Question

A wheel rotating with uniform angular acceleration covers 50 revolutions in the first five seconds after the start. If the angular acceleration at the end of five seconds is $x\pi \text{ rad/}{{\text{s}}^{2}}$, find the value of x.

Answer 1

Hint: We have been provided a wheel which is rotating in uniform angular acceleration i.e. it has constant angular acceleration while rotating it covers 50 revolutions in the first five seconds. At the initial stage, the wheel was at rest; therefore the value of initial angular velocity is zero. To find the value of x, use the kinematic equation of circular motion. Put all the values in the equation and calculate the value of angular acceleration. Now compare the calculated value of angular acceleration with the given value of angular acceleration and get the value of x.

Formula used: Expression of kinematic equation is given by,
$\theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}}$

Complete step by step answer:
The question has stated that a wheel is rotating with a uniform angular acceleration and it is covering 50 revolutions in the first five seconds but after the wheel starts. It means, initially the wheel was stopped. Since, the wheel is performing circular motion; we will use a kinematical equation for circular motion. The second kinematic equation is given as,
$\theta ={{\omega }_{0}}t+\dfrac{1}{2}\alpha {{t}^{2}}........\left( 1 \right)$
Where, θ = angular displacement of wheel in time (t)
${{\omega }_{0}}$ = initial angular velocity
$\alpha $= acceleration (angular)
t = time
At the initial stage, the wheel was stopped therefore the value of angular velocity is zero. i.e., ${{\omega }_{0}}=0$
Therefore, equation (1), can be written as,
$\begin{align}
  & \theta =0\times t+\dfrac{1}{2}\alpha {{t}^{2}} \\
 & \theta =\dfrac{1}{2}\alpha {{t}^{2}}......\left( 2 \right) \\
\end{align}$
Now we need to calculate angular acceleration at the end of five second.
Whereas,
$\begin{align}
  & \theta =50\text{ rev} \\
 & \theta \text{=2}\pi \times \text{50} \\
\end{align}$
Equation (2) implies,
$\begin{align}
  & 2\pi \times 50=\dfrac{1}{2}\alpha {{\left( 5 \right)}^{2}} \\
 & \alpha =\dfrac{2\left( 50 \right)\left( 2\pi \right)}{{{5}^{2}}} \\
 & \alpha =8\left( \pi \right)rad/{{s}^{2}}......\left( 3 \right) \\
\end{align}$
In the question it is already given that angular acceleration is
$x\pi \text{ rad/}{{\text{s}}^{2}}......\left( 4 \right)$
Compare equation (3) and (4) we get,
$x=8\text{ rad/}{{\text{s}}^{2}}$

Hence, the value of x is 8.

Note: Question itself stated that a wheel rotating with the uniform angular acceleration, it means that the wheel is performing uniform circular motion, denoted by U.C.M. In the uniform circular motion, the magnitude of velocity is constant, however the direction of velocity goes on changing continuously, hence uniform circular motion is an accelerated motion. The number of revolutions performed by particles performing uniform circular motion in unit time is nothing but frequency of revolution or revolution.