Question

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by theta = cos 2t, where theta is in radians and t is in seconds. Determine the magnitude of the accer

Answer 1

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1  to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$