Question

A particle moving along a circular path of radius 5m with constant speed $10\mathrm{m}/\mathrm{s}$. Determine its angular velocity, time period, and acceleration. Also, determine the net force acting on the particle. The mass of the particle is $2\mathrm{kg}.$

Answer 1

Step 1: Given data:
The radius of the circle $\mathrm{R}=5\mathrm{m}$
Constant speed $\mathrm{v}=10\mathrm{m}/\mathrm{s}$
Mass of the particle $\mathrm{m}=2\mathrm{kg}$

Step 2: Angular velocity:

The relationship between linear velocity and the angular velocity of the particle is,
$\mathrm{v}=\mathrm{R\omega }$
$\omega =\frac{v}{R}$

Step 3: To find angular velocity:

The angular velocity is,
$\omega =\frac{10}{5}$
$\mathrm{\omega }=2\mathrm{rad}/\mathrm{s}$
Step 4: To find the time period of the oscillation:

The time period of the oscillation is,
$T=\frac{2\mathrm{\pi }}{\omega }$
$T=\frac{2\mathrm{\pi }}{2}$
$\mathrm{T}=\mathrm{\pi }$sec

Step 5: To find the angular acceleration of the particle:

The angular acceleration of the particle is,
$\alpha ={\omega }^{2}R$
$\alpha =(2{)}^2\times 5$
$\alpha =20rad/s^2$
Step 6: To find the net force on the particle:

The net force on the particle is,
$F=\frac{m{v}^2}{R}$
$F=\frac{2kg\times {10}^2}{5\mathrm{m}}$
$F=40\mathrm{N}$

Hence, the linear and angular velocity, time period and acceleration are $10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.,2\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.,\pi \text{second},\text{and}20\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$ . The net force acting on the particle is $40\mathrm{N}$.