A fly of mass .2 g sits 12 cm from the center of a phonograph record revolving 331.3 rpm. a)What is the magnitude of the centripetal force on the fly?b)What is the minimum value of the coefficient of static friction between the fly and the record required to prevent the fly from sliding off?

To find the magnitude of the centripetal force on the fly (Part a), we can use the formula for centripetal force:

F = (m * v^2) / r

where:
F is the centripetal force,
m is the mass of the fly,
v is the linear velocity, and
r is the distance from the center of rotation.

Step 1: Convert the mass of the fly to kilograms.
0.2 g = 0.2 * 10^-3 kg (since 1 g = 10^-3 kg)

Step 2: Convert the angular velocity to linear velocity.
Angular velocity = 331.3 rpm

Linear velocity (v) = (angular velocity * 2π * r) / 60

Given that the distance from the center (r) is 12 cm = 0.12 m
v = (331.3 * 2π * 0.12) / 60

Step 3: Calculate the centripetal force (F).
F = (m * v^2) / r

Substituting the values we found:
F = (0.2 * 10^-3) * ((331.3 * 2π * 0.12) / 60)^2 / 0.12

Now, let's calculate the answer.

For part b, the minimum coefficient of static friction (µs) can be calculated using the following inequality:

µs * m * g ≥ F

where:
µs is the coefficient of static friction,
m is the mass of the fly,
g is the acceleration due to gravity, and
F is the centripetal force we calculated in part a.

We already know the values of m and F. The value of g is approximately 9.8 m/s^2.

Plug in these values and solve for the coefficient of static friction (µs):

µs * (0.2 * 10^-3) * 9.8 ≥ F

Solve for µs:

µs ≥ F / ((0.2 * 10^-3) * 9.8)

Now, you can calculate the minimum value of the coefficient of static friction required to prevent the fly from sliding off the record.