- #1
steven10137
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Homework Statement
A particle starts from rest at t=0s.
It moves along a circular path of radius 18m and has an acceleration component along its path of 6.7m/s^2.
What is the magnitude of the acceleration when t=2s
Homework Equations
[tex]\begin{array}{l}
\left| a \right| = \sqrt {a_n ^2 + a_t ^2 } \\
a_n = \frac{{v^2 }}{p} \\
v = \frac{{2\pi r}}{T} \\
\end{array}[/tex]
The Attempt at a Solution
The particle is acting in circular motion, hence:
[tex]v = \frac{{2\pi r}}{T} = \frac{{2\pi \left( {18} \right)}}{2} = 18\pi \;ms^{ - 1} [/tex]
The normal component of acceleration is then given by:
[tex]a_n = \frac{{v^2 }}{p} = \frac{{\left( {18\pi } \right)^2 }}{{18}} = \frac{{18^2 .\pi ^2 }}{{18}} = 18\pi ^2 \;ms^{ - 2} [/tex]
Then I can find the magnitude of the total acceleration:
[tex]\left| a \right| = \sqrt {a_n ^2 + a_t ^2 } = \sqrt {\left[ {18\pi ^2 } \right]^2 + 6.7^2 } \approx 178ms^{ - 2} [/tex]
Look OK? My first response would be that the figure seems quite high...