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A horizontal vinyl record of mass 0.123 kg and radius 0.0958 m rotates freely about a vertical axis through its center with an angular speed of 4.82 rad/s and a rotational inertia of 2.19 x 10-4 kg·m2. Putty of mass 0.0400 kg drops vertically onto the record from above and sticks to the edge of the record.What is the angular speed of the record immediately afterwards?

User Shohn
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1 Answer

1 vote

Answer:

The angular speed of the record is 1.8
(rad)/(s)

Step-by-step explanation:

Given:

Mass
m = 0.123 kg

Radius
r = 0.0958 m

Angular speed
\omega _(i) = 4.82 (rad)/(s)

Moment of inertia
I = 2.19 * 10^(-4 )
Kg. m^(2)

Mass of putty
M = 0.0400 Kg

For finding the final angular speed,

According to the conservation of angular momentum,


L_(i) = L_(f)


(I \omega _(i) ) = (I + Mr^(2) ) \omega _(f)


\omega _(f) = (I \omega _(i) )/(I + Mr^(2) )


\omega _(f) = (2.19 * 10^(-4) * 4.82 )/(2.19 * 10^(-4) + 0.040 * (0.0958) ^(2) )


\omega _(f) = 1.8 (rad)/(s)

Therefore, the angular speed of the record is 1.8
(rad)/(s)

User David Budworth
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