Question

A particle starts moving on a circular direction with angular speed 5 rads−1. The radius of the circle is 4 m and the angular speed at time t is given by ω =15−3t.

What is the linear speed of the particle 6 seconds after moving, and the resultant acceleration of the particle?

Answer 1

Ans:-

The relationship between angular velocity ω and linear velocity v was also defined in Rotation Angle and Angular Velocity as

$v=r\times \omega,$ $v=r\times \omega$

We have $\omega =15-3t$ and $r=4$

So, $v=4 × (15-3t)$

linear speed of the particle 6 seconds after moving is

$t=6 \Rightarrow v=4\times (15-3\times6)=-12 \ \ m/s$ $-ve$ sign means direction of speed is in opposite direction of motion.

$\alpha=\dfrac{d\omega}{dt}=15$

So, tangential acceleration

$a_t=\alpha\times r=15\times4=60$

Centripetal acceleration will be

$a_c=\omega^2\times r=5^2\times4=100$

Then resultant acceleration will be

$a=\sqrt{60^2+100^2}=116.61 m/s^2$

Hence the resultant acceleration of the particle is $116.61 \ \ m/s^2$