6.7 Non – uniform circular motion(application)

Problem : A particle, tied to a string, starts moving along a horizontal circle of diameter 2 m, with zero angular velocity and a tangential acceleration given by " $4t$ ". If the string breaks off at t = 5 s, then find the speed of the particle with which it flies off the circular path.

Solution : Here, tangential acceleration of the particle at time "t" is given as :

$$a_T=4t$$

We note here that an expression of tangential acceleration (an higher attribute) is given and we are required to find lower order attribute i.e. linear speed. In order to find linear speed, we need to integrate the acceleration function :

$$a_T=\frac{đv}{đt}=4t$$

$$\Rightarrow đv=4tđt$$

Integrating with appropriate limits, we have :

$$\Rightarrow \int đv=\int 4tđt=4\int tđt$$

$$\Rightarrow v_f-v_i=4[\frac{t^2}{2}{]}_0^5=\frac{4X{5}^2}{2}=50m∕s$$

$$\Rightarrow v_f-0=50$$

$$\Rightarrow v_f=50m∕s$$

Problem : A particle is executing circular motion. The velocity of the particle changes from (0.1 i + 0.2 j ) m/s to (0.5 i + 0.5 j ) m/s in a period of 1 second. Find the magnitude of average total acceleration.

Solution : The average total acceleration is :

$$\begin{array}{l}\mathbf{a}=\frac{\Delta \mathbf{v}}{\Delta t}=\frac{(0.5\mathbf{i}+0.5\mathbf{j})-(0.1\mathbf{i}+0.2\mathbf{j})}{1}\end{array}$$

$$\begin{array}{l}\mathbf{a}=(0.4\mathbf{i}+0.3\mathbf{j})\end{array}$$

The magnitude of acceleration is :

$$\begin{array}{l}a=\sqrt{({0.4}^2+{0.3}^2)}=\sqrt{0.25}=0.5m/s\end{array}$$

Problem : A particle starting with a speed “v” completes half circle in time “t” such that its speed at the end is again “v”. Find the magnitude of average total acceleration.

Solution : Average total acceleration is equal to the ratio of change in velocity and time interval.

$$a_{\mathrm{avg}}=\frac{v_2-v_1}{\Delta t}$$

From the figure and as given in the question, it is clear that the velocity of the particle has same magnitude but opposite directions.

$$v_1=v$$

$$v_2=-v$$

Putting in the expression of average total acceleration, we have :

$$a_{\mathrm{avg}}=\frac{v_2-v_1}{\Delta t}=\frac{-v-v}{t}$$

$$a_{\mathrm{avg}}=\frac{-2v}{t}$$

The magnitude of the average acceleration is :

$$\Rightarrow a_{\mathrm{avg}}=\frac{2v}{t}$$

Problem : The angular position of a particle (in radian), on circular path of radius 0.5 m, is given by :

$$\begin{array}{l}\theta =-0.2t^2-0.04\end{array}$$

At t = 1 s, find (i) angular velocity (ii) linear speed (iii) angular acceleration (iv) magnitude of tangential acceleration (v) magnitude of centripetal acceleration and (vi) magnitude of total acceleration.

Solution : Angular velocity is :

$$\omega =\frac{đ\theta }{đt}=\frac{đ}{đt}\left(-0.2t^2-0.04\right)=-0.2X2t=-0.4t$$

The angular speed, therefore, is dependent as it is a function in "t". At t = 1 s,

$$\omega =-0.4\mathrm{rad}/s$$

The magnitude of linear velocity, at t = 1 s, is :

$$v=\omega r=0.4X0.5=0.2m/s$$

Angular acceleration is :

$$\alpha =\frac{đ\omega }{đt}=\frac{đ}{đt}\left(-0.4t\right)=-0.4\mathrm{rad}/s^2$$

Clearly, angular acceleration is constant and is independent of time.

The magnitude of tangential acceleration is :

$$\Rightarrow a_T=\alpha r=0.4X0.5=0.2m/s^2$$

Tangential acceleration is also constant and is independent of time.

The magnitude of centripetal acceleration is :

$$\Rightarrow a_R=\omega v=0.4tX0.2t=0.08t^2$$

At t = 1 s,

$$\Rightarrow a_R=0.08m/s^2$$

The magnitude of total acceleration, at t =1 s, is :

$$a=\sqrt{\left(a_T^2+a_R^2\right)}$$

$$a=\sqrt{\{{\left(0.2\right)}^2+{\left(0.08\right)}^2\}}=0.215m/s^2$$

Problem : The speed (m/s) of a particle, along a circle of radius 4 m, is a function in time, "t" as :

$$v=t^2$$

Find the total acceleration of the particle at time, t = 2 s.

Solution : The tangential acceleration of the particle is obtained by differentiating the speed function with respect to time,

$$a_T=\frac{đv}{đt}=\frac{đ}{đt}\left(t^2\right)=2t$$

The tangential acceleration at time, t = 2 s, therefore, is :

$$\Rightarrow a_T=2X2=4m/s^2$$

The radial acceleration of the particle is given as :

$$a_R=\frac{v^2}{r}$$

In order to evaluate this expression, we need to know the velocity at the given time, t = 2 s :

$$\Rightarrow v=t^2=2^2=4m/s$$

Putting in the expression of radial acceleration, we have :

$$\Rightarrow a_R=\frac{v^2}{r}=\frac{4^2}{4}=4m/s^2$$

The total acceleration of the particle is :

$$a=\sqrt{\left(a_T^2+a_R^2\right)}=\sqrt{\left(4^2+4^2\right)}=4\sqrt{2}m∕s^2$$