We've looked at several analogies between straight-line motion and rotational motion, including:
| Variable | Straight-line motion | Rotational motion | Connection |
|---|---|---|---|
| Displacement | x | θ | θ = x/r |
| Velocity | v | ω | ω = v/r |
| Acceleration | a | α | α = a/r |
| Produces motion | F | τ | τ = r × F |
| Inertia | m | I | I = ∫ r2 dm |
| Second Law | ΣF = ma | Στ = I α | |
| Work | W = F • d | W = τ • Δθ | |
| Kinetic Energy | K = ½ mv2 | K = ½ Iω2 | |
| Power | P = F • v | P = τ • ω | |
| Momentum | p = mv | L = Iω | L = r × p |
| Impulse | Δp = F Δt | ΔL = τ Δt |
Gyroscopes can be very precise navigational tools. If they are well-balanced and there is no gravitational torque (or torque from anything else) then a gyroscope with a particular angular momentum should maintain that direction. A rotating gyroscope in a plane, for instance, will keep pointing in the same direction even when the plane changes direction.
Toy gyroscopes are generally built so that there is a torque from the force of gravity acting on the gyro. This is a good example of impulse in a rotational setting. A gyro with an angular momentum that is anything other than vertical will feel a gravitational torque directed horizontally, perpendicular to the direction of the angular momentum.
According to the impulse equation, ΔL = τ Δt, the torque will produce a change in the angular momentum, with the change in momentum and the torque pointing in the same direction. This has the effect of rotating the gyro's angular momentum about a vertical axis - this is known as precession.
A cylinder of mass M and radius R has a string wrapped around it, with the string coming off the cylinder above the cylinder. If the string is pulled to the right with a force F, what is the acceleration of the cylinder if the cylinder rolls without slipping?
One key to the problem is realizing that there is a static friction force acting on the cylinder. Let's assume this force acts to the left.
Draw the free-body diagram. Take positive to the right, and clockwise.
The normal force cancels Mg vertically. Apply Newton's Second Law for horizontal forces, and for torques:
| Forces | | | Torques | ||
|---|---|---|---|---|
| ΣFx = Ma | | | Στ = I α | ||
| F - fs = Ma | | | RF + Rfs = ½ MR2α |
For rolling without slipping, α = a/R.
The torque equation becomes:
RF + Rfs = ½ MR2a/R
R's cancel, leaving:
F + fs = ½ Ma
The force equation is:
F - fs = Ma
Subtracting the force equation from the torque equation gives:
fs = -(1/4)Ma
The minus sign means the force of friction is not to the left, as we assumed, but to the right.
The equations also show that fs = -(1/3)F
and that a = (4/3)F/M
This is rather surprising, but seeing as we've carefully applied Newton's Laws we can be confident in our answer.