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.
To solve this problem, we can use the principles of rotational motion and the laws of physics. Let's break down the questions one by one.
(A) To find the angular speed of the rod when it is horizontal, we can use the principle of conservation of energy. At the initial vertical position, the potential energy of the rod is given by mgh, where m is the mass of the rod and h is the height from which it is released.
Next, at the horizontal position, all the potential energy is converted into kinetic energy. The kinetic energy of the rod is given by (1/2)Iω^2, where I is the moment of inertia of the rod and ω is the angular speed.
Since the rod is released from rest, the initial angular speed is 0. Therefore, we can equate the initial potential energy to the final kinetic energy:
mgh = (1/2)Iω^2
The moment of inertia of a uniform rod about its end is given by I = (1/3)mL^2, where L is the length of the rod. Substituting the values given, we have:
5.48 kg * 9.8 m/s^2 * 4.76 m = (1/2) * (1/3) * 5.48 kg * 4.76 m^2 * ω^2
Solving for ω, we get:
ω = sqrt(2 * (5.48 kg * 9.8 m/s^2 * 4.76 m) / ((1/3) * (5.48 kg * 4.76 m^2)))
Calculating this value will give us the angular speed in rad/s.
(B) At the same instant, the magnitude of the angular acceleration can be found using the formula α = ω^2 / r, where α is the angular acceleration and r is the distance of the center of mass from the pivot point. Since the rod is horizontal, the center of mass is at a distance of L/2 from the pivot point.
Substituting the value of ω from part (A) into the formula, we have:
α = (ω^2) / (L/2)
Calculating this value will give us the angular acceleration in rad/s^2.
(C) To find the magnitude of the acceleration of the center of mass, we can use the formula a_cm = α * r, where a_cm is the acceleration of the center of mass and r is the distance of the center of mass from the pivot point. Again, r is equal to L/2.
Substituting the value of α from part (B) into the formula, we have:
a_cm = α * (L/2)
Calculating this value will give us the acceleration of the center of mass in m/s^2.
(D) Finally, to find the magnitude of the reaction force at the pivot, we can apply Newton's second law for rotational motion. The net torque acting on the rod is equal to the moment of inertia times the angular acceleration.
The net torque about the pivot point is provided by the gravitational force acting on the rod's center of mass. The magnitude of this torque is given by:
τ = m * g * r
Setting this equal to I * α, we have:
m * g * r = (1/3) * m * L^2 * α
Simplifying and solving for α, we get:
α = (3 * g * r) / L
Substituting the values given, we can calculate the magnitude of the reaction force at the pivot using the formula:
magnitude of reaction force = m * (α * r)
Evaluating this expression will give us the magnitude of the reaction force in Newtons.