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Chapter 8: Problem 29

A rigid body, starting at rest, rotates about a fixed axis with constant angular acceleration \(\alpha\). Consider a particle a distance \(r\) from the axis. Express \((a)\) the radial acceleration and \((b)\) the tangential acceleration of this particle in terms of \(\alpha, r\), and the time \(t .(c)\) If the resultant acceleration of the particle at some instant makes an angle of \(57.0^{\circ}\) with the tangential acceleration, through what total angle has the body rotated from \(t=0\) to that instant?

Short Answer

Expert verified
The total angle through which the body has rotated from \(t = 0\) to the given instant will be obtained from calculation in Step 4 using the time obtained from Step 3.

Step by step solution

01

Finding Radial Acceleration

The radial acceleration of a particle in rotational motion, sometimes called centripetal acceleration, can be expressed in terms of angular velocity (\(w\)) and the radial distance from the rotation axis (\(r\)). However, the problem specifies that the body starts from rest and then accelerates with a constant angular acceleration (\(α\)). Thus, the radial acceleration (\(a_r\)) can be found using the formula: \(a_{r} = \omega^{2}r\), and knowing that \(ω = ατ\), we have \(a_{r} = α^{2}t^{2}r\).
02

Finding Tangential Acceleration

The tangential acceleration (\(a_t\)) of a particle in rotational motion is given by the change in velocity with time. Because the particle starts from rest, the tangential acceleration can be expressed using the equation: \(a_{t}=αr\).
03

Calculating Resultant Acceleration

The resultant acceleration (\(a\)) is the vector sum of the radial (\(a_r\)) and tangential (\(a_t\)) accelerations. This resultant acceleration makes an angle of \(57.0^{\circ}\) with the tangential acceleration. Therefore, we can express the resultant acceleration using trigonometry as: \(a = a_{t}/cos(57.0^{\circ})\), which can be simplified to \(a = αr/sec(57.0^{\circ})\). Remember that this resultant acceleration should also be valid when squared and added to radial acceleration according to pythagorean theorem, i.e., \(a^{2}=a_{r}^{2}+a_{t}^{2}\). We may equate this equation to our previous expression and solve for \(t\).
04

Calculating Rotational Angle

We know that angular displacement (\(θ\)) is given by \(θ = ωt + 0.5αt^{2}\). As the rigid body starts from rest, ω equals 0 and the equation simplifies to \(θ = 0.5αt^{2}\). Plug in the value of \(t\) obtained from the previous step to get the total angle the body has rotated from \(t = 0\) to that instant.

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