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Chapter 50: Parallel Forces and Moments

- If two forces are used, it is possible to prevent the rotation.
- If the object is free to move, it may result.

- Two people on a seesaw are an example of parallel forces.
- The two people are sitting at the same distance from the fixed pivot point.
- The seesaw is caused by the tendency of each force to cause it to move.
- Force F g1 will cause a clockwise rotation.
- We state that a clockwise rotation is negative while a counterclockwise rotation is positive.

- The net Torque on a system is zero.
- The sum of the clockwise and counterclockwise Torques must be equal.

- Experiments can show that distances d 1 and d 2 are important since we take the weights as being constant.
- If the weights were the same, it would be no surprise to learn that d 1 and d 2 are equal.
- The weight of the seesaw is not taken into account in these examples.

- If F g2 is greater, that person must sit closer to the fulcrum.

- The force and distance products do not represent work because they are not causing a displacement.
- The product of a force and arm distance is called Torque.
- Torques and equilibrium can be seen in some more examples.

- A girl and a boy are sitting on a seesaw.

- A rotation or twist can be created when you tighten a bolt with a wrench.
- The Greek letter t represents the twisting action called Torque in physics.

- When we say that t is the distance from the center of mass to the pivot point and d is the force to the displacement, F is the force to the displacement.
- If the force is applied at a certain angle to the object, the component of the force that goes out from the pivot is taken as the force used.

- Depending on the rotation, the direction of the Torque is either positive or negative.

- Consider a string wound around a wheel.
- A 1.2- kilo mass is attached to the free end of the string.

- The falling weight causes a clockwise rotation.

- There is a dot on the axis of rotation.

- The maximum lever arm is half the length of the rotating object.

- In the following example, a rod of mass M is attached to a wall with a string.
- The string is massless and makes an angle with the horizontal.
- The end of the rod has a mass m attached to it.

- Since the system is in static equilibrium, the sum of all forces and the sum of all Torques must be zero.
- The moment arm d would be zero, all forces acting through the pivot do not contribute any Torques.
- The force reaction R- is the response of the wall to the rod.

- The net force and net Torque must be zero since this is a static problem.
- By looking at the diagram for hanging mass m, T 2 is the number.

- The rod is assumed to be the same color.
- The weight force is applied at the halfway point.
- By taking the left-hand side of the rod as our axis of rotation, the Torque from force R is zero, leaving us with the following equilibrium statement.
- The axis can be taken anywhere you want.

- The clockwise and counterclockwise Torques are negative.
- The two force statements can't be further simplified without more information.
- The length of rod L can be eliminated from the equation.

- The angle will either be determined by the geometry of the situation or by the problem.
- The unknown force R- can be determined from the two force equations once the tension T- is known.

- The rod is considered to have mass and that the Torque produced by the rod is its weight taken from the center of mass and is always negative.

- Linear inertia is simpler than rotational inertia.
- Like linear inertia, the object's moment of inertia is used to measure the object's tendency to resist changes to its motion.
- Linear inertia is the total mass of an object that we think is located at the object's center of mass.
- The rotational inertia is dependent on how the mass is distributed.

- The AP Physics 1 exam won't ask you to calculate moments of inertia, but you must understand them and be able to use them.

- The units are 2.
- The result depends on the shape and mass distribution of the object as well as the axis of rotation.
- On the AP exam, if a solid object is being used, the general moment of inertia for that object will be given or easily obtained through another relationship.
- The moment of inertia of a solid rod of length L is different when the object is about one end of the rod as opposed to the other.

- Try to hold the ruler between your fingers.
- Hold one end and oscillate it.
- You will see that rotating the ruler about its center is much easier and has a smaller moment of inertia.

- The place of mass in the linear formulas in physics is taken by rotational inertia.
- An object's position is represented by its axis of motion.
- All values are measured in radians.

- We can complete the analogy between the physics of linear motion and that of rotational motion by defining the basic quantities of Torque, Moment of inertia, and kinematic variables.
- You can simply rewrite any of our linear physics relationships.

- The inertia of the object is 15 km 2.

- Although radians are not a unit and are not formally required to be written down, it can be helpful to write them in to guide your thinking.

