A nonuniform 2.0-kg rod is 2.0 m long. The rod is mounted to rotate freely about a horizontal axis perpendicular to the rod that passes through one end of the rod. The moment of inertia of the rod about this axis is 4.0 kg m2. The center of mass of the rod is 1.2 m from the axis. If the rod is released from rest in the horizontal position, what is its angular speed as it swings through the vertical position?
Using conservation of mechanical energy: the sum of the rotational and gravitational potential energy when the rod is horizontal is equal to the sum when the rod is vertical.
When the rod is horizontal, there is no rotational kinetic energy, and the gravitational potential energy is given by:
PE_horizontal = m * g * h
where m = 2kg (mass of rod), g = 9.8m/s^2 (acceleration due to gravity), and h = 1.2m (distance of center of mass to the axis). Thus,
PE_horizontal = 2 * 9.8 * 1.2 = 23.52 Joulse (J).
When the rod is vertical, it has no gravitational potential energy (since the center of gravity is the axis), but it has rotational kinetic energy given by:
KE_vertical = 0.5 * I * ω^2
where I = 4.0kgm^2 (moment of inertia) and ω is the angular speed when the rod is vertical that we want to find.
Since the sum of energies is constant, we have:
23.52 = 0.5 * 4.0 * ω^2
ω^2 = 23.52/(0.5*4) = 11.76
ω = √(11.76) = 3.43 rad/s
So the angular speed when the rod is vertical is 3.43 rad/s.
To find the angular speed of the rod as it swings through the vertical position, we can use the principle of conservation of mechanical energy.
1. First, we need to find the potential energy of the rod at the horizontal position and the gravitational potential energy at the vertical position.
- At the horizontal position, the potential energy is zero since the rod is at the same level as the axis of rotation. Thus, the initial potential energy (UI) = 0.
- At the vertical position, when the rod is in a purely vertical orientation, the potential energy (UF) is given by the equation UF = mgh, where h is the vertical distance of the center of mass from the axis of rotation. In this case, h = 1.2 m. Therefore, UF = mgh.
2. Next, we can apply the principle of conservation of mechanical energy, which states that the sum of kinetic energy and potential energy remains constant.
- According to this principle, the initial mechanical energy (EI) = UI (since the initial kinetic energy is zero).
- The final mechanical energy (EF) = UF + kinetic energy.
3. We can equate the initial and final mechanical energies to solve for the final kinetic energy (EF - UF) and use it to find the angular speed.
- EI = EF
- Substituting the values, we get 0.5 * moment of inertia * initial angular speed^2 = UF + 0.5 * moment of inertia * final angular speed^2.
4. Rearranging the equation, we can solve for the final angular speed (final angular speed^2).
- final angular speed^2 = (UF / moment of inertia) + (initial angular speed^2).
5. Finally, we take the square root of the final angular speed^2 to find the final angular speed.
- final angular speed = √[(UF / moment of inertia) + (initial angular speed^2)].
Substitute the given values into the equation and calculate the final angular speed.
To find the angular speed of the rod as it swings through the vertical position, we can use the principle of conservation of mechanical energy.
The initial mechanical energy of the rod in the horizontal position is given by the formula:
E_initial = Translational kinetic energy + Rotational kinetic energy
Since the rod is initially at rest, its translational kinetic energy is zero. The rotational kinetic energy is given by the formula:
Rotational kinetic energy = (1/2) * (Moment of inertia) * (Angular speed)^2
In this case, the moment of inertia is given as 4.0 kg m^2. So, we have:
E_initial = 0 + (1/2) * 4.0 kg m^2 * (Angular speed)^2
Next, let's find the final mechanical energy of the rod when it swings through the vertical position. At this position, all of the initial rotational kinetic energy is converted to potential energy due to gravity.
The potential energy at the vertical position is given by:
Potential energy = Mass * Gravitational acceleration * Height
The mass of the rod is given as 2.0 kg, the gravitational acceleration is approximately 9.8 m/s^2, and the height is twice the distance from the center of mass to the axis of rotation, which is 2.4 m.
So, the final mechanical energy is:
E_final = Mass * Gravitational acceleration * Height = 2.0 kg * 9.8 m/s^2 * 2.4 m
According to the principle of conservation of mechanical energy, the initial mechanical energy is equal to the final mechanical energy:
E_initial = E_final
Substituting the expressions for E_initial and E_final, we have:
(1/2) * 4.0 kg m^2 * (Angular speed)^2 = 2.0 kg * 9.8 m/s^2 * 2.4 m
Simplifying the equation, we can solve for the angular speed:
(1/2) * 4.0 kg m^2 * (Angular speed)^2 = 47.04 kg m^2/s^2
Multiplying both sides by 2 and dividing by 4.0 kg m^2, we get:
(Angular speed)^2 = 23.52 kg m^2/s^2
Taking the square root of both sides, we find:
Angular speed = √23.52 kg m^2/s^2
Therefore, the angular speed of the rod as it swings through the vertical position is approximately 4.85 rad/s.