Question

Imagine a steel cylinder 12 ft tall, 4 ft diameter, weight 16000 lbs, CG at geometric center.

The cylinder tips over at an initial radial velocity at the tipping point of 0.6 radians/sec.

How much kinetic energy is present at the top of the cylinder just before impact?

Answer 1

So draw at the diagram and just calculate \( v = \omega r \). You are given \( \omega \), you just need to find the appropriate radial distance \( r \).

Answer 2

First, we need to find the tipping point of the cylinder so we can find the initial angle of the cylinder. When we tilt the cylinder, realize that the normal force no longer acts along the line through the center of gravity of the cylinder. Instead, it acts on the edge of the cylinder in the direction of the tip. This off-set of normal force and gravitation force creates a torque, which either sends the cylinder back to equilibrium or causes it to tip. See the illustration. It can be seen that the tipping point occurs when the CG moves "outside" the normal force. From geometry, we can find this point as the angle \(\theta\). \[\theta_{TP} = \tan^{-1} \left( 2r \over h \right)\]where \(r\) is the radius of the cylinder and \(h\) is the height of the cylinder.

Now, we know that the angle the cylinder displaces before it hits the ground must be\[\theta = 90 - \theta_{TP}\]

Second, we need to find the moment of inertia of the cylinder. The moment of inertia of a cylinder about its CG rotated about an axis parallel to its ends is\[\bar I = {1 \over 12}m (3r^2 + h^2)\]However, the cylinder doesn't rotate about its CG so we must use the parallel axis theorem to find the total moment of inertia of the cylinder. This is expressed as\[I = \bar I + m \left (h \over 2 \right)^2\]

Third, we need to establish the EOM. The driving force is the gravitational force acting at the CG. Realize that the moment (torque) created by the gravitational force depends on \(\theta\). Let's first establish this relationship. \[\tau_g = {h \over 2} g \sin(\theta)\]Therefore, the EOM can be expressed as\[I \alpha = \tau_g\]We can solve this equation for \(\alpha\), which is the angular acceleration. Let's note that\[\alpha = {d \omega \over dt} = {d^2 \theta \over dt^2}\] where \(\omega\) is the angular velocity.

We will assume, for the sake of solvability, that the angular acceleration is constant. From our kinematic relations for rotational motion, \[\theta = \theta_{TP} + \omega_i t + {1 \over 2} \alpha t^2\]We know \(\theta\), \(\theta_{TP}\), \(\omega_i\), and \(\alpha\) therefore, we can solve for \(t\).

Once we known \(t\), we can find the angular velocity of the cylinder right before impact from\[\omega = \omega_i + \alpha t\]

The kinetic energy in the tip can then be found as\[TKE = {1 \over 2} m ({\omega h})^2\]It is unclear what mass should be considered. I'll let you determine that.