An old vinyl record spins at a rate of 33 1/3 rpm and have a diameter of 12 inches (or 30.48 cm). If a ladybug sits halfway between the axis of rotation and the edge of the record, how much force must the bug exert to hold on and not go flying if the ladybug has a mass of one gram? Hint: use centripetal acceleration. a. .00035 N b. .00061 N c. .00093 N d. .00125 N e. .00222 N

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-08-22T13:02:00+02:00

Answer:

The force is 0.00093 N.

(c) is correct option.

Explanation:

Given that,

Mass = 1 g

Diameter = 30.48 cm

Radius = 15.24 cm

We need to calculate the centripetal acceleration

Position of bug is at halfway of the radius

So, radius r = 7.62 cm

Using formula of centripetal force

[tex]F=\dfrac{mv^2}{r}=mr\omega^2[/tex]

Here, [tex]v = r\omega[/tex]

[tex]\omega=2\pi f[/tex]...(I)

For frequency,

[tex]f=\dfrac{33}{60}=0.55\ Hz[/tex]

Put the value in the equation (I)

[tex]\omega=2\times3.14\times0.55[/tex]

[tex]\omega=3.49\ rad\s[/tex]

Put the value into the formula of force

[tex]F=10^{-3}\times7.62\times10^{-2}\times(3.49)^2[/tex]

[tex]F=0.00093\ N[/tex]

Hence, The force is 0.00093 N.