2015年7月12日 · $\begingroup$ Notice that after one full turn the change in position is also zero. What we are interested in here really the average value of the instantaneous acceleration, but to get it requires calculus (or at least the machinery of limits), which the OP doesn't want.

2022年4月30日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2020年9月3日 · So, from Newton's second law, we know that an inward force $\frac{mv^2}{r}$ must act on it. This inward force is gravity, whose magnitude is $\frac{GMm}{r^2}$ . Therefore, these two must be the same.

2016年11月17日 · What is the proof for kinetic energy $= \dfrac{mv^2}{2}$? Work done= mass × acceleration × displacement $$=m\times a\times s$$

2017年4月16日 · In your case, you won't find the same value because when you assume this equality $\frac{mv^2}{r} = \frac ...

2015年6月22日 · So: $ (mv^2)/r \le \mu_s mg $ In other words, with this equation you can calculate the maximum speed as you correctly observed. However this equation is used for the calculation of friction when the speed is not maximum, …

2015年6月21日 · The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$.

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2019年1月26日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2015年6月21日 · There are several ways to write centripetal acceleration $$\frac{v^2}{r} = \omega^2 r = v \omega$$ Are there intuitive explanations for any of these three forms?