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kirza4 [7]
3 years ago
6

The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func

tion in the same way.) Over a 133min playing time, the angular speed varies from 570rpm to 1600rpm .Assuming it to be constant, what is the angular acceleration (in/rads sec ) of such a DVD?
Physics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = 570 \times \dfrac{2\pi}{60}

              = 59.69 rad/s

1600 rpm =  = 570 \times \dfrac{2\pi}{60}

              = 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt    

167.6 = 59.7 + α x 7980

α  x 7980 = 107.9

α = 0.0135 rad/s²

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  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

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m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

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Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

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intensity of sound at level of microphone is 0.00139 W / m 2

sound intensity level at position of micro phone is 91.456 dB

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distance betwen microphone & speaker is   42 m

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                   I = \frac{P}{A}

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b) sound intensity level at position of micro phone is

                \beta = 10 log \frac{I}{I_o}

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                = 1 * 10^{-12} W / m 2  

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