神马作文网力矩做功相关,最高追加40分
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力矩做功相关,最高追加40分
As show in FLgure,a small ball of mass M tied with a string does a circle motion of radius R0 with the angular speed W0 on the smooth horizontal plane.Pull down the string untill the radius changes to the R0/2 .Find the new angular speed and the work done by the pull force.
As show in FLgure,a small ball of mass M tied with a string does a circle motion of radius R0 with the angular speed W0 on the smooth horizontal plane.Pull down the string untill the radius changes to the R0/2 .Find the new angular speed and the work done by the pull force.
OK
Solution:
(1)As the angle between F and r is 0 at any time,the Moment(力矩):M=0.
According to the formula:=dL/dt,the angular momentum L=L0(constant)
Becauce L=mwr^2,L0=mw0r0^2=mw(r0/2)^2,
Hence,the new angular speed:w=4w0
(2)The work done by the pull force(W) equals the change of the kinetic energy(Et-E0):W=Et-E0
E=1/2mv^2,v=wr,
So,E=1/2mw^2*r^2
Hence,W=Et-E0=1/2mw^2*r^2-1/2mw0^2*r0^2
=1/2m(w^2*r^2-w0^2-r0^2)
=1/2m[(4w0)^2*(r0/2)^2-w0^2*r0^2]
=3/2mw0^2*r0^2
需要中文的吗?
Solution:
(1)As the angle between F and r is 0 at any time,the Moment(力矩):M=0.
According to the formula:=dL/dt,the angular momentum L=L0(constant)
Becauce L=mwr^2,L0=mw0r0^2=mw(r0/2)^2,
Hence,the new angular speed:w=4w0
(2)The work done by the pull force(W) equals the change of the kinetic energy(Et-E0):W=Et-E0
E=1/2mv^2,v=wr,
So,E=1/2mw^2*r^2
Hence,W=Et-E0=1/2mw^2*r^2-1/2mw0^2*r0^2
=1/2m(w^2*r^2-w0^2-r0^2)
=1/2m[(4w0)^2*(r0/2)^2-w0^2*r0^2]
=3/2mw0^2*r0^2
需要中文的吗?