Question

At some instant, a particle traveling in a horizontal circular path of radius 5.90 m has a total acceleration with a magnitude of 18.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant and (1/8) revolution later.

Answer 1

Answer:

The final speed will be 8.02 rad/s.

Explanation:

Given that

radius ,r=5.9 m

Total acceleration = 18 $m/s^2$

Tangential acceleration = 12 $m/s^2$

Here particle is moving in circular path with varying linear speed.So there will be two acceleration ,one is radial acceleration and other one is tangential acceleration.

$tangential\ acceleration=\alpha r$

12=α x 5.9

$\alpha =2.03\ m/s^2$

$Radial\ acceleration\, a_r={\omega ^2r}$

$Total\ acceleration=\sqrt{\left({\omega ^2r}\right)^2+\left(\alpha r\right)^2}$

$18^2=12^2+\omega ^4\times 5.9^2$

ω=2.2 rad/s

θ=2πn

θ=2 x π x 1/8 rad

θ=π /4 rad

$\omega^2_f=\omega^2_i+2\alpha \theta$

$\omega^2_f=2.2^2+2\times 2.03\times \dfrac{\pi}{4}$

$\omega^2_f=8.02 rad/s$

So the final speed will be 8.02 rad/s.