- Generally speaking, objects that are rolling have both linear and rotational energy.
- If the cylindrical or spherical object is rolling without slipping, then the object's rotation speed and linear speed must be coupled by the object's radius.

- The mass cancels out when it's in every term.
- The rolling without slipping conditions cancels out the radius dependency in this expression.
- The question of who will win the race comes down to the moments of inertia for each object.
- The object with the highest moment of inertia will require more mechanical energy for rotation.
- The object will have less of the total energy for the linear energy.

- Torques cause changes in momentum over time.
- An isolated system's momentum is the same as a linear one.
- It is relatively rare for an isolated system's mass to change, but changing the moment of inertia of an isolated object is not that difficult.
- If the momentum of inertia changes due to internal forces, the rotation must change as well.
- A spinning ice-skater drawing in her arms decreases the r values in her moment of inertia calculation.
- She will increase her speed.

- As the Earth goes around the Sun, it's important to know the direction of the motion.

- The Torque is equal to the product of the lever arm.

- The lever arm and force are the same.

- The sum of the clockwise and counterclockwise Torques must be equal in equilibrium.

- The same laws apply in rotating systems as they do in linear systems.
- Everything in the world has an equation in the linear.

- In isolated systems, angular momentum is conserved.

- It is similar to solving static equilibrium problems withNewton's laws.
- The first condition for static equilibrium is provided byNewton's laws.
- You should draw a free-body diagram of the situation again.

- The sum of all the Torques must be zero.
- We take clockwise and counterclockwise Torques as negative and positive.
- There is no force going through the pivot point.
- It is wise to choose a point that eliminates the most forces.
- Only the components of forces that are in line with the direction of the pivot are responsible, a situation that usually involves the angle of the force's orientation.

- The more comfortable language of linear physics can be used to translate a rotational problem.
- Changes in mass in the middle of a problem are rare in linear problems, while changes in moment of inertia are common in rotational problems.

- A girl is sitting on a seesaw near the balance point.

- The stick is balanced.
- The 10-g mass is shown in the distance from the fulcrum.

- A solid cylinder consisting of an outer and an inner radius r 1 and r 2 is pivoted on a axle.

- Answer the questions based on the diagram.
- Massless is what the rod is considered to be.

- A net Torque of T is experienced by an object with a moment of inertia.

- A person is 2.5 m from a wall with a beam attached.
- The beam weighs 200 N and is 6 m long.
- A cable that is attached to the free end of the beam is attached to the wall.

- A ladder of length l and weight 100 N rests against a wall.
- The floor and the bottom of the ladder have the same coefficients of static friction.

- A toy top is rotating at a rate of 8 times per second and has a friction Torque of 0.2.

- The force on each side is determined by the weight.
- The arm distances are measured in m and x.

- The same factors appear on both sides of the balance equation, but we could convert all of the mass to kilograms and all of the distance to meters.

- F-2 has a counterclockwise positive Torque and F-1 has a clockwise negative Torque.
- The necessary arm distance is determined by each radius.

- The components of the remaining forces to the beam are needed.
- The 30-N force acts counterclockwise, while the 10-N force acts clockwise.

- The 10-N force will act in a clockwise direction, while the 20-N force will act in a counterclockwise direction.
- The pivot is 1.5 m from each force.
- The component of each force has to be parallel to the beam.

- We get T in the last equation.

- The ratio of these two equations gives us a tan of 1.27 and a th of 51.8deg.

- The force P is the normal force of the wall.
- The weight Fg acts from the center of the mass.
- We can write equations for the conditions of static equilibrium since we have equilibrium at this angle.

- The sum of both x and y must be zero.
- Friction is an opposing force and the reaction force.
- We can write 0 if we want to.
- Since Fg is 100 N, then N is 100 N.

- The sum of the Torques must not be more than zero.
- The component of P needs to be parallel to the ladder.
- We can see from the geometry that this is P sinth.
- The component of weight on the ladder is Fg cos th.

- The length of the ladder is irrelevant and can be canceled out.
- We know that P is 50 N and Fg is 100 N.

- An object has stopped rotating when it is not moving.

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