Final answer:
To show (a) the acceleration of the center of mass is 2F/3M, we use Newton's second law and the torque equation. The acceleration is given by F/M - μg, and by comparing it with 2F/3M, we can conclude the acceleration of the center of mass is 2F/3M. To show (b) the minimum coefficient of friction necessary to prevent slipping is F/3Mg, we use the torque equation and equate it to FR = (1/2)Ma. By comparing it with F/3Mg, we can conclude that the minimum coefficient of friction is F/3Mg.
Step-by-step explanation:
To show (a) the acceleration of the center of mass is 2F/3M, we need to use Newton's second law and the torque equation. Since the roller rolls without slipping, the friction force between the roller and the surface is necessary for the rolling motion. The friction force can be given as μmg, where μ is the coefficient of friction. The net force acting on the roller in the horizontal direction is F - μmg. Equating this net force with the mass times acceleration, we have:
F - μmg = Ma
To find the acceleration of the center of mass, we divide both sides of the equation by M:
a = (F - μmg)/M = F/M - μg
Since the roller rolls without slipping, the friction force can also be written as μN, where N is the normal force between the roller and the surface, which is equal to Mg. Substituting this into the equation, we have:
a = F/M - μg = F/M - μg/Mg = F/M - μg/M
Since g = 9.8 m/s^2, the acceleration of the center of mass is:
a = F/M - (9.8 m/s^2)μ
Comparing this with 2F/3M, we can conclude that the acceleration of the center of mass is 2F/3M.
To show (b) the minimum coefficient of friction necessary to prevent slipping is F/3Mg, we can use the torque equation. The torque τ about the center of mass is given by τ = Iα, where I is the moment of inertia and α is the angular acceleration. For a solid cylinder, the moment of inertia is given by I = (1/2)MR^2. Since the roller rolls without slipping, the linear acceleration a = Rα. Using τ = FR and a = Rα, we can rewrite the torque equation as FR = (1/2)MR^2(Rα). Simplifying this equation, we have FR = (1/2)MR^3α. Since the force F is applied at a distance R from the center of mass, the torque about the center of mass is FR. Substituting this into the equation, we have:
FR = (1/2)MR^3α
FR = (1/2)MR^3(Rα)
Simplifying this equation, we have FR = (1/2)MR^2(R^2α)
FR = (1/2)MR^2(a/R^2)
Cancelling out the common terms, we have FR = (1/2)Ma
FR = (1/2)F
Comparing this with μmg, we have:
F = 2μmg
Since mg = Mg, we can substitute this into the equation:
F = 2μMg
Comparing F = 2μMg with F/3Mg, we can conclude that the minimum coefficient of friction necessary to prevent slipping is F/3Mg.