Question

A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, determine the magnitude of the particle's acceleration when t= 2 s.

Answer 1

Answer:

4.83m/$s^{2}$

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by $a_{radial}=w^{2}$r

Applying values we get  $a_{radial}=(2t)^{2}$X0.3m

Thus $a_{radial}=1.2t^{2}$

At time = 2seconds $a_{radial}= 4.8m/s^{2}$

The tangential acceleration is given by $a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}$

$a_{tangential}=\frac{d(2tr)}{dt}$

$a_{tangential}= 2r$

$a_{tangential}=0.6m/s^{2}$

Thus the resultant acceleration is given by

$a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}$

$a_{res} =\sqrt{4.8^{2}+0.6^{2} } =4.83m/s^{2}$