A 2.50 kg disk of diameter D=42.0cm is supported by a rod of length L=76.0 cm an

Question

A 2.50 kg disk of diameter D=42.0cm is supported by a rod of length L=76.0 cm and negligible massthat is pivoted at its end. a) With the massless torsionspring unconnected, what is the period of oscillation? b) With thetorsion spring connected, the rod is vertical atequillibrium. What is the torsion constant of the spring ifthe period of oscillation has been decreased by
0.500 s? A 2.50 kg disk of diameter D=42.0cm is supported by a rod of length L=76.0 cm and negligible massthat is pivoted at its end. a) With the massless torsionspring unconnected, what is the period of oscillation? b) With thetorsion spring connected, the rod is vertical atequillibrium. What is the torsion constant of the spring ifthe period of oscillation has been decreased by
0.500 s? A 2.50 kg disk of diameter D=42.0cm is supported by a rod of length L=76.0 cm and negligible massthat is pivoted at its end. a) With the massless torsionspring unconnected, what is the period of oscillation? b) With thetorsion spring connected, the rod is vertical atequillibrium. What is the torsion constant of the spring ifthe period of oscillation has been decreased by
0.500 s?

Explanation / Answer

a)the period of oscillation T = (1/2) * (L/g)^(1/2) L = 76.0 cm = 76.0 * 10^-2 m g = 9.8 m/s^2 b)when the period of oscillation is decreased then w = (2/T - 0.500) We know from the relation w = (k/m)^(1/2) or (2/T - 0.500) = (k/m)^(1/2) or (2/T - 0.500)^2 = (k/m) or k = [(2/T - 0.500)^2/m] m = 2.50 kg

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