Question

problem # 6

• a small button placed on a horizontal rotating platform with diameter 0.320 m will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 m from the axis. (a) what is the coefficient of static friction between the button and the platform? (b) how far from the axis can the button be placed, without slipping, if the platform rotates at 60 rev/min?

Explanation:

Step1: 计算角速度

已知转速\(n = 40.0\mathrm{rev/min}\)，将其转化为角速度\(\omega\)，根据\(\omega=\frac{2\pi n}{60}\)，可得\(\omega=\frac{2\pi\times40.0}{60}\mathrm{rad/s}=\frac{4\pi}{3}\mathrm{rad/s}\)。

Step2: 分析向心力来源

当按钮刚好不滑动时，静摩擦力提供向心力，即\(f_s = m\omega^{2}r\)，又\(f_s=\mu_s N\)，在水平面上\(N = mg\)，所以\(\mu_s mg=m\omega^{2}r\)，化简可得\(\mu_s=\frac{\omega^{2}r}{g}\)。

已知\(r = 0.150\mathrm{m}\)，\(g = 9.8\mathrm{m/s^{2}}\)，\(\omega=\frac{4\pi}{3}\mathrm{rad/s}\)，代入可得\(\mu_s=\frac{(\frac{4\pi}{3})^{2}\times0.150}{9.8}\)
\(=\frac{\frac{16\pi^{2}}{9}\times0.150}{9.8}\approx0.27\)。

Step3: 计算新转速下的半径

当\(n' = 60\mathrm{rev/min}\)时，\(\omega'=\frac{2\pi n'}{60}=\frac{2\pi\times60}{60}= 2\pi\mathrm{rad/s}\)。
由\(\mu_s=\frac{\omega'^{2}r'}{g}\)，可得\(r'=\frac{\mu_s g}{\omega'^{2}}\)，将\(\mu_s = 0.27\)，\(g = 9.8\mathrm{m/s^{2}}\)，\(\omega'= 2\pi\mathrm{rad/s}\)代入，\(r'=\frac{0.27\times9.8}{(2\pi)^{2}}\approx0.067\mathrm{m}\)。

Answer:

(a) 静摩擦系数\(\mu_s\approx0.27\)；(b) 按钮不滑动的最远距离\(r'\approx0.067\mathrm{m}\)。