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The radius of the wheel is r=3 m. The tangential acceleration of the point on the edge of the wheel, as whown, is a=2t (m/s2). At t=0, the initial angular position θ0=0 and initial velocity V0=0. After the time of t=6s, determine (2) the of magnitude of the angular velocity, ω=___________ rad/s

The radius of the wheel is r=3 m. The tangential acceleration of the point on the edge of the wheel, as whown, is a=2t (m/s2). At t=0, the initial angular position θ0=0 and initial velocity V0=0. After the time of t=6s, determine (2) the of magnitude of the angular velocity, ω=___________ rad/s

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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The radius of the wheel is r=3 m. The tangential acceleration of the point on the edge of the wheel, as whown, is a=2t (m/s2). At t=0, the initial angular position θ0=0 and initial velocity V0=0. After the time of t=6s, determine

(2) the of magnitude of the angular velocity, ω=___________ rad/s



### Problem Statement The radius of the wheel is \( r = 3 \) m. The tangential acceleration of the point on the edge of the wheel, as shown, is \( a = 2t \) (m/s^2). At \( t = 0 \), the initial angular position \( \theta_0 = 0 \) and initial velocity \( V_0 = 0 \). After the time of \( t = 6 \) s, determine: 1. The magnitude of the angular velocity, \( \omega = \) __________ rad/s. ### Diagram Explanation The diagram is a circular representation of the wheel indicating its radius \( r \). There is a vector shown at the edge of the wheel representing the tangential acceleration \( a_T \). The wheel is depicted as a circle with the radius marked, indicating the tangential acceleration acting at the edge. ### Solution Approach To find the magnitude of the angular velocity \( \omega \) after \( t = 6 \) seconds, we need to integrate the given tangential acceleration to find the angular velocity. Given: - Radius \( r = 3 \) m - Tangential acceleration \( a_T = 2t \) (m/s^2) - Initial angular position \( \theta_0 = 0 \) - Initial velocity \( V_0 = 0 \) We can use the relationship between tangential acceleration \( a_T \) and angular acceleration \( \alpha \): \[ a_T = r \alpha \] Since \( a_T = 2t \): \[ 2t = 3 \alpha \] \[ \alpha = \frac{2t}{3} \] Now, integrate the angular acceleration \( \alpha \) to get the angular velocity \( \omega \): \[ \omega = \int \alpha \, dt = \int \frac{2t}{3} \, dt \] \[ \omega = \frac{2}{3} \int t \, dt \] \[ \omega = \frac{2}{3} \left( \frac{t^2}{2} \right) + C \] \[ \omega = \frac{t^2}{3} + C \] Using the initial condition \( \omega (0) = 0 \): \[ 0 = \frac{0^2}{3} + C \] \[ C = 0 \
Transcribed Image Text:### Problem Statement The radius of the wheel is \( r = 3 \) m. The tangential acceleration of the point on the edge of the wheel, as shown, is \( a = 2t \) (m/s^2). At \( t = 0 \), the initial angular position \( \theta_0 = 0 \) and initial velocity \( V_0 = 0 \). After the time of \( t = 6 \) s, determine: 1. The magnitude of the angular velocity, \( \omega = \) __________ rad/s. ### Diagram Explanation The diagram is a circular representation of the wheel indicating its radius \( r \). There is a vector shown at the edge of the wheel representing the tangential acceleration \( a_T \). The wheel is depicted as a circle with the radius marked, indicating the tangential acceleration acting at the edge. ### Solution Approach To find the magnitude of the angular velocity \( \omega \) after \( t = 6 \) seconds, we need to integrate the given tangential acceleration to find the angular velocity. Given: - Radius \( r = 3 \) m - Tangential acceleration \( a_T = 2t \) (m/s^2) - Initial angular position \( \theta_0 = 0 \) - Initial velocity \( V_0 = 0 \) We can use the relationship between tangential acceleration \( a_T \) and angular acceleration \( \alpha \): \[ a_T = r \alpha \] Since \( a_T = 2t \): \[ 2t = 3 \alpha \] \[ \alpha = \frac{2t}{3} \] Now, integrate the angular acceleration \( \alpha \) to get the angular velocity \( \omega \): \[ \omega = \int \alpha \, dt = \int \frac{2t}{3} \, dt \] \[ \omega = \frac{2}{3} \int t \, dt \] \[ \omega = \frac{2}{3} \left( \frac{t^2}{2} \right) + C \] \[ \omega = \frac{t^2}{3} + C \] Using the initial condition \( \omega (0) = 0 \): \[ 0 = \frac{0^2}{3} + C \] \[ C = 0 \
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