Question

The 500-mg balance wheel of a certain clock is made up of a thin metal ring of radius 16 mm connected by spokes of negligible mass to a fine suspension fiber as in (Figure 1) . The back-and-forth twisting of the fiber causes the wheel to move in simple harmonic motion with period T. The clock ticks four times each second, with the interval between ticks equal to T/2.

Part A

Determine the rotational inertia I of the balance wheel.

Express your answer with the appropriate units.

Part B

What is the torsional constant of the fiber that drives the balance wheel?

Express your answer with the appropriate units.

Answer 1

U new India, Part A Mass of balance wheel, m= 500 ung = 0.0005 kg Radius of the ving, r = 16 mm = 0.016 m So, rotationa inertia of the balance wheel, I = mxrl = (0.0005) x (0.016)2 = .2810+7 Kgove2 | Part B Given that the clock ticks 4 times each secondo Means, it sticks & times in 2 seconds: Internal a between the locks = 1/2 Hence, frequency of = 2 Hz So, w= 20 f = 2x3.14x2 = 12:56 rad/s. Now, wa [ka torsional constant] (12.156)² x 1,288107 | 2.02 x 105 4 = Augwer; (Inc. of All Tax