# Sphere and Bar freefall [closed]

Question The problem consists of a bar of weight $$W$$ to which a sphere of also of weight $$W$$ is fitted at one end. The conbination is released from rest, with the bar horizontal. How will the combination fall if air resistance can be neglected?

For convenience, I have marked the C.G. of the sphere as $$G_S$$ and that of the bar only as $$G_B$$. The C.G. of the combination is halfway between them, shown by the green $$G_C$$. My answer is that the combination will rotate about $$G_B$$ (centre of the bar) as it falls. I wonder if I am right.

Many thanks.

## closed as off-topic by John Rennie, Bill N, PM 2Ring, eranreches, garypMay 27 at 20:44

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• Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. – John Rennie May 27 at 10:31
• Hint: What would happen if you were weightless in a satellite, and released the ball + bar there? – PM 2Ring May 27 at 10:34
• The ball+bar would be at rest in the satellite. – BrotherBob May 27 at 10:41
• Correct. So in free fall, with no air resistance, the bar won't rotate. – PM 2Ring May 27 at 11:27
• To be possibly on-topic, you should explain your physics-based reasoning. Don't just make a guess and ask "Am I right?" Use physics concepts like force, torque, conservation of momentum, angular momentum, etc. – Bill N May 27 at 12:01

My answer is that the combination will rotate about about GB (center of the bar) as it falls. I wonder if I am right.

No you are not correct. Each section of the bar and sphere will have the same downward acceleration. Therefore there will be no rotation and the entire assembly will impact the ground horizontally.

What you are describing is what would happen if you place the assembly on a fulcrum on the ground underneath $$G_B$$. If you do that, the assembly will rotate about $$G_B$$ (counterclockwise) because the sum of the moments about $$G_B$$ will not be zero, a requirement for static equilibrium. But the fulcrum will exert an upward force preventing the assembly from downward motion. Only without the support by the fulcrum will the assembly be in free fall.

In a vacuum all objects in free fall near the surface of the earth fall with the same acceleration $$g$$. Imagine the bar and sphere being separated. In free fall each will hit the ground at the same time. Simply connecting them together doesn’t change that fact. The assembly will be horizontal when impact occurs.

Hope this helps.

• Excellent. Thank you a lot. – BrotherBob May 28 at 9:56
• @user6163 Was my answer acceptable? – Bob D May 28 at 15:23
• Yes it was excellent. It is the way an answer to a question on here should be given. Thank you very much. – BrotherBob May 28 at 19:42
• @BrotherBob Great. The only reason I asked is because you didn’t formally “accept” the answer – Bob D May 29 at 6:15
• I am new here. What do you have to click to say that you "accept" someone's answer? – BrotherBob May 29 at 11:23

Please remember that for an external force like gravity here will act on the Center of Mass. To check wether an object will rotate or not, You can check if there is any unbalanced Torques acting on the CM or any stationary point. Here since $$mg$$ acts on the CM , $$\tau_{CM}=0$$ and object doesn't rotate.

The bar will maintain a horizontal attitude until it hits the ground. If you remember the experiment conducted by an astronaut on the moon,wher a hammer & a feather were released simultaneously & both struck the ground at the same time,the result can be applied to this problem. Consider the sphere to be the hammer & the bar to be the feather. They should both strike the ground at virtually the same moment,though the shape of the sphere enables it to reach the ground very slightly sooner as it falls a slightly shorter distance. I know that in your combination the two parts are the same weight & joined together,whereas in the moon experiment they weren't,but the principle still holds good.