
An engine increases its angular speed 300rpm to 900rpm in 5 seconds. Number of revolutions made during this time by the engine is:
A. \[100\]
B. \[75\]
C. \[50\]
D. \[25\]
Answer
584.1k+ views
Hint: We use the given information about angular speed with respect to time for this solution and solve it with the help of angular acceleration formula. The angular acceleration in two dimensions shall be a pseudoscalar with a positive indicator if the angled velocity is increased in the opposite direction of the clock or decreased clockwise and negative when the angular velocity increases in the clockwise direction or decreases in the opposite direction. Angular acceleration is a pseudovector in three dimensions.
Complete step by step solution:
Given,
Initial angular velocity,
${\omega _1} = 300\,{\text{rpm}} \\
\Rightarrow {\omega _1} = \dfrac{{300 \times 2\pi }}{{60}} \\
\Rightarrow {\omega _1} = 10\pi \,{\text{rad}}\,{{\text{s}}^{ - 1}} \\$
Final angular velocity,
${\omega _1} = 900\,{\text{rpm}} \\
\Rightarrow {\omega _1}{\text{ = }}\dfrac{{900 \times 2\pi }}{{60}} \\
\Rightarrow {\omega _1} = 30\pi \,{\text{rad}}\,{{\text{s}}^{ - 1}} \\$
The given time is, \[t = 5{\text{s}}\]
So, the angular acceleration will be,
$\alpha = \dfrac{{{\omega _1} - {\omega _2}}}{t} \\
\Rightarrow \alpha = \dfrac{{\left( {30 - 10} \right)\pi }}{5} \\
\Rightarrow \alpha = \dfrac{{20\pi }}{5} \\
\Rightarrow \alpha = 4\pi \,{\text{rad}}\,{{\text{s}}^{ - 2}} \\$
And the angle is,
$\theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2} \\
\Rightarrow \theta = 10\pi \times 5 + \dfrac{1}{2} \times 4\pi \times 25 \\
\therefore \theta = 100\pi \,{\text{rad}} \\$
So, the number of revolutions is,
\[n = \dfrac{{100\pi }}{{2\pi }} = 50\]
The required answer is \[50\] revolutions. The correct option is C.
Additional information:
Angular acceleration: The angular acceleration refers to the angular velocity shift rate. As two forms of angular velocity exist, namely spin angular and orbital angular velocity, two kinds of angular acceleration are naturally also present, called spin angular acceleration and orbital angular acceleration. Spin angular speed refers to the angular speed of the rigid body around the center of rotation and orbital speed refers to the angular speed of a point particle over the fixed source. In units of angle per unit of time (which in SI units are radians per second squared), angular acceleration is measured and is typically viewed as an alpha symbol ( \[\alpha \] ).
Note: We typically consider it positive in clockwise direction. Angular displacement ( \[\theta \] ) is therefore positive in the rear direction. Thus, a "push" in anti-clockwise directions is negative angular acceleration ( \[\alpha \] ). The body speeds up, slows down (and ultimately goes backwards), whether \[\alpha \] and \[\omega \] have the same signal.
Complete step by step solution:
Given,
Initial angular velocity,
${\omega _1} = 300\,{\text{rpm}} \\
\Rightarrow {\omega _1} = \dfrac{{300 \times 2\pi }}{{60}} \\
\Rightarrow {\omega _1} = 10\pi \,{\text{rad}}\,{{\text{s}}^{ - 1}} \\$
Final angular velocity,
${\omega _1} = 900\,{\text{rpm}} \\
\Rightarrow {\omega _1}{\text{ = }}\dfrac{{900 \times 2\pi }}{{60}} \\
\Rightarrow {\omega _1} = 30\pi \,{\text{rad}}\,{{\text{s}}^{ - 1}} \\$
The given time is, \[t = 5{\text{s}}\]
So, the angular acceleration will be,
$\alpha = \dfrac{{{\omega _1} - {\omega _2}}}{t} \\
\Rightarrow \alpha = \dfrac{{\left( {30 - 10} \right)\pi }}{5} \\
\Rightarrow \alpha = \dfrac{{20\pi }}{5} \\
\Rightarrow \alpha = 4\pi \,{\text{rad}}\,{{\text{s}}^{ - 2}} \\$
And the angle is,
$\theta = {\omega _1}t + \dfrac{1}{2}\alpha {t^2} \\
\Rightarrow \theta = 10\pi \times 5 + \dfrac{1}{2} \times 4\pi \times 25 \\
\therefore \theta = 100\pi \,{\text{rad}} \\$
So, the number of revolutions is,
\[n = \dfrac{{100\pi }}{{2\pi }} = 50\]
The required answer is \[50\] revolutions. The correct option is C.
Additional information:
Angular acceleration: The angular acceleration refers to the angular velocity shift rate. As two forms of angular velocity exist, namely spin angular and orbital angular velocity, two kinds of angular acceleration are naturally also present, called spin angular acceleration and orbital angular acceleration. Spin angular speed refers to the angular speed of the rigid body around the center of rotation and orbital speed refers to the angular speed of a point particle over the fixed source. In units of angle per unit of time (which in SI units are radians per second squared), angular acceleration is measured and is typically viewed as an alpha symbol ( \[\alpha \] ).
Note: We typically consider it positive in clockwise direction. Angular displacement ( \[\theta \] ) is therefore positive in the rear direction. Thus, a "push" in anti-clockwise directions is negative angular acceleration ( \[\alpha \] ). The body speeds up, slows down (and ultimately goes backwards), whether \[\alpha \] and \[\omega \] have the same signal.
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