Circular Motion #3

Classical Mechanics Level 2

A particle moves in a circular path of radius $1$ . Its speed $v$ , at time $t$ is given by $v = 5 + 4 \sin ( \frac{ \pi }{ 6 } t )$ . What is the magnitude of the radial acceleration of the particle at $t = 5$ ?

The given information is in SI units. State your answer in SI units too.

This problem is part of the set - Circular Motion Practice

The answer is 49.

1 solution

Pranshu Gaba
Nov 4, 2015

Recall that the magnitude of radial acceleration of a particle moving in a circular path is $v^{2 } / r$ and its direction is always towards the center of the circle. At $t = 5$ , the speed of the particle is

$v = 5 + 4 \sin \left(\frac{ \pi } { 6} \times 5 \right)= 5 + 4 \times \frac{ 1 }{ 2} = 7$

Thus, the magnitude of radial acceleration at $t = 5$ is $\dfrac{ 7^{2} }{ 1 } = \boxed { 49 }$ $_ \square$

Circular Motion #3

The answer is 49.

1 solution

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