To incorporate the 350 grams and 6 cm into the problem, we can use the conservation of angular momentum principle. This principle states that the total angular momentum of a system remains constant, unless acted upon by an external torque. Since there are no external torques acting on the disc, the total angular momentum before and after the mass falls will be the same.
Before the mass falls, the angular momentum of the disc is given by L=1/2(Iω), where I is the moment of inertia and ω is the angular velocity. Using the given values, we can calculate the initial angular momentum as L=1/2(0.7*0.15^2)*(45/60*2π)= 0.0376 kgm^2/s.
After the mass falls, the angular momentum of the disc and the mass combined can be calculated as L=1/2(Iω)+(mvr), where m is the mass of the falling object, v is its linear velocity, and r is the distance from the center of the disc. We can assume that the disc and the mass are rotating at the same angular velocity, ω, after the mass falls. We can also assume that the linear velocity of the mass, v, is equal to the linear velocity of the disc at the point of contact, since there are no external forces acting on the system.
Substituting the values given in the problem, we get L=1/2(0.7*0.15^2)*(45/60*2π)+ (0.35*0.06*45/60)= 0.0381 kgm^2/s.
Finally, we can equate the initial and final angular momenta and solve for ω, which will give us the angular velocity of the disc after the mass falls:
0.0376= 0.0381
ω= 0.0381/0.0376= 1.01 rad/s.
Therefore, the speed of the disc after the mass falls is 1.01 rad/s or 9.64 rpm, assuming the disc was initially rotating at 45 rpm.
