**Problem Statement:**A turntable starts from rest and reaches 10 rad/s of angular velocity in 5.0 s. Find the angular acceleration in rad/s² (rad/s/s).**Solution:**To find the angular acceleration, we can use the formula relating angular velocity (ω), initial angular velocity (ω₀), angular acceleration (α), and time (t):\[ \omega = \omega_{0} + \alpha t \]Given that the turntable starts from rest, the initial angular velocity \(\omega_{0}\) is 0 rad/s. The final angular velocity \(\omega\) is 10 rad/s, and the time \(t\) is 5.0 seconds.Substituting these values into the formula, we get:\[ 10 \ \text{rad/s} = 0 \ \text{rad/s} + \alpha \cdot 5.0 \ \text{s} \]Solving for the angular acceleration \(\alpha\):\[ 10 \ \text{rad/s} = \alpha \cdot 5.0 \ \text{s} \]\[ \alpha = \frac{10 \ \text{rad/s}}{5.0 \ \text{s}} \]\[ \alpha = 2 \ \text{rad/s}^2 \]**Answer:**The angular acceleration of the turntable is \( 2 \ \text{rad/s}^2 \).

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A turn table starts from rest and reaches 10 rad/s of angular velocity in 5.0 s. Find the angular acceleration in rad/s/s.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

**Problem Statement:**

A turntable starts from rest and reaches 10 rad/s of angular velocity in 5.0 s. Find the angular acceleration in rad/s² (rad/s/s).

**Solution:**

To find the angular acceleration, we can use the formula relating angular velocity (ω), initial angular velocity (ω₀), angular acceleration (α), and time (t):

\[ \omega = \omega_{0} + \alpha t \]

Given that the turntable starts from rest, the initial angular velocity \(\omega_{0}\) is 0 rad/s. The final angular velocity \(\omega\) is 10 rad/s, and the time \(t\) is 5.0 seconds.

Substituting these values into the formula, we get:

\[ 10 \ \text{rad/s} = 0 \ \text{rad/s} + \alpha \cdot 5.0 \ \text{s} \]

Solving for the angular acceleration \(\alpha\):

\[ 10 \ \text{rad/s} = \alpha \cdot 5.0 \ \text{s} \]
\[ \alpha = \frac{10 \ \text{rad/s}}{5.0 \ \text{s}} \]
\[ \alpha = 2 \ \text{rad/s}^2 \]

**Answer:**

The angular acceleration of the turntable is \( 2 \ \text{rad/s}^2 \).

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