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A DVD is initially at rest. It then begins to rotate counterclockwise with constant angular acceleration . After the DVD has rotated through an angle , its angular speed is . If instead the DVD is initially at rest and its constant angular acceleration is doubled to after the DVD has rotated through the same angle, its angular speed will be

Respuesta :

Answer:

[tex]\omega' = \sqrt2\omega[/tex]

angular speed is increased by [tex]\sqrt2[/tex] factor

Explanation:

As we know that the angular acceleration is constant and initially the disc is at rest

so here we can say by kinematics

[tex]\omega^2 = \omega_0^2 + 2\alpha \theta[/tex]

here we know that angle turned by the disc is [tex]\theta[/tex]

now we have

[tex]\omega = \sqrt{2\alpha\theta}[/tex]

now the same disc start from rest with double angular acceleration and turned by same angle then final angular speed is given as

[tex]\omega' = \sqrt{2(2\alpha)\theta}[/tex]

so here we have

[tex]\omega' = \sqrt2\omega[/tex]

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