Question

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval ( in revelations)? Hint: You need to use one of the equations of constant angular acceleration that independent of angular acceleration.

Answer 1

Answer:

The appropriate solution is "23.87 rev".

Explanation:

The given values are:

Initial angular velocity,

$\omega_i=20 \ rad/s$

Final angular velocity,

$\omega_f=40 \ rad/s$

Time taken,

$t = 5.0 \ s$

If, α be the angular acceleration, then

⇒  $\omega_f=\omega_i+\alpha t$

or,

⇒  $\alpha t=\omega_f-\omega_i$

⇒  $\alpha=\frac{\omega_f-\omega_i}{t}$

On substituting the values, we get

⇒     $=\frac{40-20}{5.0}$

⇒     $=\frac{20}{5.0}$

⇒     $=4 \ rad/s^2$

If, ΔФ be the angular displacement, then

⇒  $\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2$

On substituting the values, we get

⇒        $=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2]$

⇒        $=100+50$

⇒        $=150$

On converting it into "rev", we get

⇒  $\Delta \theta=(\frac{150}{2 \pi} )$

⇒        $=23.87 \ rev$