(a) What should the distance d be so that the period of this pendulum is 2.5 s?

Question

(a) What should the distance d be so that the period of this pendulum is 2.5 s?
_____m

(b) Suppose that the pendulum clock loses 5.00 min/d. To make sure your grandmother will not be late for her quilting parties, you decide to adjust the clock back to its proper period. How far and in what direction should you move the disk to ensure that the clock will keep perfect time?
______cm upward/downward

The figure below shows the pendulum of a clock in your grandmother's house. The uniform rod of length L = 2 m has a mass m = 0.8 kg. Attached to the rod is a uniform disk of mass M = 1.2 kg and radius 0.15 m. The clock is constructed to keep perfect time if the period of the pendulum is What should the distance d be so that the period of this pendulum is 2.5 s?

Explanation / Answer

Torque = I*alpha

mg(L/2)*sin(theta) + Mgdsin(theta) = -(mL^2/3 + Mr^2/2 + Md^2)*alpha

0.8*9.8*1*(theta) + 1.2*9.8*d(theta) = -(0.8*4/3 + 1.2*0.15^2/2 + 1.2*d^2)*alpha

alpha = -(7.84+11.76d)(theta)/(1.08+1.2d^2)

w = sqrt((7.84+11.76d)/(1.08+1.2d^2))

a) T = 2.5s, w = 2pi/T = 2*3.142/2.5 = 2.5136 rad/s

sqrt((7.84+11.76d)/(1.08+1.2d^2)) = 2.5136

(7.84+11.76d)/(1.08+1.2d^2) = 6.3181

d = 1.633m

b) Current time period = 3.5-3.5*5/(24*60) = 3.48784 s

w = 2pi/T = 1.8016

(7.84+11.76d)/(1.08+1.2d^2) = 3.246

d = 3.3494m

Actual = 3.5s

(7.84+11.76d)/(1.08+1.2d^2) = 3.2235

d = 3.3741m

change = 3.3494-3.3741 = 0.0247m downwards

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