You drill a hole at the end of a solid thin rod of length 1.7 m, and then attach

Question

You drill a hole at the end of a solid thin rod of length 1.7 m, and then attach the rod through this hole onto an approximately frictionless pivot. You then release the rod from rest at a reasonably small angle from vertical, so that it executes simple harmonic motion.

a.) What will be its oscillation period?
_____ s
b.) If the bottom of the rod varies its height above the ground by 1 cm during the oscillation, what was its initial angle from the vertical?
(Note: You may assume the width of the stick is negligible to make this well defined.)
______ degrees

Explanation / Answer

a) This is just a simple pendulum. The period of oscillation is

T = 2pi sqrt (L/(2g)) = 2*pi()*sqrt(1.3/(2*9.8)) = 1.6 secs.

here you use half the length of the rod as that where the center of mass is located (assuming that the rod is uniform).

b) This part is trigonometry. You have that the difference in the projected distance in the vertical when the rod is oscillating is 2cm. This is

1.3 - 1.3* cos (angle) = 0.02

solving for angle in the equation above we get

angle = acos ((1.3-0.02)/1.3) = 10. degrees

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.