
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.
Answer
420.4k+ views
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.
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
