Question

Please help ASAP A uniform disk of mass M and radius R is pivoted such that it can rotate freely about a horizontal axis through its center and perpendicular to the plane of the disk. A small particle of mass m is attached to the rim of the disk at the top directly above the pivot. The system is given a gentle start, and the disk begins to rotate. at this point, what force must be exerted on the particle by the disk to keep it on the disk?​

Answer 1

Answer: F = mg(1 + 4m / (½M + m))

Explanation:

"At this point seems" unclear. If the particle is at the top of the disc and angular velocity is negligible, then the force would equal the weight of the particle. F = mg

The more interesting question would be what force is needed to keep the particle attached when significant angular rotation has been achieved. The maximum point would be diametrically opposed to the starting point.

I will analyze it there

The potential energy will convert to kinetic energy

   mgh = ½Iω²

mg(2R) = ½(½MR² + mR²)ω²

 4mgR = R²(½M + m)ω²

       ω² = 4mg / (R(½M + m))

With m at the lowest position, the force of attachment must support the weight of m and provide for the needed centripetal acceleration

F = m(g + ω²R)

F = m(g + 4mg / (R(½M + m))R)

F = mg(1 + 4m / (½M + m))