A long uniform rod of length 4.76 m and mass 5.48 kg is pivoted about a horizontal, frictionless pin through one end. The rod is
released from rest in a vertical position as in the figure.
(A)
At the instant the rod is horizontal, find its angular speed. The acceleration of gravity is 9.8 m/s2 . Answer in units of rad/s.
(B)
At the same instant find the magnitude of its angular acceleration. Answer in units of rad/s2.
(C)
Still at the same instant find the magnitude of the acceleration of its center of mass. Answer in units of m/s2.
(D)
Finally, find the magnitude of the reaction force at the pivot (still at the same moment). Answer in units of N.
where are the answers
To find the solutions to parts (A), (B), (C), and (D), we will use the principles of rotational motion and the conservation of energy.
(A) To find the angular speed of the rod at the instant it is horizontal, we can use the conservation of energy principle. In its initial vertical position, the potential energy of the rod is converted to kinetic energy when it becomes horizontal. The potential energy of the rod is given by PE = mgh, where m is the mass of the rod, g is the acceleration due to gravity, and h is the height of the center of mass.
In this case, h = 4.76 m, m = 5.48 kg, and g = 9.8 m/s^2.
Thus, the potential energy is PE = (5.48 kg)(9.8 m/s^2)(4.76 m).
The kinetic energy is given by KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular speed.
Considering the rod to be a thin, uniform rod pivoted at one end, the moment of inertia is given by I = (1/3)mL^2, where L is the length of the rod.
In this case, L = 4.76 m and m = 5.48 kg.
Thus, the moment of inertia is I = (1/3)(5.48 kg)(4.76 m)^2.
At the instant the rod is horizontal, all potential energy is converted to kinetic energy. So, PE = KE.
Therefore, (5.48 kg)(9.8 m/s^2)(4.76 m) = (1/2)(1/3)(5.48 kg)(4.76 m)^2(ω^2).
By solving this equation for ω, we can find the angular speed in units of rad/s.
(B) To find the magnitude of the angular acceleration at the same instant, we can use the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
In this case, we can assume that the torque exerted on the rod is caused by the force of gravity acting at the center of mass and the reaction force at the pivot.
As the rod swings to the horizontal position, the gravitational force exerts torque in the clockwise direction. The reaction force at the pivot exerts torque in the counterclockwise direction.
Thus, τ = -mg(L/2) + 0, since the reaction force at the pivot does not contribute to the torque.
Simplifying further, τ = -(5.48 kg)(9.8 m/s^2)(4.76 m/2).
Using τ = Iα, we can solve for α to find the angular acceleration in units of rad/s^2.
(C) To find the magnitude of the acceleration of the center of mass at the same instant, 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 pivot.
In this case, r = L/2.
Using the angular acceleration found in part (B), we can find the linear acceleration in units of m/s^2.
(D) To find the magnitude of the reaction force at the pivot at the same moment, we can use Newton's second law for rotational motion. The net torque acting on the rod is equal to the moment of inertia multiplied by the angular acceleration, τ = Iα.
The reaction force at the pivot counteracts the gravitational torque, so τ = -mg(L/2).
Solving for the reaction force, we can find it in units of N.
By following these steps, you can find the solutions to parts (A), (B), (C), and (D) of this problem.