ÌâÄ¿ÄÚÈÝ

6£®UÐͳص¥°å»¬Ñ©ÊÇÉîÊÜ»¬Ñ©Õßϲ°®µÄÏîÄ¿£¬Æä³¡µØ£¨ºá½ØÃ棩ÈçͼËùʾ£¬AB£¬CD¾ùΪ°ë¾¶R=3.6mµÄ$\frac{1}{4}$Ô²¹ìµÀ£¬BCΪˮƽ¹ìµÀ£¬Ò»ÖÊÁ¿m=60kgµÄÔ˶¯Ô±´ÓAµãÒÔijһËÙ¶È»¬Ï£¬¾­UÐ͹ìµÀ´ÓDµãÊúÖ±ÏòÉϷɳö£¬ÔÚDµãÉÏ·½Íê³É¶¯×÷µÄʱ¼ät=1.2s£®È»ºóÓÖ´ÓDµã»¬ÈëUÐͳأ¬Èô²»¼Æ×èÁ¦ºÍĦ²Á£¬Ô˶¯Ô±¿É¿´³ÉÖʵ㣬ÇóÔ˶¯Ô±ÔÚUÐͳػ¬Ðйý³ÌÖУ¬UÐͳضÔÔ˶¯Ô±µÄ×î´óµ¯Á¦£®

·ÖÎö ÓÉDµãÉÏ·½Íê³É¶¯×÷µÄʱ¼ät=1.2sÇóµÄDµãµÄËÙ¶È£¬Ô˶¯Ô±ÔÚUÐͳØÖ묵½Ô²»¡×îµÍµãʱѹÁ¦×î´ó£¬Óɶ¯Äܶ¨ÀíÇóµÄCµãµÄËÙ¶È£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂÉÇóµÄ

½â´ð ½â£ºÓÉÌâÒâ¿ÉÖª
${v}_{D}=g•\frac{t}{2}$
Ô˶¯Ô±ÔÚUÐͳØÖ묵½Ô²»¡×îµÍµãʱ£¬Êܵ½µÄµ¯Á¦×î´ó£¬Óɶ¯Äܶ¨Àí¿ÉµÃ
$mgR=\frac{1}{2}{mv}_{C}^{2}-\frac{1}{2}{mv}_{D}^{2}$
${F}_{N}-mg=\frac{{mv}_{C}^{2}}{R}$                   
½âµÃ    ${F}_{N}=3mg+\frac{m{g}^{2}{t}^{2}}{4R}$=2400N        
´ð£ºUÐͳضÔÔ˶¯Ô±µÄ×î´óµ¯Á¦Îª2400N

µãÆÀ ±¾Ìâ×ۺϿ¼²éÁ˶¯Äܶ¨Àí¡¢Å£¶ÙµÚ¶þ¶¨ÂÉ£¬Éæ¼°µ½Ô²ÖÜÔ˶¯Ô˶¯£¬ÔȱäËÙÖ±ÏßÔ˶¯£¬ÊúÖ±ÉÏÅ×Ô˶¯£¬×ÛºÏÐÔ½ÏÇ¿£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø