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**Physics Montwood High School R. Casao**

Rotational Dynamics Physics Montwood High School R. Casao

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**The rotational equivalent of a force is a torque**

The rotational equivalent of a force is a torque. The figure shows a small object of mass m moving in a circle of radius r. A force F is applied to the mass, tangential to the circle (perpendicular to the radius). The acceleration in the direction of this force is the tangential acceleration at. Substituting into Newton’s Second Law, F = m·at, and multiplying both sides by r to get the torque: Torque for a hoop or wheel spinning on an axis through the center: T = m·r2·α

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The angular mass is given a special name, moment of inertia, I, and therefore: T = I·, which describes rotation about a fixed axis in exactly the same manner that F = m·a. Moment of inertia is also referred to as rotational inertia. Moment of inertia (rotational inertia) is the resistance of a rotating object to changes in its angular velocity .

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An object rotating about an axis tends to remain rotating about the same axis unless interfered with by some external influence. Bodies that are rotating tend to remain rotating, while non-rotating bodies tend to remain non-rotating. If we wish to change the rate of rotation about an axis (change of angular velocity), we must apply a torque about the axis.

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The angular acceleration a this torque produces depends on the mass of the rotating object and upon the distribution of its mass with respect to the axis of rotation. If the mass remains fixed in position, torque and angular acceleration are directly proportional. If the mass is closer to the axis of rotation, the acceleration produced by the given torque is greater than if the mass is farther from the axis of rotation. Moving the mass farther from the axis of rotation changes the distribution of the mass and increases the rotational inertia. The greater the distance between the bulk of the mass of an object and the axis about which rotation takes place, the greater the rotational inertia.

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**Equation: F·r = I·; the units for rotational inertia are kg·m2/rad2.**

Industrial flywheels are constructed so that most of their mass is concentrated along the rim. Once rotating, they have a greater tendency to remain rotating than they would if their mass were concentrated nearer the axis of rotation. The greater the rotational inertia of an object, the harder it is to change the rotational state of the object. If rotating, it is difficult to stop; if at rest, it is difficult to rotate. Equation: F·r = I·; the units for rotational inertia are kg·m2/rad2. Rotational inertia and legs: short legs have less rotational inertia than long legs. An animal with shorter legs has a quicker stride than one with long legs (same is true for pendulums). When running, we bend our legs to reduce the rotational inertia so that we can rotate them back and forth more quickly.

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**The moment of inertia must be specified with respect to a chosen axis of rotation.**

See the table containing the rotational inertia equations for various shapes and axes of rotation.

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Rolling A solid cylinder will roll down an incline faster than a hollow one, whether or not they have the same mass or diameter, as illustrated. The object with the mass concentrated farthest from the axis of rotation, the hollow cylinder, has a greater rotational inertia, therefore, less acceleration. Any solid cylinder will roll down an incline with greater acceleration than any hollow cylinder, regardless of the mass or radius. The hollow cylinder has a greater rotational inertia.

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Objects of the same shape but different sizes will accelerate equally when rolled down an incline. The objects have the same rotational inertia per mass. Just as the different masses of objects in free fall do not influence the acceleration, the different masses of the rolling objects does not influence the acceleration. When rotation is introduced, the object with the larger rotational inertia compared to its own mass has the greater resistance to a change in its motion.

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**Atwood’s Machine Revisited**

When we previously examined Atwood’s machine, we ignored friction in the bearings and the mass and inertia of the pulley. Not any more! When a cord passes over a frictionless pulley of negligible mass, there is a single value for the tension throughout the cord. If the pulley is heavy, or if friction is present in the pulley’s bearings, the tension on one side of the pulley is different from the tension on the other side.

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The Atwood’s machine problems will now involve applying the F = m·a and T = I· equations to both masses as well as the pulley. Substitute as required. m1 = upward moving mass; m2 = downward moving mass For the side that accelerates downward:

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**For the side that accelerates upward:**

Replace linear acceleration a with r·. For the rotation of the pulley: Solve the linear tension equations for the tension and substitute into the torque equation for the pulley. Solve the subsequent equation for the angular acceleration. With the angular acceleration, the linear acceleration and the tensions in the system can be determined.

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**Work in Rotational Motion**

When you pedal a bicycle, you apply forces to a rotating body and do work on it; a rotating motor in a drill turns the drill bit. In both cases, work can be expressed in terms of torque and angular displacement. When the kid applies a tangential force on the merry-go-round, the force acts through the radius of the merry-go-round to produce a torque which will cause the merry-go-round to turn through an angular displacement .

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**Avoiding the calculus: the work done by the force as the merry-go-round rotates is:**

W = T or W = Fr has to be in radians If the torque remains constant while the angle changes from 1 to 2, then: If torque is in N·m and is in radians, work is in joules J. When an object is accelerating, the net torque is:

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**Maximum torque results when the force is applied tangentially to the radius.**

If the force is applied at an angle to the radius, only the tangential component will do work. A radial component would pass through the pivot and the torque would be 0 N·m.

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**Rotational Kinetic Energy**

When a torque does work on a rotating rigid body, the kinetic energy changes by an amount equal to the work done (work-energy theorem). Work = ΔKlinear + ΔKrotary Rotating bodies often have two types of kinetic energy (Work = ΔKrotary ): the kinetic energy due to the linear motion of the object and, the kinetic energy due to the rotational motion of the object.

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**Rotational Kinetic Energy**

Linear kinetic energy: Klinear = 0.5·m·v2 Rotational kinetic energy: Krotational = 0.5·I·w2 Total kinetic energy: Ktotal = Klinear +Krotational Many problems ask for the translational velocity of a rolling object, using v = ·r, solve for = v/r and substitute this into the rotational kinetic energy equation. See the example at the end of the notes.

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**Conservation of Energy for Rotational Motion**

Energy before = Energy after Ugi + Ki linear + Ki rotational = Ugf + Kf linear + Kf rotational

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**Power and Rotational Motion**

When a torque acts on a body that rotates with angular velocity , the power (rate of doing work) is: P = T· If torque is in N·m and is in rad/s, power is in watts W. Once an object is rotating at constant angular velocity ω, the only torque needed to keep the object rotating is the frictional torque; P = TFf·ω

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**Translational Velocity of Solid Ball at the Bottom of an Incline**

Potential energy at the top of an incline is transformed into kinetic energy at the bottom of the incline. Kinetic energy = 0.5·m·v ·I·2 Rotational inertia for a solid ball: 0.4·m·r 2 Relationship between linear and angular velocity: v = ·r; = v/r

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Substituting: Ug top = K bottom:

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**(for each object involved)**

Angular Momentum Rotating objects have angular momentum L. Angular momentum is the rotational equivalent of linear momentum. Equation: L = I·w w in rad/s Units: Conservation of Angular Momentum: Lbefore = Lafter SI1·w1 = SI2·w2 (for each object involved)

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