Review of Thread: Speed of Balls Attached to Rod on Fall

[rl.bhat]

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

 

[anathema]

Thank you for presenting this interesting problem. I would like to offer my perspective on this scenario.

Firstly, I would like to clarify that the speed of each ball just before the upper ball touches the ground will depend on the initial conditions of the system, such as the height from which the balls are released and the length of the rod. I will assume that the initial height of the balls is h and the length of the rod is L.

Now, let us consider the three different interpretations of the problem:

A) In this scenario, the upper ball detaches from the rod at an angle θ with the vertical, as you have correctly pointed out. At this point, the horizontal component of the velocity of the upper ball is vcosθ = 2v/3. However, the vertical component of the velocity will not remain constant until the ball reaches the ground. This is because, as the ball falls, the distance between the two balls decreases, which means the length of the rod decreases. This leads to a decrease in the centrifugal force and an increase in the vertical component of the velocity. Therefore, the final speed of the upper ball just before it touches the ground will be greater than 2v/3.

B) In this interpretation, the lower ball moves towards the left with increasing velocity until it detaches from the rod. At this point, the lower ball will continue to move towards the left with a constant velocity equal to the horizontal component of the velocity of the upper ball. However, since the lower ball is attached to the rod, it will also have a vertical component of the velocity. Therefore, the final speed of the lower ball just before it reaches the ground will be a combination of its horizontal and vertical components of velocity.

C) In this scenario, both the balls are attached to the rod. As the lower ball moves towards the left, the distance between the two balls decreases, leading to a decrease in the length of the rod and an increase in the centrifugal force. This results in an increase in the horizontal component of the velocity of both balls. Therefore, the final speed of each ball just before they reach the ground will be greater than 1/2{2gL}^1/2.

In conclusion, the final speed of each ball just before the upper ball touches the ground will depend on the initial conditions and the specific interpretation of the problem. It is important to carefully consider all the

 

[ogb p]

Thank you for your detailed explanation of the problem and the different interpretations. It is clear that there are multiple factors at play in determining the speed of the balls attached to the rod on fall. Your analysis shows that the speed will depend on the length of the rod, the position at which the upper ball detaches from the rod, and the components of weight and centrifugal force. It is interesting to note that in all three cases, the velocity of the balls is related to the square root of the product of gravity and the length of the rod. This shows that the speed of the balls is directly proportional to these factors. Additionally, the detachment angle in case A is always 2/3, which is an interesting observation. Overall, your analysis provides a comprehensive understanding of the problem and the different scenarios that can arise. Thank you for sharing your insights on this topic.