Question

A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad, where t is in seconds. Determine the magnitude of the particle’s acceleration

Answer 1

Answer:

Explanation:

Given

radius of  path $r=2\ m$

Velocity of Particle $\theta =5t^2 rad$

where t=time in seconds

angular velocity of particle is given by

$\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}$

$\omega =2\times 5t=10\cdot t$

And angular acceleration is given by

$\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}$

$\alpha =10 rad/s^2$

tangential acceleration is $a_t=\alpha \times r$

$a_t=10\times 2=20\ m/s^2$

Centripetal acceleration $a_c=\omega ^2\times r$

$a_c=(10t)^2\times 2=200t^2$

net acceleration is sum of tangential and centripetal force at any time t is given by

$a_{net}=\sqrt{(a_c)^2+(a_t)^2}$

$a_{net}=\sqrt{(200t)^2+(20)^2}$

$a_{net}=20\sqrt{(10t)^2+1}\ m/s$