Question Number 103072 by mr W last updated on 12/Jul/20

Commented byajfour last updated on 12/Jul/20

Commented bymr W last updated on 12/Jul/20

a wedge with mass M has the shape  of a quadrant. a small block with  mass m is released at the top of the  wedge as shown and can slide down  frictionlessly.  find the velocity of both objects at the  instant when the small block leaves  the wedge.  (the friction between wedge and   ground is small enough so that the  wedge can always move)

Commented byDwaipayan Shikari last updated on 12/Jul/20

v=(√(2gR))  If there is no friction ,so Work done by nonconservative force  is zero  So  (1/2)mv_m ^2 =mgR⇒v_m =(√(2gR))  The wedge can move sir?

Commented bymr W last updated on 12/Jul/20

it is said that the wedge can move,  but with friction μ!

Commented byajfour last updated on 13/Jul/20

let radius of quadrant be r.  System: Wedge_(−)   Nsin θ+Mg=R       ......(i)  Ncos θ−μR=mA   .......(ii)  block (relative to wedge):_(−)   mAcos θ+N−mgsin θ=((mu^2 )/r)   (iii)  mAsin θ+mgcos θ=((mudu)/(rdθ))     ..(iv)  ________________________  from (i) & (ii)  N(cos θ−μsin θ)−μMg=mA  now using (iii)  (mgsin θ+((mu^2 )/r)−mAcos θ)×  (cos θ−μsin θ)−μMg=mA  ⇒  A=(((cos θ−μsin θ)(gsin θ+(u^2 /r))−μ((M/m))g)/(1+cos θ(cos θ−μsin θ)))  substituting this in (iv)  ((udu)/(rdθ))=(([(cos θ−μsin θ)(gsin θ+(u^2 /r))−μ((M/m))g]sin θ)/(1+cos θ(cos θ−μsin θ)))          +gcos θ  If we are able to solve this D.E.  we get u(θ).  Let   (u^2 /(rg))=t   ;  sin θ=s  ⇒  ((udu)/(rg))=(dt/2)   &   cos θdθ=ds  (1/2)(dt/ds)=((t[(1−((μs)/(√(s^2 +1))))(s+t)−μ((M/m))])/((1−s^2 −((μs)/(s^2 +1)))))+1  then we obtain A(θ)  then  V(θ)   and  v(θ)  we then set   θ=(π/2) ,   angular  position of block when it leaves  contact with wedge;   (for m≪M)

Commented bymr W last updated on 12/Jul/20

if the ground is smooth, i.e. μ=0,  then we can get V and u without  D.E., right?

Commented byajfour last updated on 12/Jul/20

the diff. eq. would be much  simpler then Sir....

Commented bymr W last updated on 12/Jul/20

that′s true...

Commented bymr W last updated on 12/Jul/20

if we only want to determine the  velocities V and u, we don′t need to  solve the D.E., because we have  MV=mu_x .

Answered by ajfour last updated on 13/Jul/20

Yes Sir, you are right!  For μ=0 ;  m≪M  (1/2)MV^2 +(1/2)m(v−V)^2 =mgr     &  MV=m(v−V)