Thank you for this video. At 3:25, isn't the acceleration always directed towards the centre in circular motion? Surely therefore there is no component in the direction of s to plug into F=ma?
What you're referring to here is the centripetal force which facilitates circular motion. The centripetal force here is provided by the tension in the string. The gravitational acceleration is provided by the gravitational force.
If theta is small, then you can let sin(theta) = theta. That makes the differential equation solvable by guessing a solution. It will look just like a simple harmonic oscillator.
I found a different way to derive, but im not sure if its correct. force of gravity = centripetal force? mg = mw^2 L w^2 = g/L w = (g/L)^1/2 is it a correct way to derive?
like when should consider the gradient of gravity, like a tennis ball falling the same distance from the moon to earh, since the gravity will change in some rate, and obviously the acceleration will not be as the same as earth surface. i think would be very nice see how it is done.
@@HigorMadeira97 You'd need to use differential equations and g (now a variable) would be GM / R where G is the universal gravitation constant, M is the mass of the earth and R is the distance between the point mass (assume tennis ball to be a point mass) and Earth's centre . Essentially, we are having g to vary with distance here.
not theta(t), but f(t) - right? It's a function that satisfies the differential equation. If you take the derivative twice, you get the same function with a negative constant out front.
angular frequency (ω) is defined as the amount of radians an oscillator undergoes per second, which would be 2πf because frequency is the amount of cycles and there are 2π radians/cycle
If you really want to understand this.. you need to know how to solve linear differential equations with constant coefficients and complex numbers. Requires more than just physics.
