Question

A particle moves in a circle of radius of$0.5m$ at a speed that uniformly increases. Find the angular acceleration of a particle if its speed changes from$2.0m/s$ to$4.0m/s$ in$4.0s$ .

Answer 1

To solve this question, we must be aware that change in normal speed is only given. So, firstly we must find the tangential acceleration from the given parameters using the formula for tangential acceleration${{a}_{t}}$ . After finding tangential acceleration, we can find angular acceleration by using the formula for angular acceleration and already found tangential acceleration.

Formula used: ${{a}_{t}}=\dfrac{\delta v}{\delta t}$ -- (equation for tangential acceleration)
$\alpha =\dfrac{{{a}_{t}}}{r}$ -- (equation for angular acceleration)

Complete step-by-step answer:
In this question, we need to find the angular acceleration of a given particle. First of all, let us understand what is meant by angular acceleration.
Angular acceleration is the rate of change of angular velocity with respect to time. i.e. $\alpha =\dfrac{\delta \omega }{\delta t}$ . Where $\delta \omega$ is a change in angular momentum. $\delta \omega$ is given by $\delta \omega =\dfrac{\delta v}{r}$ where $r$ is the radius of the circular path which the particle rotates.
From this we can understand that $\alpha =\dfrac{\delta v}{\delta t\times r}$
Where $\dfrac{\delta v}{\delta t}$ is known as tangential acceleration. That means acceleration towards the tangent drawn to appoint the circular path.
In the question the change in velocity is given as well as the change in time as,
$\delta v=4.0-2.0=2m/s$
And $\delta t=4.0\sec$
$\Rightarrow {{a}_{t}}=\dfrac{\delta v}{\delta t}=\dfrac{2m/s}{4.0\sec }=0.5m/{{s}^{2}}$
Therefore, ${{a}_{t}}=0.5m/{{s}^{2}}$ . Now to find angular acceleration, $\alpha =\dfrac{{{a}_{t}}}{r}$
Where, $r$ is given as $0.5m$
$\Rightarrow \alpha =\dfrac{{{a}_{t}}}{r}=\dfrac{0.5m/{{s}^{2}}}{0.5m}=1rad/{{\sec }^{2}}$
Where $rad/{{\sec }^{2}}$ is the S.I. unit of angular acceleration.
So we have found the angular acceleration of the given particle as$\alpha =1rad/{{\sec }^{2}}$ .

Note: As we discussed during the solving of this question, we arrived at the point where we derived angular acceleration$\alpha =\dfrac{\delta v}{\delta t\times r}$ . So, above we mentioned tangential acceleration. But, we can directly substitute the given values here itself and can solve this problem quicker.