What is the minimum coefficient of static
friction necessary to allow a penny to rotate
along a 𝟑𝟑.3rpm record (diameter= 0.30 m),
when the penny is placed at the outer edge
of the record?
We can start by using the formula for centripetal force:
Fc = mv^2/r
where Fc is the centripetal force, m is the mass of the penny, v is the tangential velocity of the penny (which is related to the rotational speed of the record), and r is the radius of the record (which is half of its diameter).
We can also use the formula for frictional force:
Ff = μsN
where Ff is the frictional force, μs is the coefficient of static friction, and N is the normal force (which is equal to the weight of the penny in this case).
In order for the penny to rotate along the record, the frictional force must be equal to the centripetal force:
Ff = Fc
μsN = mv^2/r
We can solve for the coefficient of static friction:
μs = mv^2/(Nr)
Now we just need to plug in the given values and solve:
m = 0.001 kg (mass of a penny)
v = 2πr/T = 2π(0.15 m)/(33.3 s) = 0.028 m/s (tangential velocity of the penny, where T is the period of the record's rotation)
r = 0.15 m (radius of the record)
N = mg = 0.001 kg × 9.81 m/s^2 = 0.00981 N (weight of the penny)
μs = (0.001 kg × (0.028 m/s)^2)/(0.00981 N × 0.15 m) ≈ 0.636
Therefore, the minimum coefficient of static friction necessary to allow a penny to rotate along a 33.3 rpm record with a diameter of 0.30 m, when the penny is placed at the outer edge of the record, is approximately 0.636.