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A 1.80- kgkg monkey wrench is pivoted 0.250 mm 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?

Respuesta :

superdele4eva

Answer:

A. 1.125×10^-7 kgm^2

B. 6.64875 rad/s

Explanation:

The moment of inertia is defined as a quantity expressing a body's tendency to resist angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

A. Moment of inertia = m1✖r1^2

=1.80 x (2.5x10^-4)^2

= 6.25x10^8 x 1.80

= 1.125 x 10^-7 kgm^2.

B. w is represented as Angular speed.

V is velocity, T is time in period.

Velocity= distance / time.

V= 2.5x10^-4 / 0.940

V= 2.6595 metre per seconds

w= v/r

w= 2.6595 / 0.400

w= 6.64875 rad/s.

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