Q9: Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the surface and the cylinder is 0.12. If the cylinder is 0.20 m from the center of the turntable, what is the maximum speed that the cylinder can move along its circular path without slipping off the turntable

0.48 m/s

Explanation:

Given

Mass of the cylinder, m = 0.1 kg

Coefficient of static friction, μ = 0.12

Radius of the turntable, r = 0.2 m

It is known that centrifugal forces always acts on the cylinders and tends to move away the rotating platform from the rotational axis. The centripetal force provide by the frictional force is given by

mv²/r = μmg

Making v the subject of formula, we get

v² = μrg

v = √μrg

If we substitute the values, we have

v = √(0.12 * 0.2 * 9.8)

v = √0.2352

v = 0.4849 m/s

Therefore, the maximum speed the cylinder can move without slipping is 0.48 m/s

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

it may be wrong but i think its a