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

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

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

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

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its a 5.5 n

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