均质杆AB长为L,质量为m,角速度为ω,则AB杆的动能和动量为
dm=(m/L)dx
动能为:
(1/2)∫(0,x)dm*(xω)²
=(1/2)∫(0,x)dm*(xω)²
=(1/2)∫(0,x)(m/L)dx*(xω)²
=(1/2)(m/L)(ω)²∫(0,x)x²dx
=mω²/(2L)(1/3)x³
=mω²/(6L)*x³
所以所求动能为:
T=mω²/(6L)*(L/4)³+mω²/(6L)*(3L/4)³
=mω²/(6L)L³(28/64)
=(7/96)mL²ω²
所以答案是B
动能为:
(1/2)∫(0,x)dm*(xω)²
=(1/2)∫(0,x)dm*(xω)²
=(1/2)∫(0,x)(m/L)dx*(xω)²
=(1/2)(m/L)(ω)²∫(0,x)x²dx
=mω²/(2L)(1/3)x³
=mω²/(6L)*x³
所以所求动能为:
T=mω²/(6L)*(L/4)³+mω²/(6L)*(3L/4)³
=mω²/(6L)L³(28/64)
=(7/96)mL²ω²
所以答案是B
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