What is the Speed of the Upper End of a Falling Pole?

In summary, the question is about finding the speed of the upper end of a 3.30m long pole that is balanced vertically and given a small push. The formula to use is MgH = 1/2 MV^2 + 1/2 Iw^2, where "w" is the rotational velocity. Since there is no slippage, all energy is rotational and the moment of inertia is 1/3mL^2. The potential energy can be found by using mgh, with the center of mass at h = L/2. When plugging in v/r for w, the length of the pole should be used for "r".
  • #1
HPVic03
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Please help me get this, it's due tonigh tonight on webassign. The question is:

A 3.30 m long pole is balanced vertically on its tip. It is given a tiny push. What will be the speed of the upper end of the pole just before it hits the ground? Assume the lower end does not slip.

I know to use MgH = 1/2 MV^2 + 1/2 Iw^2

but when you plug in for w, what do you use? w is rotational velocity and you have to know the time frame, and it doesn't give you that. how do you do this problem??
 
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  • #2
My editing caused double posted for some reason. :\
 
  • #3
Since there is no slippage, all energy is rotational (PE = RE.) Your moment of inertia in this case is [itex]\frac{1}{3}mL^2[/itex], where "L" is the length of the pole. You can find the potential energy when the pole is standing up (mgh) but remember that the center of mass is at h = L/2. And remember that [itex]v = r\omega[/itex]

Oh, and let me guess, Physics - Giancoli :-p
 
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haha yeah its giancoli. thanks so much for your help, but i still have one question though. you you are plugging in v/r for w, what do you use for r?
 
  • #5
HPVic03 said:
haha yeah its giancoli. thanks so much for your help, but i still have one question though. you you are plugging in v/r for w, what do you use for r?

The length of the pole because it asks for the speed of the top end of the pole which is 3.30m from the pivot point.
 
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  • #6
hey thanks i got it
 
  • #7
No Problem :smile:
 

FAQ: What is the Speed of the Upper End of a Falling Pole?

1. What is rotational kinetic energy?

Rotational kinetic energy is the energy possessed by an object due to its rotational motion. It is calculated using the formula KE = 1/2 Iω^2, where I is the moment of inertia and ω is the angular velocity of the object.

2. How is rotational kinetic energy different from linear kinetic energy?

Rotational kinetic energy is the energy associated with the rotational motion of an object, while linear kinetic energy is the energy associated with the linear motion of an object. They are different because rotational motion involves movement around an axis, while linear motion involves movement in a straight line.

3. What factors affect rotational kinetic energy?

The factors that affect rotational kinetic energy are the moment of inertia, the angular velocity, and the mass of the object. A larger moment of inertia or angular velocity will result in a higher rotational kinetic energy, while a larger mass will result in a lower rotational kinetic energy.

4. How is rotational kinetic energy related to work and power?

Rotational kinetic energy is related to work and power through the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Power, which is the rate at which work is done, is also related to rotational kinetic energy through the formula P = τω, where τ is the torque applied to the object and ω is the angular velocity.

5. How is rotational kinetic energy used in real-life applications?

Rotational kinetic energy is used in various real-life applications, such as in the operation of machines, vehicles, and sports equipment. It is also used in the generation of electricity, as in hydroelectric dams and wind turbines, where the rotational motion of turbines is converted into electrical energy.

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