A uniform rod of mass 444 g and length 1.41 m is pivoted about one end and oscillates in a vertical plane. Find the period of oscillations if the amplitude of the motion is small. The acceleration of gravity is 9.8 m/s^2.
Answer in units of s.
To find the period of oscillations for a uniform rod pivoted about one end, we can use the formula for the period of a simple pendulum.
The formula for the period of a simple pendulum is:
T = 2π√(L/g)
Where:
T = period of oscillation
L = length of the pendulum
g = acceleration due to gravity
In this case, the length of the pendulum is equal to the length of the uniform rod, which is given as 1.41 m. The acceleration due to gravity is given as 9.8 m/s^2.
Plugging these values into the formula, we get:
T = 2π√(1.41/9.8)
Calculating this expression, we find:
T ≈ 2π√(0.1439)
T ≈ 2π(0.3795)
T ≈ 2.384 seconds
Therefore, the period of oscillations for the given uniform rod is approximately 2.384 seconds.