How Do You Calculate the Moment of Inertia for a Swinging Monkey Wrench?

[cb]

I need hlep with this question...

A 1.8-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 osciallations 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 throught the equilibrium position?

I think it did this right, but hopefully someone can check my work.

$$d\sin(\theta) = 0.250 m$$

$$\theta\rightarrow 0 \therefore d= 0.250m$$

$$T = 2\pi\sqrt{I / mgd}$$

$$I = -9.87 X 10^{-1}$$.

Now, I'm not sure how to start part b. I would use energy methods, but I'm not sure how the use it in this situation. Could someone help me with that part.

 

[Doc Al]

I think it did this right, but hopefully someone can check my work.

$$d\sin(\theta) = 0.250 m$$

$$\theta\rightarrow 0 \therefore d= 0.250m$$

I don't know what you are doing here. d in the formula below for period is not a displacement, but the distance from pivot to center of mass. Nonetheless, d = 0.25m, as given.

$$T = 2\pi\sqrt{I / mgd}$$

$$I = -9.87 X 10^{-1}$$.

Does a negative rotational inertia make sense? And don't forget units.

Now, I'm not sure how to start part b. I would use energy methods, but I'm not sure how the use it in this situation. Could someone help me with that part.

Mechanical energy is conserved. As the pendulum swings down from its initial point, gravitational PE is transformed to rotational KE.

 

[wais]

For part a, your solution is correct. To solve part b, we can use the conservation of energy principle. At the equilibrium position, the wrench has only potential energy, which is equal to its initial potential energy at the maximum displacement. So, we can write the equation:

PE = mgh = \frac{1}{2}I\omega^2

Where m is the mass of the wrench, g is the acceleration due to gravity, h is the maximum displacement of the wrench, I is the moment of inertia and \omega is the angular speed.

Solving for \omega, we get:

\omega = \sqrt{\frac{2mgh}{I}}

Substituting the values given in the question, we get:

\omega = \sqrt{\frac{2(1.8)(9.8)(0.250)(0.400)}{-9.87 X 10^{-1}}}

\omega = 0.934 rad/s

Therefore, the angular speed of the wrench as it passes through the equilibrium position is 0.934 rad/s.