Question

A plumb-line is set up on a rotating disk and makes an angle of a with the vertical, as in Fig. The distance r from the point of suspension to the axis of rotation is known, and so is the length l of the thread. Find the angular velocity of rotation.

• A. w = $\sqrt{\frac{g\tan\alpha}{r\mathcal{+ l}\sin\alpha}}$
• B. w = $\sqrt{\frac{g\sin\alpha}{r + \mathcal{l}\tan\alpha}}$
• C. w = $\sqrt{\frac{g}{r\mathcal{+ l}\sin\alpha}}$
• D. w = $\sqrt{\frac{g\cos\alpha}{r + \mathcal{l}\sin\alpha}}$

Answer 1

Answer: (A)

w = $\sqrt{\frac{g\tan\alpha}{r\mathcal{+ l}\sin\alpha}}$

Answer 2

The plumb-line is so set up that the resultant of its weight mg and the tension in the thread T produces a centripetal force F = mw²R (fig.). Clearly R = r + l sin a. Therefore,

w² = $\frac{g\tan\alpha}{r\mathcal{+ l}\sin\alpha}$, w = $\sqrt{\frac{g\tan\alpha}{r + \mathcal{l}\sin\alpha}}$.