**Problem Statement:**A turntable starts from rest and reaches 11 rad/s of angular velocity in 9.5 s. Find the angular acceleration in rad/s².**Solution:**To find the angular acceleration, we use the formula:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where:- \(\alpha\) is the angular acceleration,- \(\Delta \omega\) is the change in angular velocity,- \(\Delta t\) is the time interval.In this problem:- The initial angular velocity \(\omega_0 = 0 \, \text{rad/s}\) (since the turntable starts from rest),- The final angular velocity \(\omega = 11 \, \text{rad/s}\),- The time interval \(\Delta t = 9.5 \, \text{s}\).The change in angular velocity \(\Delta \omega\) is:\[\Delta \omega = \omega - \omega_0 = 11 \, \text{rad/s} - 0 \, \text{rad/s} = 11 \, \text{rad/s}\]Substitute the values into the angular acceleration formula:\[\alpha = \frac{11 \, \text{rad/s}}{9.5 \, \text{s}} \approx 1.16 \, \text{rad/s}^2\]Therefore, the angular acceleration is approximately \(1.16 \, \text{rad/s}^2\).

Given Infomation:- Initial angular velocity of the turn table is ωi=0. Final angular velocity of the turn table is ωf=11 rad/s. Time t=9.5 s. Required:- We have to find the angular acceleration of the turntable. Formula:- We will use the kinematic equation ωf=ωi+αt to find the angular acceleration of the turn table. Where α is the angular acceleration, ωf and ωi are initial and final angular velocity respectively.Calculation:- Substituting the values of ωf, ωi, t in the above kinematic equation we get, 0+9.5α=11 ⇒α=119.5 rad/s2 ⇒α=1.2 rad/s2 Answer:- Angular acceleration of the turn table is1.2 rad/s2 .

Answered: A turn table starts from rest and…

Science

Physics

A turn table starts from rest and reaches 11 rad/s of angular velocity in 9.5 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

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

**Problem Statement:**

A turntable starts from rest and reaches 11 rad/s of angular velocity in 9.5 s. Find the angular acceleration in rad/s².

**Solution:**

To find the angular acceleration, we use the formula:

\[
\alpha = \frac{\Delta \omega}{\Delta t}
\]

where:
- \(\alpha\) is the angular acceleration,
- \(\Delta \omega\) is the change in angular velocity,
- \(\Delta t\) is the time interval.

In this problem:
- The initial angular velocity \(\omega_0 = 0 \, \text{rad/s}\) (since the turntable starts from rest),
- The final angular velocity \(\omega = 11 \, \text{rad/s}\),
- The time interval \(\Delta t = 9.5 \, \text{s}\).

The change in angular velocity \(\Delta \omega\) is:

\[
\Delta \omega = \omega - \omega_0 = 11 \, \text{rad/s} - 0 \, \text{rad/s} = 11 \, \text{rad/s}
\]

Substitute the values into the angular acceleration formula:

\[
\alpha = \frac{11 \, \text{rad/s}}{9.5 \, \text{s}} \approx 1.16 \, \text{rad/s}^2
\]

Therefore, the angular acceleration is approximately \(1.16 \, \text{rad/s}^2\).

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