Question

A wheel starts rotating with an angular speed 5pi rad/sec and acceleration 12.5 rad/sec?. Find the number of revolutions made by the wheel before the wheel comes to rest.​

Answer 1

Given:

A wheel starts rotating with an angular speed 5pi rad/sec and acceleration 12.5 rad/sec²

To find:

Revolutions performed before coming to rest?

Calculation:

• Angular acceleration will be negative in sign as it is slowing down rotation.

Applying EQUATIONS OF Rotational Kinematics:

${ \omega}^{2} = { \omega_{0} }^{2} + 2 \alpha \theta$

$\implies {0}^{2} = { (5\pi)}^{2} + 2( - 12.5) \theta$

$\implies 0 = 25 {\pi}^{2} - 2(12.5) \theta$

$\implies 25 \theta = 25 {\pi}^{2}$

$\implies \theta = {\pi}^{2} \: rad$

So, angular displacement is π² radians.

Answer 2

Explanation:

Given:

A wheel starts rotating with an angular speed 5pi rad/sec and acceleration 12.5 rad/sec²

To find:

Revolutions performed before coming to rest?

Calculation:

Angular acceleration will be negative in sign as it is slowing down rotation.

Applying EQUATIONS OF Rotational Kinematics:

{ \omega}^{2} = { \omega_{0} }^{2} + 2 \alpha \thetaω2=ω02+2αθ

\implies {0}^{2} = { (5\pi)}^{2} + 2( - 12.5) \theta⟹02=(5π)2+2(−12.5)θ

\implies 0 = 25 {\pi}^{2} - 2(12.5) \theta⟹0=25π2−2(12.5)θ

\implies 25 \theta = 25 {\pi}^{2}⟹25θ=25π2

\implies \theta = {\pi}^{2} \: rad⟹θ=π2rad

So, angular displacement is π² radians.