Inclined plane problem help please

[Vatsa]

If the angle of a frictionless inclined plane 'a' is such that a=wt

where w is angular velocity and a is a function of t
and an object is sliding down that plane then...
if m is the mass of the object

m(dv/dt)=mgsin(wt)

solving for v i got

v=(-gcos(wt)/w)

Why is the velocity negative and Why is that when i substitute t=0 v is greater than 0?

 

[kushan]

Hello Vatsa
What is the question asking for?
Can you please post the complete question , ?

 

[Vatsa]

Well...i just like solving problems i come up with...

So suppose you have a rotating inclined plane and the angle of inclination is a function of time such that...angle 'x'=angular velocity(w)*time(t)

now mgsin(x) is the horizotal component of g so the body on the frictionless inclined plane experiences a horizontal force=mgsin(x)..let the velocity along the incline plane be v
where m is the mass of the body
so...

m(dv/dt)=mgsin(x)

solving for v i got the above result...

my question is that even at t=0 why is there velocity..because at t=0 angle=0 and why is the velocity negative?

 

[CWatters]

As the slope and mass are rotating about the bottom of the slope isn't there also a Centrifugal force acting up the slope?

 

[kushan]

and when you frame the equation , integrate it voila!

 

[Nogueira]

I'm not really sure, but I think you are not using the exact formula, the equation: mgsin(x), is not giving you the velocity, instead, is giving you the potencial energy of the body, that is why with t=0 gives you a value, because, while stopped, the body has potential energy instead of cinectic energy. If you want to find the velocity, you have to do the variation of the mechanichal energy so it will be: Em= Emf-Emi⇔Em= (Ecf+Epf)-(Eci+Epi)⇔Em=Ecf-Emi (because Epf=0 and Eci=0). The value is negative, because g=-9,8m/s.