
The angular speed of the truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck wheel during this time is ______.
Answer
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Hint: Angular motion is defined as the motion of an object about a fixed point or fixed axis drawn by a line to an object. To solve this problem, we use a relation between the number of revolutions with the angular displacement. Here we use \[\alpha \]is the angular acceleration and calculated in terms of \[rad/{\sec ^2}\], \[\omega \] is the angular velocity and calculated in terms of rad/s and t is the time taken and calculated in terms of seconds.
Formula used:
Angular displacement equation is given as,
\[\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}\]
Where, \[{\omega _0}\]is the initial angular velocity, t is the time taken and \[\alpha \] is the angular acceleration.
Final angular velocity is given as,
\[\omega = {\omega _0} + \alpha t\]
Where, \[\omega \]is the final angular velocity and \[{\omega _0}\] is the initial angular velocity.
Complete step by step solution:
Initial angular speed, \[{\omega _i} = 900\,rpm\]
\[{\rm{ }}{\omega _i} = 900 \times \dfrac{{2\pi }}{{60}}\,rad/\sec \]
\[\Rightarrow{\omega _i} = 30\pi {\rm{ }}\,rad/\sec \]
Final angular speed, \[{\omega _f} = 2460\,rpm\]
\[{\rm{ }}{\omega _f} = 2460 \times \dfrac{{2\pi }}{{60}}\,rad/\sec \]
\[\Rightarrow {\omega _f} = 82\pi {\rm{ }}\,rad/\sec \]
As we know that \[\omega = {\omega _0} + \alpha t\]
So angular acceleration is given as,
\[\alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}\]
\[\Rightarrow \alpha = \dfrac{{82\pi - 30\pi }}{{26}}\]
\[\Rightarrow \alpha = 2\pi {\rm{ rad/se}}{{\rm{c}}^2}\]
As we know displacement equation is given as,
\[\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}\]
Substituting the values,
\[\theta = 30\pi \times 26 + \dfrac{1}{2} \times 2\pi \times {(26)^2}\]
\[\Rightarrow \theta = 1456\pi \]
Let the number of revolutions be N, then we can write
\[\theta = 2\pi N\]
\[\Rightarrow {\rm{ }}N = \dfrac{\theta }{{2\pi }}\]
Substituting the values, we get
\[\therefore N = \dfrac{{1456\pi }}{{2\pi }} = 728\]
Therefore, the number of revolutions by the truck wheel during this time is 728.
Note: Angular velocity is a vector quantity. It is described as the rate of change of angular displacement (\[\theta \]) which describes the angular speed or we can say the rotational speed of an object and the axis about which the object is rotating.
In a circular motion, the angular acceleration of a body is defined as the rate with which its angular velocity changes with time. It is also called rotational acceleration.
Formula used:
Angular displacement equation is given as,
\[\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}\]
Where, \[{\omega _0}\]is the initial angular velocity, t is the time taken and \[\alpha \] is the angular acceleration.
Final angular velocity is given as,
\[\omega = {\omega _0} + \alpha t\]
Where, \[\omega \]is the final angular velocity and \[{\omega _0}\] is the initial angular velocity.
Complete step by step solution:
Initial angular speed, \[{\omega _i} = 900\,rpm\]
\[{\rm{ }}{\omega _i} = 900 \times \dfrac{{2\pi }}{{60}}\,rad/\sec \]
\[\Rightarrow{\omega _i} = 30\pi {\rm{ }}\,rad/\sec \]
Final angular speed, \[{\omega _f} = 2460\,rpm\]
\[{\rm{ }}{\omega _f} = 2460 \times \dfrac{{2\pi }}{{60}}\,rad/\sec \]
\[\Rightarrow {\omega _f} = 82\pi {\rm{ }}\,rad/\sec \]
As we know that \[\omega = {\omega _0} + \alpha t\]
So angular acceleration is given as,
\[\alpha = \dfrac{{{\omega _f} - {\omega _i}}}{t}\]
\[\Rightarrow \alpha = \dfrac{{82\pi - 30\pi }}{{26}}\]
\[\Rightarrow \alpha = 2\pi {\rm{ rad/se}}{{\rm{c}}^2}\]
As we know displacement equation is given as,
\[\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}\]
Substituting the values,
\[\theta = 30\pi \times 26 + \dfrac{1}{2} \times 2\pi \times {(26)^2}\]
\[\Rightarrow \theta = 1456\pi \]
Let the number of revolutions be N, then we can write
\[\theta = 2\pi N\]
\[\Rightarrow {\rm{ }}N = \dfrac{\theta }{{2\pi }}\]
Substituting the values, we get
\[\therefore N = \dfrac{{1456\pi }}{{2\pi }} = 728\]
Therefore, the number of revolutions by the truck wheel during this time is 728.
Note: Angular velocity is a vector quantity. It is described as the rate of change of angular displacement (\[\theta \]) which describes the angular speed or we can say the rotational speed of an object and the axis about which the object is rotating.
In a circular motion, the angular acceleration of a body is defined as the rate with which its angular velocity changes with time. It is also called rotational acceleration.
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