Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider an object moving according to the position function \(\mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j}\) Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathbf{T}},\) and \(a_{\mathbf{N}}\)

Short Answer

Expert verified
\(\mathbf{T}(t) = \mathbf{v}(t)/||\mathbf{v}(t)||\), \(\mathbf{N}(t) = \mathbf{T'}(t)/||\mathbf{T'}(t)||\), \(a_{\mathbf{T}} = \mathbf{a}.\mathbf{T}\), and \(a_{\mathbf{N}} = \mathbf{a}.\mathbf{N}\)

Step by step solution

01

Find Velocity

Differentiate the given position vector function \(\mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j}\) with respect to time \(t\) to find the velocity \(\mathbf{v}(t)\).
02

Compute Tangential Vector

Now find the tangential vector \(\mathbf{T}(t)\) by normalizing the velocity vector, i.e., \(\mathbf{T}(t) = \mathbf{v}(t)/||\mathbf{v}(t)||\).
03

Find Acceleration

Next, compute the acceleration \(\mathbf{a}(t)\) by differentiating the velocity vector \(\mathbf{v}(t)\) with respect to time \(t\).
04

Compute Normal Vector

Find the normal vector \(\mathbf{N}(t)\) that is orthogonal to the tangent vector \(\mathbf{T}(t)\). This is done by taking the cross product of \(\mathbf{T}(t)\) and its derivative, and normalizing the result. i.e., \(\mathbf{N}(t) = \mathbf{T'}(t)/||\mathbf{T'}(t)||\).
05

Calculate Tangential and Normal Accelerations

Finally, calculate the tangential acceleration \(a_{\mathbf{T}}\) and normal acceleration \(a_{\mathbf{N}}\) by taking the dot product of the acceleration vector \(\mathbf{a}(t)\) with the unit tangent vector \(\mathbf{T}(t)\) and the unit normal vector \(\mathbf{N}(t)\) respectively. i.e., \(a_{\mathbf{T}} = \mathbf{a}.\mathbf{T}\) and \(a_{\mathbf{N}} = \mathbf{a}.\mathbf{N}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free