Question

a uniform rod of mass m and length l is free to rotate about one end. if the rod is released from rest at an angle of 60 with respect to the horizontal, what is the tangential acceleration of the tip of the rod at this moment?

Answer 1

The tangential acceleration of the tip of the rod at the moment of release is approximately 8.49 m/s^2.

Step-by-step explanation:

To find the tangential acceleration of the tip of the rod at the moment it is released, we can use the equation:

α = g * sin(θ)

where α is the angular acceleration, g is the acceleration due to gravity, and θ is the angle the rod makes with the horizontal.

In this case, θ is 60°. Plugging in the values, we get:

α = 9.8 m/s^2 * sin(60°)

Simplifying, we find that the tangential acceleration of the tip of the rod is approximately 8.49 m/s^2.