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question 10
problem reference 8 - 1
a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 12 rad/s over the next 4.0 seconds. assume that the angular acceleration is constant.
through how many revolutions does the grindstone turn during the 4.0 - second interval?
63 rev
1.3 rev
6.4 rev
40 rev

Explanation:

Step1: Recall the rotational kinematic equation for angular displacement

The formula for angular displacement \(\theta\) when angular acceleration \(\alpha\) is constant is \(\theta=\omega_0t + \frac{1}{2}\alpha t^2\), or we can also use the average angular velocity formula \(\theta=\frac{\omega_0+\omega}{2}t\), where \(\omega_0\) is the initial angular velocity, \(\omega\) is the final angular velocity, and \(t\) is the time.

Given \(\omega_0 = 8.0\space rad/s\), \(\omega=12\space rad/s\), \(t = 4.0\space s\).

Using the average angular velocity formula: \(\theta=\frac{\omega_0+\omega}{2}\times t\)

Substitute the values: \(\theta=\frac{8.0 + 12}{2}\times4.0=\frac{20}{2}\times4.0 = 10\times4.0=40\space rad\)

Step2: Convert angular displacement from radians to revolutions

We know that \(1\) revolution \(= 2\pi\) radians. So the number of revolutions \(N=\frac{\theta}{2\pi}\)

Substitute \(\theta = 40\space rad\): \(N=\frac{40}{2\pi}=\frac{20}{\pi}\approx6.4\space rev\)

Answer:

6.4 rev