Explanation:
If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let [tex]\omega[/tex] 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 :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]\alpha =\dfrac{\omega-0}{t}[/tex]
[tex]\alpha =\dfrac{\omega}{t}[/tex]
[tex]t=\dfrac{\omega}{\alpha }[/tex]............(1)
Using second equation of kinematics as :
[tex]\theta=\omega_i t+\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=\dfrac{1}{2}\alpha t^2[/tex]
Using equation (1) in above equation
[tex]\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }[/tex]
In one revolution, [tex]\theta=4\pi[/tex] (in 2 revolutions)
[tex]4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }[/tex]
[tex]\omega=\sqrt{8\pi \alpha}[/tex]
[tex]\omega=2\sqrt{2\pi \alpha}[/tex]
Hence, this is the required solution.
Answer:
The answer is: [tex]w=\sqrt{4\pi \alpha }[/tex]
Explanation:
Please look at the solution in the attached Word file.
