Final answer:
The minimum angular velocity (ω) at which the 235 g block on the rotating platform will slide away can be calculated using the relationship between the maximum static friction and the required centripetal force. This is obtained by taking the square root of the quotient of the product of the coefficient of static friction and gravity, and the radius of the circular path.
Step-by-step explanation:
To determine when the block will slide away from the rotating platform, we can use the formula for the force of static friction (Ffriction) and set it equal to the required centripetal force (Fcentripetal) needed to keep the block moving in a circular path without sliding. The force of static friction is given by Ffriction = μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force N is equal to the gravitational force on the block (mg). Hence, Ffriction = (0.437)(235 g)(9.8 m/s2) = (0.437)(2.35 kg)(9.8 m/s2).
The centripetal force required to keep the block in circular motion is Fcentripetal = mω2r, where m is the mass of the block, ω is the angular velocity, and r is the radius of the circular path (distance from the center of the platform). Setting Ffriction equal to Fcentripetal gives us: (0.437)(2.35 kg)(9.8 m/s2) = (2.35 kg)ω2(2.8 m). To find the minimum angular velocity at which the block starts sliding, ω, we solve for ω = √{ (μsg) / r }.