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Answer 1

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$kmg=m\frac{{v}_{m}^{2}}{R}$
¼´${v}_{mr}^{\;}=\sqrt{kgr}=\sqrt{2.25¡Á10¡Á40}m/s=30m/s$
${v}_{mR}^{\;}=\sqrt{kgR}=\sqrt{2.25¡Á10¡Á90}m/s=45m/s$
£¨2£©Ö±µÀµÄ³¤¶È$x=\sqrt{{L}_{\;}^{2}-£¨R-r£©_{\;}^{2}}=50\sqrt{3}m$
ÔÚÖ±µÀÉÏÔ˶¯µÄƽ¾ùËÙ¶È$\overline{v}=\frac{{v}_{mr}^{\;}+{v}_{mR}^{\;}}{2}=\frac{45+30}{2}m/s=\frac{75}{2}m/s$
ʱ¼ä${t}_{1}^{\;}=2\frac{x}{\overline{v}}=2¡Á\frac{50\sqrt{3}}{\frac{75}{2}}s=4.61s$
Óɼ¸ºÎ¹Øϵ֪СԲ»¡¶ÔÓ¦µÄÔ²ÐĽÇ${¦È}_{1}^{\;}=120¡ã$£¬´óÔ²»¡¶ÔÓ¦µÄÔ²ÐĽÇ${¦È}_{2}^{\;}=240¡ã$
ÔÚСԲ»¡ÉÏÔ˶¯µÄʱ¼ä${t}_{2}^{\;}=\frac{2¦Ðr}{3{v}_{mr}^{\;}}=\frac{2¡Á3.14¡Á40}{3¡Á30}s=2.79s$
ÔÚ´óÔ²»¡ÉÏÔ˶¯µÄʱ¼ä${t}_{3}^{\;}$=$\frac{4¦ÐR}{3{v}_{mR}^{\;}}=\frac{4¡Á3.14¡Á90}{3¡Á45}s=8.37s$
×Üʱ¼ä$t={t}_{1}^{\;}+{t}_{2}^{\;}+{t}_{3}^{\;}=15.77s$
´ð£º£¨1£©ÔÚÁ½Ô²»¡ÍäµÀÉϵÄ×î´óËÙÂÊ·Ö±ðÊÇ30m/s¡¢45m/s
£¨2£©Èü³µµÀÈÆÐÐһȦÔ˶¯µÄ×Üʱ¼äÊÇ15.77s

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