Consider the figure above consisting of three particles of mass m attached to a massless rod. Given an axis of rotation through point P, the rod rotates as shown in the figure. If the rod is released from rest in the horizontal position at t = 0. What is the angular acceleration of the system (rod and three particles) immediately after being released? Let d = 2.25 m.
The distance between particles 1 and 2 is d. The distance between particles 2 and 3 is also d. Point p is between particles 2 and 3 and is a distance of (2d/3) from particle 3.
Thanks!
I total= I1+I2+I3 where Ii= 1/2 Mi distance^2, where distance is the distance from each Mi to P.
NetMoments=Itotal*angacceleration
now on writing Net moments, each of the three moments will be mass*g*distancefromP
the MUST here is to make clockwise moments+, ccw-.
solve for ang acceleration.
Why would clockwise moments be positive in this case?
To determine the angular acceleration of the system immediately after being released, we need to consider the torque acting on the system.
Torque is the rotational equivalent of force and is given by the equation:
τ = r x F
where τ is the torque, r is the position vector from the axis of rotation to the point where the force is applied, and F is the force.
In this case, we can consider the gravitational force acting on each particle and calculate the torque exerted by that force.
For particle 1, the torque exerted by the gravitational force will be zero since its position vector from the axis of rotation is perpendicular to the force vector.
For particle 2, the torque exerted by the gravitational force will be:
τ2 = (d/2) * mg * sinθ
where θ is the angle between the position vector and the vertical axis. Since the rod is released from rest in the horizontal position, θ is 90 degrees, so sinθ = 1.
For particle 3, the torque exerted by the gravitational force will be:
τ3 = ((2d/3) + d) * mg * sinθ
The total torque exerted on the system will be the sum of the torques exerted by each particle:
τ_total = τ2 + τ3
To calculate the angular acceleration, we need to use the rotational analog of Newton's second law:
τ_total = I * α
where τ_total is the total torque acting on the system, I is the moment of inertia of the system, and α is the angular acceleration.
The moment of inertia of the system can be calculated as the sum of the moments of inertia of each particle:
I = I1 + I2 + I3
The moment of inertia of each particle can be calculated using the equation:
I = m * r^2
where m is the mass of the particle and r is the perpendicular distance from the particle to the axis of rotation.
In this case, since all three particles have the same mass, m, the moment of inertia of each particle will be:
I = m * ((d/2)^2)
Therefore, I1 = I2 = I3 = (m * (d^2))/4
Substituting the values into the equation:
τ_total = (m * (d^2))/4 * α
Since the rod is released from rest, its initial angular velocity is zero, so we can use the relation between angular acceleration, angular velocity, and time:
α = ω / t
where ω is the angular velocity and t is the time.
Since α = (2π / T) and T is the period of rotation, we can rewrite the equation as:
α = (2π / T) = (2π / t)
Given that the initial angular position is horizontal, the period of rotation can be calculated as the time it takes for particle 2 to reach the vertical position and then return to the horizontal position.
The time it takes for particle 2 to reach the vertical position can be calculated using kinematic equations. The vertical distance traveled by particle 2 is d, and the acceleration in the vertical direction is g. Therefore, using the equation:
d = (1/2) * g * t^2
we can solve for t:
t = sqrt((2d) / g)
Finally, substituting the value of t into the equation for α, we can calculate the angular acceleration of the system immediately after being released.