A uniform cylinder of radius R and mass M is mounted so as to rotate freely about a horizontal axis that is parallel to, and a distance h from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position? State your answers in terms of the given variables, using g when appropriate.

I = M*(0.5R^2 + h^2)

w = sqrt [2*g*h / (0.5R^2 + h^2)]

Explanation:

Given:

- The radius of the cylinder R

- The mass of the cylinder M

- The distance between horizontal axis h

- The moment of inertia of cylinder I_c

Solution:

- When the axis of rotation and axis with respect to rotation inertia is required do not coincide then we have to apply parallel axis theorem to calculate the moment of inertia about the required axis which is at a distance h from body rotational axis as follows:

                                    I = I_c + M*h^2

- Where I_c of a cylinder is = 0.5*M*R^2

                                    I = 0.5*M*R^2 + M*h^2

                                    I = M*(0.5R^2 + h^2)

- From conservation of total mechanical energy of the cylinder, the change in gravitational potential energy ΔP.E plus the change in kinetic energy ΔK.E of the cylinder must be zero:

                                   ΔP.E = ΔK.E

                                  M*g*h = 0.5*I*w^2

                                  M*g*h = 0.5*M*(0.5R^2 + h^2)*w^2

                                  w^2 = 2*g*h / (0.5R^2 + h^2)

                                  w = sqrt [2*g*h / (0.5R^2 + h^2)]

Where,

             w: The angular speed of the cylinder

the force on the thing and the distance the force moves the thing

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west