Let’s take a look at the example of a wrench turning a bolt. A force is applied at a distance from the axis of rotation. Call this distance r. When you apply forces at 90 degrees to the imaginary line leading from the axis of rotation to the point where the force is applied (known as the line of action), you obtain maximum torque. As the angle at which the force applied decreases (θ), so does the torque causing the bolt to turn. Therefore, you can calculate the torque applied as: In some cases, physicists will refer to the rsinθ as the lever arm, or moment arm, of the system. The lever arm is the perpendicular distance from the axis of rotation to the point where the force is applied. Alternately, you could think of torque as the component of the force perpendicular to the lever multiplied by the distance r. Units of torque are the units of force × distance, or Newton-meters (N·m).

Question: A pirate captain takes the helm and turns the wheel of his ship by applying a force of 20 Newtons to a wheel spoke. If he applies the force at a radius of 0.2 meters from the axis of rotation, at an angle of 80° to the line of action, what torque does he apply to the wheel?

Answer:

Question: A mechanic tightens the lugs on a tire by applying a torque of 110 N·m at an angle of 90° to the line of action. What force is applied if the wrench is 0.4 meters long?

Answer: Question: How long must the wrench be if the mechanic is only capable of applying a force of 200N?

Answer: Objects which have no rotational acceleration, or a net torque of zero, are said be in rotational equilibrium. This implies that any net positive (counter-clockwise) torque is balanced by an equal net negative (clockwise) torque.

Moment of Inertia Previously, the inertial mass of an object (its translational inertia) was defined as that object’s ability to resist a linear acceleration. Similarly, an object’s rotational inertia, or moment of inertia, describes an object’s resistance to a rotational acceleration. The symbol for an object’s moment of inertia is I. Objects that have most of their mass near their axis of rotation have a small rotational inertia, while objects that have more mass farther from the axis of rotation have larger rotational inertias. For common objects, you can look up the formula for their moment of inertia. For more complex objects, the moment of inertia can be calculated by taking the sum of all the individual particles of mass making up the object multiplied by the square of their radius from the axis of rotation. This can be quite cumbersome using algebra, and is therefore typically left to calculus-based courses or numerical approximation using computing systems.

Question: Calculate the moment of inertia for a solid sphere with a mass of 10 kg and a radius of 0.2 m.

Answer: Question: Calculate the moment of inertia for a hollow sphere with a mass of 10 kg and a radius of 0.2 m.

Answer:

Newton's 2nd Law for Rotation In the chapter on dynamics, you learned about forces causing objects to accelerate. The larger the net force, the greater the linear (or translational) acceleration, and the larger the mass of the object, the smaller the translational acceleration.

The rotational equivalent of this law, Newton’s 2nd Law for Rotation, relates the torque on an object to its resulting angular acceleration. The larger the net torque, the greater the rotational acceleration, and the larger the rotational inertia, the smaller the rotational acceleration:

Question: What is the angular acceleration experienced by a uniform solid disc of mass 2 kg and radius 0.1 m when a net torque of 10 N·m is applied? Assume the disc spins about its center.

Answer:

2

Question: A Round-A-Bout on a playground with a moment of inertia of 100 kg·m starts at rest and is accelerated by a force of 150N at a radius of 1m from its center. If this force is applied at an angle of 90° from the line of action for a time of 0.5 seconds, what is the final rotational velocity of the Round-A-Bout? Answer: Start by making our rotational kinematics table:

Since you only know two items on the table, you must find a third before you solve this with the rotational kinematic equations. Since you are given the moment of inertia of the Round-A-Bout as well as the applied force, you can solve for the angular acceleration using Newton’s 2nd Law for Rotational Motion.

Now, use your rotational kinematics to solve for the final angular velocity of the Round-A-Bout.

Honors Physics - Angular Momentum Previously, you learned that linear momentum, the product of an object’s mass and its velocity, is conserved in a closed system. In similar fashion, spin angular momentum L, the product of an object’s moment of inertia and its angular velocity about the center of mass, is also conserved in a closed system with no external net torques applied.

This can be observed by watching a spinning ice skater. As an ice skater launches into a spin, she generates rotational velocity by applying a torque to her body. The skater now has an angular momentum as she spins around an axis which is equal to the product of her moment of inertia (rotational inertia) and her rotational velocity. To increase the rotational velocity of her spin, she pulls her arms in close to her body, reducing her moment of inertia. Angular momentum is conserved, therefore rotational velocity must increase. Then, before coming out of the spin, the skater reduces her rotational velocity by move her arms away from her body, increasing her moment of inertia. Question: Angelina spins on a rotating pedestal with an angular velocity of 8 radians per second. Bob 2 2 throws her an exercise ball, which increases her moment of inertia from 2 kg·m sup> to 2.5 kg·m . What is Angelina’s angular velocity after catching the exercise ball? (Neglect any external torque from the ball.) Answer: Since there are no external torques, you know that the initial spin angular momentum must equal the final spin angular momentum, and can therefore solve for Angelina’s final angular velocity:

2

Question: A disc with moment of inertia 1 kg·m spins about an axle through its center of mass with angular velocity 10 rad/s. An identical disc which is not rotating is slid along the axle until it makes contact with the first disc. If the two discs stick together, what is their combined angular velocity?

Answer: Once again, since there are no external torques, you know that spin angular momentum must be conserved. When the two discs stick together, their new combined moment of inertia must be the sum 2 of their individual moments of inertia, for a total moment of inertia of 2 kg·m .

Honors Physics - Rotational Kinetic Energy Kinetic Energy When kinematics was first introduced, kinetic energy was defined as the abil- ity of a moving object to move another object. Then, translational kinetic energy for a moving object was calculated using the formula:

Since an object which is rotating also has the ability to move another object, it, too, must have kinetic energy. Rotational kinetic energy can be calculated using the analog to the translational kinetic energy formula -- all you have to do is replace inertial mass with moment of inertia, and translational velocity with angular velocity!

If an object exhibits both translational motion and rotational motion, the total kinetic energy of the object can be found by adding the translational kinetic energy and the rotational kinetic energy:

Because you’re solving for energy, of course the answers will have units of Joules.

Question: Gina rolls a bowling ball of mass 7 kg and radius 10.9 cm down a lane with a velocity of 6 m/s. Find the rotational kinetic energy of the bowling ball, assuming it does not slip. Answer: To find the rotational kinetic energy of the bowling ball, you need to know its moment of inertia and its angular velocity. Assume the bowling ball is a solid sphere to find its moment of inertia.

Next, find the ball’s angular velocity.

Finally, solve for the rotational kinetic energy of the bowling ball.

Question: Find the total kinetic energy of the bowling ball from the previous problem. Answer: The total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy of the bowling ball.

Formula Equivalencies

Putting all this information together, rotational physics mirrors translational physics in terms of both variables and formulas. These equivalencies and relationships are summarized below.