6.7 Examples of solutions to problems involving motion of rigid bodies
The best way to learn how to use the equations in section 6.6 is just to work through a series of examples.
6.7.1 Solutions to 2D problems
Example 1: A solid of revolution (eg a cylinder or sphere) with mass M and mass moment of inertia about its COM IGzz is released from rest at the top of a ramp. It rolls without slip. Calculate its velocity at the bottom of the ramp.
This formula predicts that an object with a smaller inertia IGzz will move faster than an object with a large inertia. A sphere rolls down the ramp more quickly than a cylinder, for example, and a solid cylinder rolls more quickly than a ring.
Example 2: For the problem treated in the preceding section, calculate the critical value of friction coefficient necessary to prevent slip at the contact.
If we want to learn about forces, we have to use the linear and angular momentum equations. This problem can be solved with the 2D formulas in terms of accelerations:
The formula shows that objects with large values of IGzz / MR2 are more likely to slip. If the inertia is very small, slip will never occur. A ring will slip on a lower slope than a cylinder, which will slip on a lower slope than a sphere.
Example 3: A vertical mast can be idealized as a slender rod with length L and mass M, which is held in an inverted position by a torsional spring with stiffness κ at its base. Find the equation of motion for the angle θ in the figure, and hence determine the natural frequency of vibration of the mast.
This is a conservative system. Also, the mast rotates about a fixed point. We can analyze the problem using energy methods, and use the special formulas for rotation about a fixed point.
Example 4: A thin uniform disk of radius R, mass m and mass moment of inertia mR2 /2 is placed on the ground with a positive velocity v0 in the horizontal direction, and a counterclockwise rotational velocity (a backspin) ω0 . The contact between the disk and the ground has friction coefficient µ . The disk initially slips on the ground, and for a suitable range of values of ω0 and v0 its direction of motion may reverse. The goal of this problem is to calculate the conditions where this reversal will occur.