Question

A particle is moving on a circle of radius 1 km with constant speed 10π m/s. Find the change in velocity vector of particle in a time duration of 1 second (in m/s). (change in velocity = vf - vi)​

Answer 1

Given that,

Radius of circle = 1 km

Speed = 10π m/s

Time = 1 sec

The velocity is negative in opposite direction in same level .

The velocity at half revolution is

$v_{f}=-10\pi\ m/s$

We need to calculate the arc

arc = distance

Using formula of arc

$D=v\times t$

Put the value into the formula

$D=10\pi\times1$

$D=10\pi\ m$

We need to calculate the angle

Using formula of angle

$\theta=\dfrac{D}{R}$

Where, D = arc

R = radius

Put the value into the formula

$\theta=\dfrac{10\pi}{1000}$

$\theta=\dfrac{\pi}{100}\ rad$

We need to calculate the change in velocity

Using formula of change in velocity

$\Delta v=\sqrt{v_{i}^2+v_{f}^2-2v_{i}v_{f}\cos\theta}$

Put the value into the formula

$\Delta v=\sqrt{(10\pi)^2+(10\pi)^2+2\times10\pi\times10\pi\times\cos\theta}$

$\Delta v=10\pi\sqrt{2(1+\cos\theta)}$

$\Delta v=10\pi\sqrt{2\times2\sin^2\dfrac{\theta}{2}}$

Put the value of angle

$\Delta v=20\pi\sin(\dfrac{\pi}{200})\ m/s$

Hence, The change in velocity is $20\pi\sin(\dfrac{\pi}{200})\ m/s$