Rotation of rigid bodies – problems and solutions

**The m****oment of force **

1. Three forces act on a beam with a length of 6 meters, as shown in the figure below. What is the net torque rotates the beam about the point O as the axis of rotation?

__Known :__

*The axis of rotation at point O. *

Force 1 (F_{1}) = F

The distance between the line of action of F_{1 }with the axis of rotation (r_{1}) = 3 meters

Force 2 (F_{2}) = 2F

The distance between the line of action of F_{2} with the axis of rotation (r_{2}) = 2 meters

Force 3 (F_{3}) = 2F

The distance between the line of action of F_{3} with the axis of rotation (r_{3}) = 3 meters

__Wanted:__ The magnitude of the moment of force

__Solution :__

The moment of force 1 :

τ_{1 }= F_{1} r_{1} = (F)(3) = -3F

*The **moment of force **1** rotates beam clockwise so we assign **a negative **sign.*

The moment of force 2 :

τ_{2 }= F_{2} r_{2} = (2F)(2) = 4F

*The moment of force 2 rotates beam counterclockwise so we assign positive sign.*

The moment of force 3 :

τ_{3 }= F_{3} r_{3 }sin 30^{o }= (2F)(3)(0.5) = 3F

*The moment of force 2 rotates beam counterclockwise so we assign positive sign.*

The resultant of the moment of force :

Στ = τ_{1 }+ τ_{2 }+ τ_{3}

Στ = -3F + 4F + 3F

Στ = 4F

The magnitude of the moment of force is 4F Newton-meter. The resultant of the moment of force rotates beam counterclockwise so we assign a positive sign.

2. α = 30^{o}, length of AB = BC = 1 meter. What is the moment of force about the axis of rotation at point A?

__Known :__

*The axis of rotation at point **A. *

Force 1 (F_{1}) = 10 N

*The distance between the line of action of F _{1 }with the axis of rotation (r_{1}) = 1 meter *

Force 2 (F_{2}) = 10 N

*The distance between the line of action of F _{2 }with the axis of rotation (r_{2}) = 1 meter *

Force 3 (F_{3}) = 20 N

*The distance between the line of action of F _{3 }with the axis of rotation (r_{3}) = 2 meters*

__Wanted:__ The resultant of the moment of force

__Solution :__

The moment of force 1 :

τ_{1 }= F_{1} r_{1} sin 30^{o }= (10)(1)(0.5) = 5 N m

*The **moment of force **1** rotates beam **counter**clockwise so we assig**n a positive **sign.*

The moment of force 2 :

τ_{2 }= F_{2} r_{2 }sin 30^{o }= (10)(1)(0.5) = -5 Newton meter

*The **moment of force 2 **rotates bea**m **clockwise so we assig**n negative **sign.*

The moment of force 3 :

τ_{3} = F_{3} r_{3 }sin 60^{o }= (20)(2)(0.5√3) = -20√3 Newton meter

*The **moment of force 3 **rotates bea**m **clockwise so we assig**n a negative **sign.*

The resultant of the moment of force :

Στ = τ_{1} + τ_{2 }+ τ_{3}

Στ = 5 – 5 – 20√3

Στ = – 20√3 N m

The magnitude of the moment of force is 20√3 N m. The resultant of the moment of force rotates beam clockwise so we assign negative sign.

**What is a rigid body, and how is it different from a non-rigid body?****Answer:**A rigid body is an idealized object in which the distance between any two given points within the body remains constant, regardless of external forces or torques. In contrast, a non-rigid body can deform, allowing the distance between points within the body to change.

**How is the moment of inertia of a rigid body related to its mass distribution?****Answer:**The moment of inertia of a rigid body is a measure of its resistance to rotational motion about a given axis and depends on both the mass of the body and its distribution relative to the axis of rotation. It is calculated by the sum of the products of each mass element’s mass and the square of its distance from the axis of rotation.

**What is the significance of the rotational kinetic energy of a rotating rigid body?****Answer:**Rotational kinetic energy is a measure of the energy due to the rotation of a rigid body. It depends on both the moment of inertia and the angular velocity of the body, given by 1/2 $ Iω$, where $I$ is the moment of inertia, and $ω$ is the angular velocity.

**What happens to the angular momentum of a system of particles if no external torques act on it?****Answer:**If no external torques act on a system of particles, the total angular momentum of the system is conserved. This is the principle of conservation of angular momentum.

**How does the parallel-axis theorem help in finding the moment of inertia of a rigid body?****Answer:**The parallel-axis theorem allows one to calculate the moment of inertia of a rigid body about any axis parallel to and a distance $d$ away from an axis through its center of mass. It states that $I=I_{cm}+md$, where $I_{cm}$ is the moment of inertia about the center of mass, $m$ is the total mass, and $d$ is the distance between the two axes.

**What is the difference between rolling without slipping and rolling with slipping?****Answer:**Rolling without slipping occurs when a rigid body rotates about a fixed axis while also translating, without any relative motion between the body and the surface. Rolling with slipping means there is relative motion or sliding between the body and the surface.

**How does the radius of gyration relate to the moment of inertia?****Answer:**The radius of gyration is a measure that describes how the mass of a body is distributed about its axis of rotation. It is defined as the square root of the ratio of the moment of inertia to the mass and provides an equivalent distance from the axis where all the mass could be concentrated without changing the moment of inertia.

**What effect does increasing the moment of inertia have on the angular acceleration of a rigid body for a given torque?****Answer:**For a given torque, increasing the moment of inertia will decrease the angular acceleration, as $α=τ/I $, where $α$ is angular acceleration, $τ$ is torque, and $I$ is the moment of inertia.

**Can a force applied to a rigid body cause both translational and rotational motion? Explain how.****Answer:**Yes, a force applied to a rigid body can cause both translational and rotational motion. If the force is applied at a point that does not coincide with the center of mass, it can cause the body to translate (move linearly) and rotate. The translational motion is determined by the net force, while the rotational motion depends on the torque created by the force about the center of mass.

**Why does a figure skater spin faster when they pull their arms close to their body?**

**Answer:**By pulling their arms close to their body, a figure skater reduces their moment of inertia. According to the conservation of angular momentum, if the moment of inertia decreases and no external torque is applied, the angular velocity must increase. Thus, the skater spins faster.

These questions and answers provide an understanding of key concepts related to the rotation of rigid bodies.