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3 years ago

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A piece of gum is stuck to the outer edge of a horizontal turntable, which is revolving at a constant speed. The shadow of the g

um is projected onto a wall behind the turntable. How does the frequency of rotation of the turntable compare to the frequency of oscillation of the gum's shadow?

1 answer:

cestrela7 [59]3 years ago

7 0

Answer:

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

Explanation:

Since we know that the projection on the wall is the vertical component of the position of the gum on the rotating table

So here we will say

$y = R sin\theta$

so the angle made by the radius vector depends on the angular frequency of the disc by which it is rotating

So we can say

$\theta = \omega t$

so here we can say

$y = R sin(\omega t)$

so here we can say that

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

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Solution:

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Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

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