a windmill blade experiences the same magnitude for the centripetal acceleration and tangential acceleration is 1 second.

Step by step solution

01

Understand and Express Tangential Acceleration

Tangential acceleration is defined as the rate of change of tangential speed. Here, the tangential acceleration \(a_t\) is given as \(0.236 \mathrm{rad} / \mathrm{s}^{2}\). This value will also be equivalent to the magnitude of centripetal acceleration \(a_c\) when both accelerations become equal.
02

Understand and Express Centripetal Acceleration

Centripetal acceleration is a measure of how quickly the velocity of a point moving in a circular path is changing. It's given by the formula \(a_c = r \theta'^2\) where \(r\) is the radius (distance from the center to the point on the blade) and \(\theta'\) is the angular velocity. However, since the blade starts from rest, initial angular velocity is zero. As time passes, the angular velocity increases due to angular acceleration (alpha) which is constant and equal to the tangential acceleration. Hence after some time 't', \(\theta' = \alpha \cdot t\). Substituting this in the formula for \(a_c\), we get \(a_c = r \cdot (\alpha \cdot t)^2\)
03

Set Up Equation and Solve

Since at the instant when the point on the blade experiences the same magnitude for the centripetal acceleration and tangential acceleration, we can set \(a_t = a_c\). Hence, \(0.236 = r \cdot (\alpha \cdot t)^2\). We see that radius 'r' will cancel out when we divide both sides by 'r' (as 'r' is a nonzero term since the point lies on the blade not at the center). We get \(\frac{0.236}{r} = \alpha \cdot t^2\). Since \(\alpha = \frac{0.236}{r}\), we have \(\frac{0.236}{r} = \frac{0.236}{r} \cdot t^2\). Solving for 't', we get \(t = \sqrt{1}\) Therefore, \(t = 1 \mathrm{s}\)

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