Find Initial Acceleration of Rod's Center of Mass with Newton's 2nd Law

[jstealth03]

here's the question:

A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle of 27 degrees above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.

So far I've only been able to figure out that I have to use Newton's Second Law of rotation, F x r = I x alpha. How do i derive that to figure out acceleration? Do i break down F into ma, and cancel the m's on each side that result from inertia?

 

[Doc Al]

Since the rod is constrained to rotate about the hinge, all you need is Newton's 2nd law for rotation. That's: Τ = I α.

The only force creating a torque about the pivot is the object's weight. You'll also need to know the rotational inertia of the rod.

So, solve for α. The linear acceleration is rα.

Do things algebraically and you'll find that things will cancel nicely.

 

[M98Ranger]

To find the initial acceleration of the rod's center of mass, we can use Newton's Second Law of rotation, which states that the net torque acting on an object is equal to its moment of inertia times its angular acceleration. In this case, the net torque acting on the rod is due to the force of gravity, which is equal to the weight of the rod acting at its center of mass. Therefore, we can write the equation as:

τ = Iα

Where τ is the net torque, I is the moment of inertia, and α is the angular acceleration.

To find the moment of inertia for a uniform rod rotating about one end, we can use the formula I = (1/3)ML^2, where M is the mass of the rod and L is its length. Plugging in the values given in the question, we get I = (1/3)(M)(1.15)^2.

Now, we can substitute this value into our equation and solve for α:

τ = Iα
Mg(1.15/2)sin(27) = (1/3)(M)(1.15)^2α
(0.575M)sin(27) = (0.0391M)α
α = (0.575sin(27))/(0.0391) ≈ 1.23 rad/s^2

This is the angular acceleration of the rod. To find the linear acceleration of the center of mass, we can use the formula a = rα, where r is the distance from the pivot to the center of mass. In this case, r = 1.15/2 = 0.575 m.

Thus, the initial acceleration of the rod's center of mass is:

a = (0.575)(1.23) ≈ 0.707 m/s^2

In conclusion, the magnitude of the initial acceleration of the rod's center of mass is approximately 0.707 m/s^2.