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Suppose you have a mass m hanging from the end of a rod. take the rod to be massless, take its length to be r, and let it be able to swing freely about the other end. in the beginning the rod is horizontal and the mass is moving down with an initial speed vo. what initial kinetic energy is required for the mass to pivot three quarters of the way around a complete circle to the vertical position?

a. mgr
b. mg(2r)
c. mg(3r)
d. none of these

User ShiyamTJ
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2 Answers

3 votes

Answer:

mgr

Step-by-step explanation:

Ei = Ef

Ei = Ek

Ef = Eg

Ek (m)

Eg (r)

User Sandeep Modak
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2 votes

Final answer:

The initial kinetic energy required for a mass to pivot from a horizontal position three quarters around a circle to the vertical is equal to the potential energy it would have at the top, which is mgr.

Step-by-step explanation:

The question asks about the initial kinetic energy required for a mass m to pivot three quarters of a way around a circle starting from a horizontal position until it reaches the vertical. To solve this problem, we can use the conservation of mechanical energy principle. The initial kinetic energy plus the initial potential energy of the system must equal the final potential energy at the vertical position because at that instant the kinetic energy is zero (the mass is momentarily at rest).

At the beginning, when the rod is horizontal to the ground, the potential energy of the mass is zero, and it has an initial kinetic energy. When the mass swings up to three quarters of the circular path (which is equivalent to the mass being at the top), it will have a potential energy equal to mgr because it has risen by a distance r (the length of the rod), and it must have enough initial kinetic energy to do so. Therefore, the answer is: a. mgr.

User Frank Pearson
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