Question

15£®ÈçͼËùʾ£¬ÔÚת¶¯µÄˮƽԲÅÌÉÏ£¬Ñذ뾶·½Ïò·ÅÖÃÁ½¸öÓÃϸÏßÏàÁ¬µÄÖÊÁ¿¾ùΪm=1kgµÄСÎïÌåA¡¢B£¬ËüÃǵ½×ªÖá¾àÀë·Ö±ðΪr_A=20cm£¬r_B=30cm£®A¡¢BÓëÅÌÃæ¼äµÄ×î´ó¾²Ä¦²ÁÁ¦¾ùΪÖØÁ¦µÄ0.4±¶£¬ÊÔÇó£º
£¨1£©µ±Ï¸ÏßÉÏ¿ªÊ¼³öÏÖÕÅÁ¦Ê±£¬Ô²Å̵ĽÇËٶȦØ₀£»
£¨2£©µ±A¸ÕÒª»¬¶¯Ë²¼ä£¬Ô²Å̵ĽÇËٶȦأ»
£¨3£©µ±Ô²Å̵ĽÇËٶȴӣ¨1£©ËùÇóµÃµÄ¦Ø₀¿ªÊ¼»ºÂýÔö´ó£¬Ö±µ½A¸ÕÒª»¬¶¯Ë²¼ä£¬Çó´Ë¹ý³ÌÖÐĦ²ÁÁ¦Ëù×öµÄ¹¦£®

Answer 1

·ÖÎö £¨1£©ÓÉÌâÒâ¿ÉÖªBÓëÅ̼äÒÑ´ïµÄ×î´ó¾²Ä¦²ÁÁ¦£¬¹Ê¾²Ä¦²ÁÁ¦³äµ±ÏòÐÄÁ¦£¬ÓÉÏòÐÄÁ¦¹«Ê½¿ÉÇóµÃ½ÇËٶȣ»
£¨2£©A¿ªÊ¼»¬¶¯Ê±£¬ËµÃ÷AÒÑ´ïµ½×î´ó¾²Ä¦²ÁÁ¦£¬ÓÉÏòÐÄÁ¦¹«Ê½¿ÉÇóµÃ½ÇËٶȣ»
£¨3£©ÒÀ¾Ý¶¯Äܶ¨Àí¿É¼ÆË㹦£®

½â´ð ½â£º£¨1£©µ±Çá¸Ë¿ªÊ¼³öÏÖµ¯Á¦Ê±£¬±íÃ÷BÓëÅ̼äµÄ¾²Ä¦²ÁÁ¦ÒÑ´ï×î´ó£¬Ôò£º
  0.4mg=m¦Ø₀²r_B
½âµÃ£º¦Ø₀=$\sqrt{\frac{0.4g}{{r}_{B}}}=\sqrt{\frac{0.4¡Á10}{0.3}}$=3.61rad/s£®
£¨2£©µ±A¿ªÊ¼»¬¶¯Ê±£¬±íÃ÷AÓëÅ̼äµÄ¾²Ä¦²ÁÁ¦Ò²ÒÑ´ï×î´ó£¬Ôò£º
¶ÔA£º0.4mg-F_T=m¦Ø²r_A
¶ÔB£ºF_T+0.8mg=m¦Ø²r_B
ÓÉÒÔÉÏÁ½Ê½Áª²¢´úÈëÊý¾Ý½âµÃ£ºF_T=0.8N
Ôòµ±A¡¢B¿ªÊ¼»¬¶¯Ê±£¬Çá¸Ë¶ÔAµÄµ¯Á¦´óСΪ0.8N£¬·½ÏòÔ¶ÀëÔ²ÐÄ£¬Ô²ÅÌת¶¯½ÇËٶȣº
¦Ø=$\sqrt{\frac{0.4mg-{F}_{T}}{m{r}_{A}}}=\sqrt{\frac{0.4¡Á10-0.8}{1¡Á0.2}}$=4rad/s£®£®
A¿ªÊ¼»¬¶¯Ê±Ô²Å̵ĽÇËÙ¶ÈΪ4rad/s£®
£¨3£©Ä¦²ÁÁ¦×ö¹¦Ê¹µÃϵͳ¶¯ÄÜÔö¼Ó£¬Óɴ˿ɵãº
$W=\frac{1}{2}{m}_{A}£¨{r}_{A}¦Ø£©^{2}$+$\frac{1}{2}{m}_{B}{£¨{r}_{B}¦Ø£©}^{2}$-$\frac{1}{2}{m}_{A}{£¨{r}_{A}{¦Ø}_{0}£©}^{2}$-$\frac{1}{2}{m}_{B}{£¨{r}_{B}{¦Ø}_{0}£©}^{2}$=0.02J£®
´ð£º£¨1£©µ±Ï¸ÏßÉÏ¿ªÊ¼³öÏÖÕÅÁ¦Ê±£¬Ô²Å̵ĽÇËÙ¶È3.61rad/s£®
£¨2£©µ±A¸ÕÒª»¬¶¯Ë²¼ä£¬Ô²Å̵ĽÇËÙ¶È4rad/s£»
£¨3£©µ±Ô²Å̵ĽÇËٶȴӣ¨1£©ËùÇóµÃµÄ¦Ø₀¿ªÊ¼»ºÂýÔö´ó£¬Ö±µ½A¸ÕÒª»¬¶¯Ë²¼ä£¬Çó´Ë¹ý³ÌÖÐĦ²ÁÁ¦Ëù×öµÄ¹¦0.02J£®

µãÆÀ ±¾Ì⿼²éÔ²ÖÜÔ˶¯ÖÐÁ¦ÓëÔ˶¯µÄ¹Øϵ£¬×¢Òâ±¾ÌâÖÐΪ¾²Ä¦²ÁÁ¦ÓëÉþ×ÓµÄÀ­Á¦³äµ±ÏòÐÄÁ¦£¬¹ÊӦעÒ⾲Ħ²ÁÁ¦ÊÇ·ñÒÑ´ïµ½×î´ó¾²Ä¦²ÁÁ¦£®