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A thin rod of mass M and length l is suspended vertically from a frictionless pivot at its upper end. A mass m of putty traveling horizontally with a speed v srikes the rod at its CM and sticks there. How high does the bottom of the rod swing?

User Squash
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1 Answer

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Final answer:

The bottom of the rod swings to a height of (m * h) / (M + m) when the putty sticks to the rod.

Step-by-step explanation:

The height to which the bottom of the rod swings can be determined by conserving energy. Initially, the system has potential energy due to the elevated position of the putty. As the putty sticks to the rod and the system swings, this potential energy is converted to gravitational potential energy when the bottom of the rod reaches its highest point.

Setting up the conservation of energy equation, we have:

E_initial = E_final,

where E_initial is the initial total energy and E_final is the final total energy.

The initial energy is given by:

E_initial = m * g * h,

where m is the mass of the putty, g is the acceleration due to gravity, and h is the initial height of the putty.

The final energy is given by:

E_final = M * g * H,

where M is the mass of the rod (including the putty), g is the acceleration due to gravity, and H is the maximum height reached by the bottom of the rod.

Equating the initial and final energy equations, we have:

m * g * h = (M + m) * g * H.

Simplifying the equation, we can solve for H:

H = (m * h) / (M + m).

Therefore, the height to which the bottom of the rod swings is (m * h) / (M + m).

User Victorwoo
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