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Chapter 8: Problem 22

A point on the rim of a \(0.75\) -m-diameter grinding wheel changes speed uniformly from \(12 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.2 \mathrm{~s}\). What is the angular acceleration of the grinding wheel during this interval?

Short Answer

Expert verified
The angular acceleration of the grinding wheel during this interval is \(5.6rad/s^2\).

Step by step solution

01

Calculate the Initial Angular Velocity

First convert the initial linear velocity (\(12 ms^{-1}\)) to angular terms using the equation \(\omega = \frac{v}{r}\), where \(v\) represents linear velocity and \(r\) represents radius. Remember the radius is half the diameter of the grinding wheel, thus \(r = 0.75/2 = 0.375m\). Therefore, initial angular velocity \(\omega_i = \frac{12}{0.375} = 32rad/s\).
02

Calculate the Final Angular Velocity

Similarly, convert the final linear velocity (\(25 ms^{-1}\)) to angular terms using the equation \(\omega = \frac{v}{r}\), where \(v\) represents linear velocity and \(r\) represents radius. So, final angular velocity \(\omega_f = \frac{25}{0.375} = 66.7rad/s\).
03

Calculate the Change in Angular Velocity

Change in angular velocity \(\Delta \omega\) is the final angular velocity minus the initial angular velocity, i.e. \(\Delta \omega = \omega_f - \omega_i = 66.7 - 32 = 34.7rad/s\).
04

Calculate Angular Acceleration

Once the initial and final angular velocities are calculated, angular acceleration can be obtained using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\). Subsituting the values \(\Delta \omega = 34.7rad/s\) and \(\Delta t = 6.2s\) in the formula, we obtain \(\alpha = \frac{34.7}{6.2} = 5.6rad/s^2\) which is the required solution of the problem.

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Most popular questions from this chapter

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?

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.33 \mathrm{~s}\), the angular velocity of the wheel is \(4.96 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=23.0 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=46.0 \mathrm{~s}\) ?

What are \((a)\) the angular speed, \((b)\) the radial acceleration, and \((c)\) the tangential acceleration of a spaceship negotiating a circular turn of radius \(3220 \mathrm{~km}\) at a constant speed of \(28,700 \mathrm{~km} / \mathrm{h} ?\)

A certain wheel turns through 90 rev in \(15 \mathrm{~s}\), its angular speed at the end of the period being \(10 \mathrm{rev} / \mathrm{s} .(a)\) What was the angular speed of the wheel at the beginning of the 15 -s interval, assuming constant angular acceleration? \((b)\) How much time had elapsed between the time the wheel was at rest and the beginning of the 15 -s interval?

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