applied mechanics !! Sad
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Posted by: ravindra.patil ®

01/16/2006, 01:51:23

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i have a vry simple questn frm applied mechanics!!

a pully is driven by a motor !!!

a rope is suspended on pully !!

on both ends of rope two same masses are attached !!!

suppose pully is rotated by motor so one mass move upward & ader down ward with velocity V !!!

den what is the minimum power is required by motor to stop the masses in time T ????








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Re: applied mechanics !!
Re: applied mechanics !! -- ravindra.patil Post Reply Top of thread Forum
Posted by: zekeman ®

01/19/2006, 22:33:28

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To absorb the energy, utilizing a motor you don't supply power, you absorb power;this is called dynamic braking. For example, if you use a DC PM motor in a generator configuration, you shunt the terminals with a resistance, R, and the rotational opposing torque is proportional to v/R, where v is the instantaneous angular velocity of the pulley. Then, since the energy differential equation is
Td@=-d(mv^2/2)=-mvdv
where
@= angular displacement of pulley
m=sum of the masses on pulley
and since d@=vdt, T=kv
v= instantaneous pulley angular velocity
k= constnant
t= time
The differntial equation becomes
dt=-m/kg(dv/v)
which after integration yields a theoretical value of time of infinity. But practically, since coulomb and windage friction are present comes to rest more quickly. The more usual paramete r of interest is the stopping distance, @, which can be obtained using the first equation
Td@=-mvdv
kvd@=-mvdv
d@=-m/kdv which very simply integrates to
@=mV/k
V is the intial angular velocity of the pulley= linear vel/pulley radius.








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Re: applied mechanics !!
Re: applied mechanics !! -- ravindra.patil Post Reply Top of thread Forum
Posted by: Yates ®

01/16/2006, 04:35:48

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2mvsquared/tfifth power







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