This question is incomplete, the complete question is;
1.80 kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. the period for small angle oscillations is 0.940 s.
a) what is the moment of inertia of the wrench about an axis through the pivot?
b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position
Answer:
a) the moment of inertia of the wrench about an axis through the pivot is 0.0987 Kg.m²
b) the angular speed of the wrench as it passes through the equilibrium position is 2.6559 rad/s
Step-by-step explanation:
Given that
mass m = 1.80 kg
L or d = 0.250 m
period T = 0.940 sec
g = 9.8
a)
we take a look at the expression for the moment of inertia
I = T²mgL / 4π²
we substitute
I = ((0.940)² × 1.80 × 9.8 × 0.250) / 4 × π²
I = 3.896676 / 4 × π²
I = 0.0987 Kg.m²
Therefore the moment of inertia of the wrench about an axis through the pivot is 0.0987 Kg.m²
b)
If the wrench is initially displaced 0.400 rad from its equilibrium position,
the angular speed of the wrench as it passes through the equilibrium position = ?
Using conservation of energy;
mg × d × ( 1 - cos∅) = 1/2×I×w²
we substitute
1.80 × 9.8 × 0.250 × ( 1 - cos(0.400 rad)) = 1/2 × 0.0987 × w²
4.41 × ( 1 - 0.921060994 ) = 0.04935 × w²
0.000441 = 0.04935w²
w² = 0.3481 / 0.04935
w² = 7.0536
w = √7.0536
w = 2.6559 rad/s
Therefore, the angular speed of the wrench as it passes through the equilibrium position is 2.6559 rad/s