Question

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 1

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.