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In this forum there was one problem. Two heavy balls of equal mass M are attached to long light rod. The rod is held vertical and allowed to fall, say towards right. What will be the speed of each ball just before the upper ball touches the ground.
I can interpret this problem in three different ways.
A) The lower ball is fixed to the end of the rod and is allowed to move smoothly in a socket fixed on the ground. The upper ball is detachable. (Attach a small platform and keep the ball on it.) As the rod falls towards right, we can see the following changes in the ball. 1) The KE increases. 2) PE decreases. 3) Component of the weight of upper ball the along the rod decreases. And 4) Centrifugal force on the ball increases. At one particular position of the ball,
component of the weight along the rod is equal the centrifugal force θ. At that point the ball detaches from rod. In this position if the rod makes an angle θ with the vertical, the component of the weight along the rod is Mgcosθ. In this position if the ball has fallen a height h vertically and has acquired a velocity v, then Mgh = ½*Mv^2 or 2Mgh = Mv^2. The centrifugal force on the ball is Mv^2/L, where L is the length of the rod. Hence Mgcosθ = Mg(L-h)/L = Mv^2/L = 2Mgh/L. Or
L –h = 2h or 3h = L or h = L/3. Irrespective of the length of the rod, the ball detaches from the rod at an angle whose cosine is always 2/3. At this point the horizontal component of the velocity is vcos(θ) =2v/3 = 2/3[(2gh)^1/2] and this velocity remains the same till it reaches the ground. The vertical component goes on increasing and reaches the value
(2gL)^1/2
B) The lower ball is allowed to move smoothly on the ground. In this case the component of the weight along the rod pushes it towards left until the ball detaches from the rod. Later on it continues to moves in the same direction with constant velocity equal to the horizontal component of the velocity of the upper ball.
C) Both the balls are attached to the rod. In this case, the lower ball moves towards the left with first increasing and then decreasing velocity until the component of the weight is equal to the centrifugal force. Later on both the ball will move towards right with increasing velocity until it reaches the ground. At that instant the velocity of each ball will be
1/2{2gL}^1/2
I can interpret this problem in three different ways.
A) The lower ball is fixed to the end of the rod and is allowed to move smoothly in a socket fixed on the ground. The upper ball is detachable. (Attach a small platform and keep the ball on it.) As the rod falls towards right, we can see the following changes in the ball. 1) The KE increases. 2) PE decreases. 3) Component of the weight of upper ball the along the rod decreases. And 4) Centrifugal force on the ball increases. At one particular position of the ball,
component of the weight along the rod is equal the centrifugal force θ. At that point the ball detaches from rod. In this position if the rod makes an angle θ with the vertical, the component of the weight along the rod is Mgcosθ. In this position if the ball has fallen a height h vertically and has acquired a velocity v, then Mgh = ½*Mv^2 or 2Mgh = Mv^2. The centrifugal force on the ball is Mv^2/L, where L is the length of the rod. Hence Mgcosθ = Mg(L-h)/L = Mv^2/L = 2Mgh/L. Or
L –h = 2h or 3h = L or h = L/3. Irrespective of the length of the rod, the ball detaches from the rod at an angle whose cosine is always 2/3. At this point the horizontal component of the velocity is vcos(θ) =2v/3 = 2/3[(2gh)^1/2] and this velocity remains the same till it reaches the ground. The vertical component goes on increasing and reaches the value
(2gL)^1/2
B) The lower ball is allowed to move smoothly on the ground. In this case the component of the weight along the rod pushes it towards left until the ball detaches from the rod. Later on it continues to moves in the same direction with constant velocity equal to the horizontal component of the velocity of the upper ball.
C) Both the balls are attached to the rod. In this case, the lower ball moves towards the left with first increasing and then decreasing velocity until the component of the weight is equal to the centrifugal force. Later on both the ball will move towards right with increasing velocity until it reaches the ground. At that instant the velocity of each ball will be
1/2{2gL}^1/2