Question

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angular velocity of the turntable after it has made one sommer (irs359) – Rotation Quiz 1 – craig – (2019-1A AP12) 2 complete revolution? 1. k~ωk = 2 π α 2. k~ωk = √ 2 π α 3. k~ωk = 4 π α 4. k~ωk = √ 2 α 5. k~ωk = 2 α 6. k~ωk = 2 √ π α

Answer 1

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

$\omega=\sqrt{8\pi \alpha}$

$\omega=2\sqrt{2\pi \alpha}$

Hence, this is the required solution.

Answer 2

Answer:

The answer is: $w=\sqrt{4\pi \alpha }$

Explanation:

Please look at the solution in the attached Word file.

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