Question

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question 9
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.
what is the magnitude of the angular acceleration of the grindstone?
1.0 rad/s²
3.0 rad/s²
2.0 rad/s²
5.0 rad/s²

Explanation:

Step1: Recall angular acceleration formula

The formula for angular acceleration \(\alpha\) when angular speed changes from \(\omega_0\) to \(\omega\) in time \(t\) is \(\alpha=\frac{\omega - \omega_0}{t}\).

Step2: Identify given values

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

Step3: Substitute values into formula

Substitute the values into the formula: \(\alpha=\frac{12 - 8.0}{4.0}=\frac{4.0}{4.0}=1.0\space rad/s^2\)? Wait, no, wait, 12 - 8 is 4, 4 divided by 4 is 1? Wait, no, wait the options have 1,2,3,5. Wait, maybe I made a mistake. Wait, 12 - 8 is 4, time is 4 seconds. So \(\alpha=\frac{12 - 8}{4}=\frac{4}{4} = 1\)? But wait, no, wait the options: 1,3,2,5. Wait, maybe I misread the numbers. Wait, the initial angular speed is 8 rad/s, final is 12 rad/s, time is 4 seconds. So change in angular speed is \(12 - 8 = 4\) rad/s, time is 4 s. So \(\alpha=\frac{4}{4}=1\space rad/s^2\)? But wait, the option 1.0 rad/s² is there. Wait, but let me check again. Wait, maybe the numbers are different? Wait, no, the problem says initial angular speed 8.0 rad/s, final 12 rad/s, time 4.0 seconds. So \(\alpha=\frac{\Delta\omega}{\Delta t}=\frac{12 - 8}{4}=\frac{4}{4}=1\space rad/s^2\). So the correct answer should be 1.0 rad/s²? Wait, but let me check the options again. The options are 1.0, 3.0, 2.0, 5.0. So according to the calculation, it's 1.0.

Wait, maybe I made a mistake. Wait, 12 - 8 is 4, divided by 4 is 1. Yes. So the angular acceleration is 1.0 rad/s².

Answer:

1.0 rad/s² (Option: 1.0 rad/s²)