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Chapter 54: Angular Kinematics
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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