Determine the linear acceleration of the tip of the rod

[shimizua]

Homework Statement

Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown

Homework Equations

So there is a first part to this problem that i already got
A uniform rod of mass M = 5.02kg and length L = 1.08m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below.The rod is held horizontally and then released. At the moment of release, determine the angular acceleration of the rod. Use units of rad/s^2. So i got the right answer for this and it was 13.6 rad/s^2. i just don't know what to do for the second part

The Attempt at a Solution

 

[horatio89]

Linear acceleration is related to angular acceleration by the eqn. r(\alpha) = a

 

[thomate1]

To determine the linear acceleration of the tip of the rod, we can use the equation a = αr, where a is the linear acceleration, α is the angular acceleration, and r is the distance from the center of mass to the tip of the rod.

Since we know the angular acceleration from the first part of the problem, we can simply plug it into the equation along with the distance from the center of mass to the tip of the rod. However, we also need to take into account the force of gravity acting on the rod.

Using Newton's second law, F = ma, we can set up an equation for the forces acting on the rod. The only force acting on the rod is the force of gravity, which can be represented as mg, where m is the mass of the rod and g is the acceleration due to gravity. Therefore, we can write:

F = mg = ma

Solving for acceleration, we get:

a = g

Since gravity is acting at the center of mass of the rod, the distance from the center of mass to the tip of the rod is simply half of the length of the rod, or L/2. Plugging this into our original equation, we get:

a = α(L/2)

Substituting in the value of α that we found in the first part of the problem, we get:

a = (13.6 rad/s^2)(1.08m/2) = 7.344 m/s^2

Therefore, the linear acceleration of the tip of the rod is 7.344 m/s^2.