Question

2£®ÈçͼËùʾ£¬ÖÊÁ¿ÎªmµÄСÇòÖÃÓÚÁ¢·½ÌåµÄ¹â»¬ºÐ×ÓÖУ¬ºÐ×ӵı߳¤ÂÔ´óÓÚÇòµÄÖ±¾¶£®Ä³Í¬Ñ§ÄÃןúÐ×ÓÔÚÊúֱƽÃæÄÚ×ö°ë¾¶ÎªRµÄÔÈËÙÔ²ÖÜÔ˶¯£¬ÒÑÖªÖØÁ¦¼ÓËÙ¶ÈΪg£¬¿ÕÆø×èÁ¦²»¼Æ£¬ÒªÊ¹ÔÚ×î¸ßµãʱºÐ×ÓÓëСÇòÖ®¼ä×÷ÓÃÁ¦Ç¡Îª0.5mg£¬Ôò£¨¡¡¡¡£©

• A£® ¸ÃºÐ×Ó×öÔÈËÙÔ²ÖÜÔ˶¯µÄÖÜÆÚÒ»¶¨²»´óÓÚ2¦Ð$\sqrt{\frac{2R}{3g}}$
• B£® ¸ÃºÐ×Ó×öÔÈËÙÔ²ÖÜÔ˶¯µÄÖÜÆÚÒ»¶¨µÈÓÚ2¦Ð$\sqrt{\frac{2R}{g}}$
• C£® ºÐ×ÓÔÚ×îµÍµãʱºÐ×ÓÓëСÇòÖ®¼äµÄ×÷ÓÃÁ¦´óС¿ÉÄÜСÓÚ2.5mg
• D£® ºÐ×ÓÔÚ×îµÍµãʱºÐ×ÓÓëСÇòÖ®¼äµÄ×÷ÓÃÁ¦´óС¿ÉÄÜ´óÓÚ2.5mg

Answer 1

·ÖÎö ÔÚ×î¸ß¶È£¬ÈôСÇòÓëºÐ×ÓµÄÉϱßÔµ´æÔÚ×÷ÓÃÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öÖÜÆڵĴóС£¬ÈôСÇòÓëºÐ×ÓϱßÔµÓÐ×÷ÓÃÁ¦£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öÖÜÆڵĴóС£®
ͬÀíͨ¹ýÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³ö×î¸ßµãµÄËٶȣ¬ÔÚ×îµÍµã£¬×¥×¡ºÐ×Ó¶ÔСÇòµÄ×÷ÓÃÁ¦·½ÏòÏòÉÏ£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³ö×÷ÓÃÁ¦µÄ´óС£®

½â´ð ½â£ºA¡¢ÔÚ×î¸ßµã£¬ÈôСÇòÓëºÐ×ÓµÄÉϱßÔµ´æÔÚ×÷ÓÃÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÓУºmg+F=m$R\frac{4{¦Ð}^{2}}{{T}^{2}}$£¬½âµÃT=$2¦Ð\sqrt{\frac{2R}{3g}}$£¬
ÈôСÇòÓëºÐ×ÓϱßÔµ´æÔÚ×÷ÓÃÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÓУº$mg-F=mR\frac{4{¦Ð}^{2}}{{T}^{2}}$£¬½âµÃT=$2¦Ð\sqrt{\frac{2R}{g}}$£¬¹ÊAÕýÈ·£¬B´íÎó£®
C¡¢ÔÚ×î¸ßµã£¬ÈôСÇòÓëºÐ×ÓµÄÉϱßÔµ´æÔÚ×÷ÓÃÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÓУºmg+F=m$\frac{{v}^{2}}{R}$£¬½âµÃv=$\sqrt{1.5gR}$£¬ÔÚÔÚ×îµÍµã£¬$F¡ä-mg=m\frac{{v}^{2}}{R}$£¬½âµÃ$F¡ä=mg+m\frac{{v}^{2}}{R}=2.5mg$£¬
ÔÚ×î¸ßµã£¬ÈôСÇòÓëºÐ×ÓϱßÔµ´æÔÚ×÷ÓÃÁ¦£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÓÐ$mg-F=m\frac{{v}^{2}}{R}$£¬½âµÃv=$\sqrt{0.5gR}$£¬ÔÚ×îµÍµã£¬$F¡ä-mg=m\frac{{v}^{2}}{R}$£¬½âµÃ$F¡ä=mg+m\frac{{v}^{2}}{R}=1.5mg$£¬¹ÊCÕýÈ·£¬D´íÎó£®
¹ÊÑ¡£ºAC£®

µãÆÀ ½â¾ö±¾ÌâµÄ¹Ø¼üÖªµÀÔ²ÖÜÔ˶¯ÏòÐÄÁ¦µÄÀ´Ô´£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂɽøÐÐÇó½â£¬×¢ÒâÔÚ×î¸ßµã£¬Ð¡Çò¿ÉÄÜÓëÉϵ×ÓÐ×÷ÓÃÁ¦£¬Ò²¿ÉÄÜÓëϵ×ÓÐ×÷ÓÃÁ¦£®