Hello. My son is in pre-school, but he's already starting his Undergraduate for college. Never too early to start, right? So he got this problem for homework. Can anyone help? He supposedly needed "nap time" and slept through the entire lecture the professor was giving, so he doesn't understand any of it. Anyways, here's the question:
A pendulum consists of a mass m at the end of a massless stick of length l. The
other end of the stick is made to oscillate vertically with a position given by y(t) =
A cos(ωt), where A << l. It turns out that if ω is large enough, and if the pendulum
is initially nearly upside-down, then it will, surprisingly, not fall over as time goes
by. Instead, it will (sort of) oscillate back and forth around the vertical position.
Explain why the pendulum doesn’t fall over, and find the frequency of the back and
forth motion.
Interesting problem.
The vertical motion exerts an acceleration of d²y/dy² on the mass.
Intuition tells me the mass will move on each side on successive vertical strokes, hence frequency = ω/2.
Next time ask him to drill his calculus and give the bambin a mug of strong coffee before class!
* d²y/dt²
To understand why the pendulum doesn't fall over and find the frequency of its back and forth motion, we need to apply the concepts of centripetal force and restoring force.
Let's break it down step by step:
1. The pendulum consists of a mass m at the end of a massless stick of length l. The other end of the stick is made to oscillate vertically with a position given by y(t) = A cos(ωt), where A << l. This equation represents the vertical position of the oscillating end of the stick as a function of time.
2. When the pendulum is nearly upside-down, the gravitational force acting on the mass m provides a restoring force. This restoring force is directed toward the equilibrium position (vertical position) and prevents the pendulum from falling over completely. The restoring force is given by F_restoring = -mgl sin(θ), where θ is the angular displacement of the pendulum from its equilibrium position (θ = 0 for the vertical position).
3. The gravitational force acts as the centripetal force in this case. Due to the pendulum's motion, there is a certain velocity and acceleration associated with it. The centripetal force is given by F_centripetal = mω²A sin(ωt). This force is directed towards the center of the circular path followed by the oscillating end of the pendulum.
4. For the pendulum to remain in equilibrium and not fall over, the restoring force and the centripetal force should balance each other. In other words, F_restoring = F_centripetal.
5. Equating the two forces, -mgl sin(θ) = mω²A sin(ωt). The m (mass) cancels out, and we have -gl sin(θ) = ω²A sin(ωt). Since A is much smaller than l (A << l), we can consider sin(θ) ≈ θ, where θ is in radians.
6. Simplifying the equation, we have -glθ = ω²A sin(ωt). Dividing both sides by l, we get -gθ = ω²A (sin(ωt)/l).
7. As sin(ωt)/l represents a dimensionless term, let's denote it as B. So, we have -gθ = ω²A B.
8. Ignoring the negative sign for simplicity, we have gθ = ω²A B. Rearranging the equation, we get θ = (ω²A B)/g.
9. The frequency of the back and forth motion can be found by considering that the angular displacement θ is given by θ = A cos(ωt). We can differentiate this equation twice with respect to time (t) to find the angular acceleration. Recall that angular acceleration is proportional to the square of angular frequency, so we have θ'' = -ω²A cos(ωt). Comparing this equation to θ = (ω²A B)/g, we can see that ω²A = -g.
10. Substituting ω²A = -g into θ = A cos(ωt), we have θ = A cos(ωt) = -gB/g. Simplifying further, we get θ = -B.
11. Since B is a constant term, we can consider it as the angular displacement at t = 0, denoted as θ_0. So, θ = θ_0.
12. The back and forth motion of the pendulum corresponds to the oscillation of the angular displacement around its equilibrium position. The frequency of this motion is given by f = ω/(2π), where ω is the angular frequency.
13. Substituting -g = ω²A into f = ω/(2π), we have f = (-g/A)/(2π) = (-1/A)(g/2π). Since A << l, we can approximate the frequency as f ≈ (g/2πA).
In conclusion, the pendulum doesn't fall over because the gravitational force acting as the centripetal force provides a restoring force that balances the centripetal force due to the pendulum's motion. The frequency of the back and forth motion is approximately given by f ≈ (g/2πA), where g is the acceleration due to gravity and A is the amplitude of the oscillating end of the stick.