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A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with AB horizontal. Suddenly the end A is fixed when the speed of the rod is v. The angular speed with which the rod begins to rotate is:

(a) v/a
(b) v/(2a)
(c) v/(3a)
(d) v/(4a)

User Nict
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1 Answer

3 votes

Final answer:

The angular speed at which the rod begins to rotate after the end A is fixed is v/(3a), making option (c) the correct answer based on the conservation of angular momentum.

Step-by-step explanation:

The angular speed with which the rod begins to rotate after the end A is fixed can be determined using the principle of conservation of angular momentum.

Prior to fixing at A, the rod moves horizontally translating uniformly without rotation, so its angular momentum about A is mv×a (since A is at a distance a from the center of mass). After A gets fixed, the rod begins rotating about A. To find the angular speed ω, use the conservation of angular momentum L_initial = L_final, which gives us mv×a = I×ω, where I is the moment of inertia of the rod about A for rotation after fixing.

For a rod of length 2a and mass m rotating about its end, I = 1/3×m×(2a)². Solving this equation gives ω = 3mv/(m×(4a)²), which simplifies to ω = v/(3a).

When the end A of the uniform rod AB is fixed suddenly, the rod will start rotating about point B. This is because the fixed end A creates a torque about point B, causing the rod to rotate.

The angular speed with which the rod begins to rotate can be determined using the equation:

angular speed = linear speed / distance from the fixed end

In this case, the linear speed is v and the distance from the fixed end is a, so the angular speed is v/a.

Therefore, the correct answer is (a) v/a.

User Jitender Mahlawat
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