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applied mechanics !! | |||
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Posted by: ravindra.patil ® 01/16/2006, 01:51:23 Author Profile eMail author Edit |
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 Author Profile eMail author Edit |
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 !! -- ravindra.patil | Post Reply | Top of thread | Forum |
Posted by: Yates ® 01/16/2006, 04:35:48 Author Profile eMail author Edit |
2mvsquared/tfifth power |
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