Question

Find the tangential and normal components of the acceleration vector of a particle with position function. \({\rm{r(t) = ti + 2tj + }}{{\rm{t}}^{\rm{2}}}{\rm{k}}\). Exercise 56 in Section 10.

Short answer

Therefore, the value is, \(\begin{aligned}{*{20}{c}}{{a_T} = \frac{{4t}}{{\sqrt {5 + 4{t^2}} }}}\\{{a_N} = \frac{{2\sqrt 5 }}{{\sqrt {5 + 4{t^2}} }}}\end{aligned}\) .

Step-by-step solution

Formula of the tangential acceleration of the particle

The tangential acceleration of a particle moving in a circular motion is defined by at =2tm/s2.

Evaluating the value of tangential component

The tangential component is given by, \({a_T} = \frac{{r'(t) \cdot r''(t)}}{{\left| {r'(t)} \right|}}\)

\(\begin{aligned}{*{20}{c}}{r'(t) \times r''(t) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\1&2&{2t}\\0&0&2\end{aligned}} \right| = 4i - 2j}\\{\left| {r'(t) \times r''(t)} \right| = \sqrt {{4^2} + {{( - 2)}^2}} = 2\sqrt 5 }\\{{a_N} = \frac{{2\sqrt 5 }}{{\sqrt {5 + 4{t^2}} }}}\end{aligned}\)

The magnititude of the first derivative is, \(\left| {r'(t)} \right| = \sqrt {{1^2} + {2^2} + {{(2t)}^2}} = \sqrt {5 + 4{t^2}} \)

Therefore,

 Cocept of the normal acceleration of the particle

The normal component, also known as the circular path component, is always directed toward the curve's center of the sphere and therefore is denoted by. It tells us how much of the acceleration is orthogonal to the tangential acceleration.

Evluation the value of normal component

The normal component is given by,

\(\begin{aligned}{*{20}{c}}{{a_N} = \frac{{\left| {r'(t) \times r''(t)} \right|}}{{\left| {r'(t)} \right|}}}\\{r'(t) \times r''(t) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\1&2&{2t}\\0&0&2\end{aligned}} \right| = 4i - 2j}\\{\left| {r'(t) \times r''(t)} \right| = \sqrt {{4^2} + {{( - 2)}^2}} = 2\sqrt 5 }\\{{a_N} = \frac{{2\sqrt 5 }}{{\sqrt {5 + 4{t^2}} }}}\end{aligned}\)

Therefore, the value is, \(\begin{aligned}{*{20}{c}}{{a_T} = \frac{{4t}}{{\sqrt {5 + 4{t^2}} }}}\\{{a_N} = \frac{{2\sqrt 5 }}{{\sqrt {5 + 4{t^2}} }}}\end{aligned}\) .