Question

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle${{\theta }_{1}}$; in the next 2 sec, it rotates through an additional angle${{\theta }_{2}}$. The ratio of ${{\theta }_{2}}$and${{\theta }_{1}}$ is-

Answer 1

The equation that represents the relation between the angular velocity, the angular acceleration, the time and the angle should be used to solve this problem. The question represents the two situations, one being the initial condition with no angular velocity and the other being the final condition with time 2 seconds. Using these equations, we will solve this problem.

Formula used:
$\theta =\omega t+\dfrac{1}{2}\alpha {{t}^{2}}$

Complete step-by-step solution: -
From the given information, we have the data as follows.
A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle${{\theta }_{1}}$; in the next 2 sec, it rotates through an additional angle${{\theta }_{2}}$.
As we are asked to find the ratio of the angles, so, using the equation that relates the parameters such as the angular velocity, the angular acceleration, the time and the angle should be used, as the values of the other parameters are given.
The formula that relates the relation between the angular velocity, the angular acceleration, the time and the angle is given as follows.
$\theta =\omega t+\dfrac{1}{2}\alpha {{t}^{2}}$
Where $\theta$ is the angle, $\omega$ is the angular velocity, $t$ is the time and $\alpha$ is the angular acceleration.
Consider the first situation.

& {{\theta }_{1}}={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}} \\\ & \Rightarrow {{\theta }_{1}}=(0)t+\dfrac{1}{2}\alpha {{(2)}^{2}} \\\ & \therefore {{\theta }_{1}}=2\alpha \\\ \end{aligned}$$ Let the angular velocity changes from $${{\omega }_{0}}$$to $$\omega '$$ Then the final angular velocity is given as follows. $$\begin{aligned} & \omega '={{\omega }_{0}}+\alpha t \\\ & \Rightarrow \omega '=0+\alpha (2) \\\ & \therefore \omega '=2\alpha \\\ \end{aligned}$$ Consider the second situation. $$\begin{aligned} & {{\theta }_{2}}=\omega 't+\dfrac{1}{2}\alpha {{t}^{2}} \\\ & \Rightarrow {{\theta }_{2}}=(2\alpha )2+\dfrac{1}{2}\alpha {{(2)}^{2}} \\\ & \therefore {{\theta }_{2}}=6\alpha \\\ \end{aligned}$$ Now we will compute the ratio of the angles. $$\begin{aligned} & \dfrac{{{\theta }_{2}}}{{{\theta }_{1}}}=\dfrac{6\alpha }{2\alpha } \\\ & \Rightarrow \dfrac{{{\theta }_{2}}}{{{\theta }_{1}}}=\dfrac{6}{2} \\\ & \therefore \dfrac{{{\theta }_{2}}}{{{\theta }_{1}}}=3 \\\ \end{aligned}$$ $$\therefore $$ The ratio of the angles made by the rotating wheel is, 3. **Note:** The units of the parameters should be taken care of. In this case, the ratio of angles is asked, so the remaining parameters get cancelled out. The relation between the angular velocity, the angular acceleration, the time and the angle should be used to solve this problem.