A long uniform rod of length 6.66 m and mass 6.74 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.
Question 1:
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.
Question 2:
At the same instant find the magnitude of its angular acceleration. Answer in units of rad/s2.
Question 3:
Still at the same instant find the magnitude of the acceleration of its center of mass. Answer in units of m/s2.
Question 4:
Finally, find the magnitude of the reaction force at the pivot (still at the same moment). Answer in units of N.
I do not see the picture but assume that the rod falls with one end fixed.
The easy way to approach this is to use conservation of energy. The center of gravity of the rod has fallen 3.33 meters by the time it reaches horizontal.
It has therefore lost potential energy equal to m g (3.3) = 6.74 * 9.8 * 3.33
= 220 Joules
Therefore since the hinge is frictionless I have 220 J of kinetic energy
Call the angular velocity w, then the velocity, v, of the center of mass is 3.33 w
Ke = (1/2) m v^2 + (1/2) I w^2
I for a rod about its center is (1/12) m L^2
so
220 = (1/2)(6.74)(3.33w)^2 +(1/24)(6.74)(6.66)^2 w^2
or
220 = 49.8 w^2
w= 2.1 rad/s
(check my math!)
Now what is the angular acceleration?
The moment about the hinge is 3.33* mg
= 3.33 * 6.74 * 9.8
= 220 N m
That equals the moment of inertia about the hinge times A, the angular acceleration. I about hinge = (1/3) m L^2
so
220 = (1/3)(6.74)(6.66^2)A
so A = 2.21 rad/s^2
acceleration of middle = 3.33A = 7.35 m/s^2
force = mass * acceleration
force up = F at hinge
Force down = m g = 6.74*9.8 = 66 N
so
66 - F = 6.74*7.35
F = 16.51 N
-----------------------
check
that force * 3.33 should be (1/12) 6.74 (6.66^2) A about center
That gives A = 2.21, sure enough
To solve these questions, we need to apply the principles of rotational motion and the torque equation. Let's go through each question step by step.
Question 1:
To find the angular speed of the rod at the instant it is horizontal, we need to use conservation of mechanical energy. When the rod is released from rest in the vertical position, it has only gravitational potential energy. At the instant it becomes horizontal, it has only rotational kinetic energy.
The gravitational potential energy at the initial vertical position is given by:
PE = mgh = (6.74 kg) * (9.8 m/s^2) * (6.66 m)
At the instant it becomes horizontal, the potential energy is converted into rotational kinetic energy:
KE = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular speed.
The moment of inertia of a uniform rod rotating about one end is given by:
I = (1/3) * m * L^2
where L is the length of the rod.
Substituting the values, we can calculate the angular speed (ω).
Question 2:
To find the magnitude of the angular acceleration at the same instant, we need to use the torque equation. Torque (τ) is defined as the product of the force and the distance from the pivot.
In this case, the only torque acting on the rod is due to its weight. The torque (τ) is given by:
τ = m * g * (L/2)
The torque (τ) is also equal to the moment of inertia (I) multiplied by the angular acceleration (α):
τ = I * α
Setting the two expressions for torque equal to each other, we can find the angular acceleration (α).
Question 3:
To find the magnitude of the acceleration of the center of mass at the same instant, we need to use the kinematic equation for rotational motion. The acceleration of the center of mass (a_cm) is related to the angular acceleration (α) and the distance from the pivot to the center of mass (r_cm) by the equation:
a_cm = α * r_cm
The distance from the pivot to the center of mass is half the length of the rod, so we can calculate the acceleration of the center of mass (a_cm).
Question 4:
To find the magnitude of the reaction force (F_pivot) at the pivot at the same moment, we need to consider the rotational equilibrium. At the instant the rod becomes horizontal, the torque due to the weight of the rod is balanced by the reaction force at the pivot.
The equation for torque (τ) is given by:
τ = F_pivot * L
Setting the torque equal to zero (since the rod is in equilibrium), we can solve for the reaction force at the pivot (F_pivot).
By following these steps and performing the necessary calculations, you can find the answers to each of the questions.