The challenge is to find the instantaneous acceleration of a point on a sphere when it touches the larger sphere it is rolling inside of. Before finding the acceleration of a point on the circumference of the circle, let’s find that of the centre. The radius of the larger circle is 1 metre. The centre of the small circle is not on the circumference though. It is 25cm away because the small circle has radius 25cm. This means the centre of the smaller centre is 100-75 = 25 centimetres away from that of the larger one. The smaller circle completes a rotation whose circumference is 2π*75cm = 150πcm every 4 seconds. It’s velocity is thus 375π millimetres per second sideways. This velocity changes, and it’s vector’s endpoint will naturally also complete a full circular rotation each second. This circle has radius 1.5π m/s and thus circumference 2π*375π mm/s = 75π2 centimetres per second inward. To complete this circle every second, the acceleration must be 75π2 cm/s2. The same rule applies to smaller circle. The acceleration equals 4π2r/s2 inward times the rate of rotation. For the smaller circle, this gives us 1π2 m/s2 toward it’s centre. When the point touches the rim, these vectors will align, and the point will feel their cumulative acceleration. It’s counterintuitive that the point experiences the most force when it stops, but it’s true. Add 1π2 m/s2 inward and 75π2 cm/s2 inward together to get our answer, 175π2 cm/s2 inward.
Day: September 15, 2019
Solution to Fascinating Frequencies 2
Before trying to find the apoapsis of a geostationary orbit, first remember what a geostationary orbit and apoapsis are. Can a path always at the same height have a highest point? No. They are all equal! The speed necessary to complete a revolution every 20 hours is dπ/ 20 per hour, where d is the distance from the object. The orbital speed necessary at any point for a geostationary orbit is that at which centrifugal force equals gravity. Centrifugal force’s acceleration equals 4π2dv/sm outward. And that of gravity equals Gm/d2 inward. We are left with a system of equations that has too many units. We will solve it in Ubiquitous Units.
\vec v=\frac{\vec d\pi}{72000s}\\[8pt] \frac{4\pi\vec v\vec d}{sm}=\frac{Gm_1}{\vec d^2}\\[8pt] G=\frac{6.67408m^3}{10^{11}s^2kg}\\[8pt]