Particles 1 and 2, each of mass m, are attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 75 cm. The rod is held horizontally on the fulcrum and then released. What is the magnitude of the initial acceleration of particle 1? What is the magnitude of the initial acceleration of particle 2?

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I = mL^2, L=L1+L2
I am not given mass and the answer needs to be numerical.
Tnet=I * alpha

How do I find the initial acceleration?

The rod is massless, so ignore it in I calculations. You know the I of each particle (1/2 m distance^2).

Now net torque= torque on log rod - torque on short end.

Torque around fulcrum= Itotal*angularacc
so solve for angular acceleration, then intial acceleration2= angularacceleration*distance2

To find the initial acceleration of particle 1, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, particle 1 experiences a net force due to the torque applied by the rod and particle 2.

To determine the torque on particle 1, we need to consider the rotational motion of the system. The moment of inertia, denoted by "I," is a measure of an object's resistance to changes in its rotational motion. In this case, since the rod is massless, we only need to consider the moments of inertia of particle 1 and particle 2.

The moment of inertia of a point mass rotating about an axis at a distance "r" is given by I = m * r^2, where "m" is the mass of the object and "r" is the distance of the object from the axis of rotation.

For particle 1, the moment of inertia is (1/2) * m * (L1)^2, since it is located at the end of the rod with length L1.

Now, we need to calculate the torque acting on particle 1. The torque is defined as the product of the moment of inertia and the angular acceleration, denoted by "alpha." Therefore, the torque on particle 1 is equal to I1 * alpha, where I1 is the moment of inertia of particle 1.

Let's denote the initial angular acceleration as alpha0.

The net torque acting on the system is equal to torque on the long end minus the torque on particle 1:

Net torque = I_total * alpha0 - I1 * alpha0

Since the rod is initially held horizontally, the net torque is zero, as there is no angular acceleration at that moment.

Therefore, we can set the equation equal to zero:

0 = I_total * alpha0 - I1 * alpha0

Solving for alpha0, we get:

alpha0 = 0 / (I_total - I1)

As the numerator is zero, alpha0 is equal to zero as well. This indicates that particle 1 does not experience any initial angular acceleration.

Since particle 1 is at the fulcrum, it remains stationary when the rod is released. Therefore, the magnitude of the initial acceleration of particle 1 is zero.

To find the initial acceleration of particle 2, we can use the relationship between linear acceleration and angular acceleration.

The linear acceleration of a point on the rim of a rotating object is given by a = alpha * r, where "a" is the linear acceleration, "alpha" is the angular acceleration, and "r" is the perpendicular distance from the axis of rotation to the point.

In this case, particle 2 is located at the end of the rod with length L2. Therefore, the distance "r" for particle 2 is L2.

We already found that the angular acceleration, alpha0, is zero for particle 1. Hence, the linear acceleration of particle 2 is also zero.

Therefore, the magnitude of the initial acceleration of particle 2 is zero as well.