
When a ceiling fan is switched on, it makes $10$ revolution in the first $3$$s$. Assuming a uniform angular acceleration, how many rotations it will make in the next $3$$s$?
A. $10$
B. $20$
C. $30$
D. $40$
Answer
555.3k+ views
Hint:Split the whole situation in two parts. First part will deal with the motion occurring in the first $3$ seconds and the second part will deal with the motion occurring in the next $3$ seconds. Use the kinematics equations for rotational motion and find the final unknowns in the first part and use them to find the final unknowns in the second part.
Complete step by step answer:
First part: When the ceiling fan is switched on, the initial angular velocity will be zero ${\omega _0} = 0$.The time taken by the ceiling fan to complete $10$ revolutions is $3\,s$. Let the final angular velocity in the first part be ${\omega _1}$, then we can use the formulae mentioned above to find this,
$
{\omega _1} = 0 + \alpha t \\
\Rightarrow\alpha = \dfrac{{{\omega _1}}}{t} \\
$
Substituting this value in equation $\left( 3 \right)$
$
\theta = {\omega _0}t + \dfrac{1}{2}\left( {\dfrac{{{\omega _1}}}{t}} \right){t^2} \\
\Rightarrow\theta = \left( {\dfrac{{2{\omega _0} + {\omega _1}}}{2}} \right)t \\
$
$
\Rightarrow\theta = \left( {\dfrac{{\left( 2 \right)\left( 0 \right) + \left( {{\omega _1}} \right)}}{2}} \right)\left( 3 \right) \\
\Rightarrow\theta = \dfrac{{3{\omega _1}}}{2} \\ $
$10$ revolutions means that the angle rotated will be
$10 \times 2\pi $\[rad\]$ = 20\pi $$rad$,
${\omega _1} = \dfrac{{2\theta }}{3}\\
\Rightarrow{\omega _1} = \dfrac{{\left( 2 \right)\left( {20\pi } \right)}}{3}\\
\Rightarrow{\omega _1} = \dfrac{{40\pi }}{3}$$\dfrac{{rad}}{s}$.
From here we can get the acceleration of the fan is,
$ \alpha = \dfrac{{\left( {\dfrac{{40\pi }}{3}} \right)}}{3}\\
\therefore\alpha = \dfrac{{40\pi }}{9}$$\dfrac{{rad}}{{{s^2}}}$
Second part:
Here, the initial velocity will be ${\omega _1} = \dfrac{{40\pi }}{3}$$\dfrac{{rad}}{s}$ and the time is $3$seconds. Using the equation $\left( 3 \right)$ we can get the angle rotated by the fan in the next $3$ seconds and then the number of revolutions can be obtained.
$
\theta = \left( {\dfrac{{40\pi }}{3}} \right)\left( 3 \right) + \dfrac{1}{2}\left( {\dfrac{{40\pi }}{9}} \right){\left( 3 \right)^2} \\
\Rightarrow\theta = 40\pi + 20\pi\\
\therefore\theta = 60\pi \\
$
The angle rotated by the fan in the next $3$$s$ is $60\pi $$rad$. The fan revolves $1$ time and rotates by an angle of $2\pi $$rad$, then for \[60\pi \]$rad$ it will rotate $\dfrac{{60\pi }}{{2\pi }} = 30$ times.
Thus, it will make $30$ rotations in the next $3\,s$.Hence,option C is correct.
Note: Always split the questions into parts according to the given data. Find the final angular velocity, find the final velocity in the current part to find the rest unknowns in the next part as the body will be carrying the final angular velocity in some part as initial angular velocity in the next part. Remember the kinematics equations for rotational motion.
Complete step by step answer:
First part: When the ceiling fan is switched on, the initial angular velocity will be zero ${\omega _0} = 0$.The time taken by the ceiling fan to complete $10$ revolutions is $3\,s$. Let the final angular velocity in the first part be ${\omega _1}$, then we can use the formulae mentioned above to find this,
$
{\omega _1} = 0 + \alpha t \\
\Rightarrow\alpha = \dfrac{{{\omega _1}}}{t} \\
$
Substituting this value in equation $\left( 3 \right)$
$
\theta = {\omega _0}t + \dfrac{1}{2}\left( {\dfrac{{{\omega _1}}}{t}} \right){t^2} \\
\Rightarrow\theta = \left( {\dfrac{{2{\omega _0} + {\omega _1}}}{2}} \right)t \\
$
$
\Rightarrow\theta = \left( {\dfrac{{\left( 2 \right)\left( 0 \right) + \left( {{\omega _1}} \right)}}{2}} \right)\left( 3 \right) \\
\Rightarrow\theta = \dfrac{{3{\omega _1}}}{2} \\ $
$10$ revolutions means that the angle rotated will be
$10 \times 2\pi $\[rad\]$ = 20\pi $$rad$,
${\omega _1} = \dfrac{{2\theta }}{3}\\
\Rightarrow{\omega _1} = \dfrac{{\left( 2 \right)\left( {20\pi } \right)}}{3}\\
\Rightarrow{\omega _1} = \dfrac{{40\pi }}{3}$$\dfrac{{rad}}{s}$.
From here we can get the acceleration of the fan is,
$ \alpha = \dfrac{{\left( {\dfrac{{40\pi }}{3}} \right)}}{3}\\
\therefore\alpha = \dfrac{{40\pi }}{9}$$\dfrac{{rad}}{{{s^2}}}$
Second part:
Here, the initial velocity will be ${\omega _1} = \dfrac{{40\pi }}{3}$$\dfrac{{rad}}{s}$ and the time is $3$seconds. Using the equation $\left( 3 \right)$ we can get the angle rotated by the fan in the next $3$ seconds and then the number of revolutions can be obtained.
$
\theta = \left( {\dfrac{{40\pi }}{3}} \right)\left( 3 \right) + \dfrac{1}{2}\left( {\dfrac{{40\pi }}{9}} \right){\left( 3 \right)^2} \\
\Rightarrow\theta = 40\pi + 20\pi\\
\therefore\theta = 60\pi \\
$
The angle rotated by the fan in the next $3$$s$ is $60\pi $$rad$. The fan revolves $1$ time and rotates by an angle of $2\pi $$rad$, then for \[60\pi \]$rad$ it will rotate $\dfrac{{60\pi }}{{2\pi }} = 30$ times.
Thus, it will make $30$ rotations in the next $3\,s$.Hence,option C is correct.
Note: Always split the questions into parts according to the given data. Find the final angular velocity, find the final velocity in the current part to find the rest unknowns in the next part as the body will be carrying the final angular velocity in some part as initial angular velocity in the next part. Remember the kinematics equations for rotational motion.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

