Đáp án:
a) Ta có
+\({T_{giay}} = 1p = 60s \Rightarrow {\omega _{giay}} = \dfrac{{2\pi }}{{{T_{giay}}}} = \dfrac{\pi }{{30}}rad/s\)
+\({T_{phut}} = 1h = 60.60 = 3600s \Rightarrow {\omega _{phut}} = \dfrac{{2\pi }}{{{T_{phut}}}} = \dfrac{\pi }{{1800}}rad/s\)
+\({T_{gio}} = 12h = 12.60.60 = 43200s \Rightarrow {\omega _{gio}} = \dfrac{{2\pi }}{{{T_{gio}}}} = \dfrac{\pi }{{21600}}rad/s\)
b) Tốc độ dài \({v_{giay}} = {\omega _{giay}}.l\)
c) Gia tốc hướng tâm: \(a = {\omega ^2}l \Rightarrow \dfrac{{{a_{phut}}}}{{{a_{gio}}}} = {\left( {\dfrac{{{\omega _{phut}}}}{{{\omega _{gio}}}}} \right)^2}.\dfrac{{{l_{phut}}}}{{{l_{gio}}}} = {\left( {\dfrac{{\dfrac{\pi }{{1800}}}}{{\dfrac{\pi }{{21600}}}}} \right)^2}.\dfrac{4}{3} = 192\)