Question

A 5.0 kg, 54-cm-diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. (Figure 1)What is the magnitude of the cylinder's initial angular acceleration?What is the magnitude of the cylinder's angular velocity when it is directly below the axle?

Answer 1

Answer: a) 24.2 rad/s² b) 6.95 rad/s

Explanation:

So the cylinder's moment of inertia is $I = \frac{MR^2}{2}$, where M is the mass of the cylinder and R is its radius.

The problem you have here is that the axle is modifying it, so we use the parallel axis theorem:

Therefore

$I = \frac{MR^2}{2} + MD^2$ where D is the distance from the center of mass to the axis which is a radius R

$\=> I = \frac{MR^2}{2} + M(R)^2 = \frac{3MR^2}{2}$

Next the Momentum with respect to the center of mass of the cylinder is

$M_{c} = I\alpha = F*R = m*g*R$ because there are no other forces rather than the cylinder's weight, now $I = \frac{3*5*0.27^2}{2} = 0.54675$ kg*m^2

a ) replace and find $\alpha = (m*g*R)/(I )$

$\alpha = (5.0*9.8*0.27)/(0.54675 ) = 24.2 rad/s^2$

b) we start by assuming that the energy is conserved because the cylinder is rolling without sliding:

so the potential energy is given by $E = M*g*R = 5*9.8*0.27 = 13.23J$

now you have that the same amount is going to be equal to the kinetic rotational energy of the cylinder so $13.23J = \frac{I*\omega^2}{2} => \omega = \sqrt(\frac{2*13.23J}{0.54675} ) = 6.95 rad/s$