Physics Help Forum [SOLVED] circular + parabolic motion

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jul 20th 2009, 08:12 PM #1 Member   Join Date: Jul 2009 Posts: 86 [SOLVED] circular + parabolic motion hi all please help me A point object of mass m is connected to an inertialess string of length L. The other end of which is fixed to a point O. At time t = 0, the object is assumed to begin to move horizontally in a vertical plane from the bottom point A (OA = L) in the clockwise direction with an initial speed Vo. If $\displaystyle \sqrt{2gL}$ < Vo < $\displaystyle \sqrt{5gL}$(g=acceleration due to gravity), then at a point B the magnitude of the force acting on the object from the string becomes zero, where OB = L and the velocity of the object is perpendicular to OB. We restrict ourselves to the case 0 < $\displaystyle \theta$< $\displaystyle \pi$/2. From the point B, for a while, the object takes a parabolic orbit till a point C, where OC = L. In the case $\displaystyle \theta$ = $\displaystyle \pi$/3, find the angle $\displaystyle \varphi$ I've found the speed V = and the initial speed Vo = . I also got the maximum elevation (with respect to the location B) = What should i do next? thx
 Jul 21st 2009, 09:31 AM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 292 When the tension in the string becomes zero, the particle starts falling freely. Since it has a initial horizontal velocity it travels in the parabolic path. Let us see the method finding the angle. -y = -Lsin60 is the displacement of particle from B to the reference diameter. θ = 30 degree is the angle of projection. x is the range of the projectile on the reference line. vsinθ is the vertical component of the velocity v = sqrt(gLsin60). vcosθ is the horizontal component of the velocity v. t = x/vcosθ Using the formula y = ut + 1/2*gt^2 we get -Lsqrt3/2 = vsin30*x/vcos30 - 1/2*g*x^2/v^2cos^230 -L(sqrt3)/2 = x/sqrt3 - 1/2*g*x^2/gLsin60*3/4 -L(sqrt3)/2 = x/sqrt3 - 1/2*g*x^2/gL*(sqrt3)/2*3/4 -9L^2 = 6Lx - 8x^2 Or 8x^2 - 6Lx - 9L^2 = 0. Solve for x x = [6L + sqrt( 36L^2 + 4*8*9L^2)]/2*8 x = [ 6L + 18L]/16 = 3L/2 So the particle crosses the reference diameter at a distance 3L/2 from the foot of y. If P is the foot of the y, then OP = Lcos60 = L/2 and distance from P to the circumference of the circle along the diameter is L/2 + L = 3L/2 = x. Hence the point C must be on the reference diameter.Hence φ =π.
 Jul 22nd 2009, 02:40 AM #3 Member   Join Date: Jul 2009 Posts: 86 wow thx a lot

 Tags circular, motion, parabolic, solved