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A disc rotates about the fixed axis. Its angular velocity $\omega$ varies with time according to the equation $\omega = at + b$. Initially, at t=0, its angular velocity is 1.0 rad/sec, and the angular position is 2 radians; at the instant t=2seconds, angular velocity is 5.0 rad/second. Determine angular position $\theta$ and angular acceleration $\alpha$ when t=4 seconds.

Last updated date: 20th Jun 2024
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Hint: In this question, we need to determine the angular position and the angular acceleration of the rotating disc at 4 seconds. For this, we will follow the relation between the angular position, angular velocity, and the angular acceleration of the rotating body by differentiation and integration process.

The angular velocity of the disc rotating about the fixed axis has been given as $\omega = at + b$ where, $\omega$ is the angular velocity, ‘t’ is the time and ‘a’ and ‘b’ are the constants.
According to the question, at t=0, the angular velocity of the rotating disc is 1 radian per second. So, substitute $t = 0{\text{ and }}\omega = 1{\text{ rad/sec}}$ in the equation $\omega = at + b$ to establish a relation between the constants ‘a’ and ‘b’.
$\Rightarrow \omega = at + b \\ \Rightarrow 1 = a(0) + b \\ \therefore b = 1 - - - - (i) \\$
Again, at t=2 seconds, the angular velocity of the rotating disc is 1 radian per second. So, substitute $t = 2{\text{ sec and }}\omega = 5{\text{ rad/sec}}$ in the equation $\omega = at + b$ to establish a relation between the constants ‘a’ and ‘b’.
$\Rightarrow \omega = at + b \\ \Rightarrow 5 = a(2) + 1 \\ \Rightarrow 2a = 5 - 1 \\ \Rightarrow a = \dfrac{4}{2} \\ \therefore a = 2 - - - - (ii) \\$
Substitute the values for the equations (i) and (ii) in the equation $\omega = at + b$ we get
$\Rightarrow\omega = at + b \\ \Rightarrow\omega = 2t + 1 - - - - (iii) \\$
The rate of change in the angular velocity of the rotating body results in the angular acceleration of the rotating body, Mathematically, $\dfrac{{d\omega }}{{dt}} = \alpha$ where, $\omega$ is the angular velocity and $\alpha$ is the angular acceleration.
So, differentiate the equation $\omega = 2t + 1$ with respect to time to determine the angular acceleration of the rotating disc.
$\Rightarrow\omega = 2t + 1 \\ \Rightarrow\dfrac{{d\omega }}{{dt}} = \dfrac{d}{{dt}}\left( {2t + 1} \right) \\ \Rightarrow\alpha = 2{\text{ rad/se}}{{\text{c}}^2} \\$
Also, the integration of the angular velocity of the rotating disc results in the angular position of the rotating disc. Mathematically, $\int\limits_{{\theta _0}}^\theta {\omega dt} = \theta$.
So, integrate the equation $\omega = 2t + 1$ to determine the expression for the angular position of the rotating disc.
$\Rightarrow\omega = 2t + 1 \\ \Rightarrow\int\limits_{{\theta _0}}^\theta {\omega dt} = \int\limits_0^t {\left( {2t + 1} \right)} dt \\ \Rightarrow \left[ \theta \right]_{{\theta _0}}^\theta = \left[ {{t^2} + t} \right]_0^t \\ \Rightarrow\theta - {\theta _0} = {t^2} + t - - - - (iv) \\$
It is given in the question that the initial angular position of the rotating disc is 2 radians, and we need to determine the angular position at t= 4 seconds. So, substitute ${\theta _0} = 2{\text{ rad and }}t = 4{\text{ sec}}$ in the equation (iv), we get
$\Rightarrow\theta - {\theta _0} = {t^2} + t \\ \Rightarrow\theta - 2 = {(4)^2} + 4 \\ \Rightarrow\theta = 16 + 4 + 2 \\ \Rightarrow\theta= 22{\text{ rad}} \\$

Hence, the angular position and the angular acceleration of the rotating disc at 4 seconds are 22 radians and 2 radians per square seconds, respectively.

Note:It is interesting to note here that, the angular acceleration of the rotating disc is independent of the variable ‘t’, and so, we can say that the disc is rotating with the constant angular acceleration and is not changing with respect to time.