Презентация - Opening a door: physics in disguise. Why are most door handles placed far from the hinges?

Opening a Door: Physics in Disguise
Why are most door handles placed far from the hinges?

Describing Rotational Motion
Rotational motion is the motion of objects that spin about an axis.
Translational motion is the motion of objects that move from point to another.

Translational vs. Rotational (or Angular) Quantities
.TRANSLATIONAL.ANGULAR (OR ROTATIONAL)
Position..
Displacement..
Velocity..
Acceleration..
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Angular Position
We define the particle’s angle θ in terms of arc length and radius of the circle:
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Translational vs. Rotational (or Angular) Quantities
.TRANSLATIONAL.ANGULAR (OR ROTATIONAL)
Position..
Displacement..
Unit: meter
Unit: radians
Unit: meter
Unit: radians
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Translational vs. Rotational (or Angular) Quantities
.TRANSLATIONAL.ANGULAR (OR ROTATIONAL)
Velocity..
Acceleration..
Unit: m/s
Unit: rad/s
Unit: rad/s2
Unit: m/s2

SAMPLE PROBLEM
The rotor on a helicopter turns at an angular velocity of revolutions per minute. (a) Express the angular velocity in radians per second. (b) The pilot opens the throttle and the angular velocity of the blade increases while rotating 26 times in 3.60 sec. Calculate the angular velocity at that time.
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SAMPLE PROBLEM
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular velocity of the wheel is 2.00 rad/sec at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2 s? (b) What is the angular velocity of the wheel at t = 2.0 sec?
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SAMPLE PROBLEM
A dentist’s drill starts from rest. After 3.20 sec of constant angular acceleration, it turns at a rate of rpm. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.
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SAMPLE PROBLEM
An electric motor rotating a workshop grinding wheel at a rate of 100 rpm is switched off. Assume that the wheel has a constant negative acceleration of magnitude 2.00 rad/s2. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)?
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QuickCheck 7.9
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle will it have turned after time 2t? 25 rad 50 rad 75 rad 100 rad 200 rad
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QuickCheck 7.9
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. Through what angle will it have turned after time 2t? 25 rad 50 rad 75 rad 100 rad 200 rad
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Example Problem
A high-speed drill rotating counterclockwise takes 2.5 s to speed up to 2400 rpm. What is the drill’s angular acceleration? How many revolutions does it make as it reaches top speed?
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QuickCheck 7.8
Starting from rest, a wheel with constant angular acceleration spins up to 25 rpm in a time t. What will its angular velocity be after time 2t? 25 rpm 50 rpm 75 rpm 100 rpm 200 rpm
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QuickCheck 7.8
Starting from rest, a wheel with constant angular acceleration spins up to 25 rpm in a time t. What will its angular velocity be after time 2t? 25 rpm 50 rpm 75 rpm 100 rpm 200 rpm
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Rotational Dynamics and Moment of Inertia
A torque causes an angular acceleration. The translational and angular accelerations are
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Rotational Dynamics and Moment of Inertia
We compare with torque: We find the relationship with angular acceleration:
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Newton’s Second Law for Rotational Motion
The quantity Σmr2 in Equation 7.20, which is the proportionality constant between angular acceleration and net torque, is called the object’s moment of inertia I: The units of moment of inertia are kg  m2. The moment of inertia depends on the axis of rotation.
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Newton’s Second Law for Rotational Motion
A net torque is the cause of angular acceleration.
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Interpreting the Moment of Inertia
The moment of inertia is the rotational equivalent of mass. An object’s moment of inertia depends not only on the object’s mass but also on how the mass is distributed around the rotation axis.
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Interpreting the Moment of Inertia
The moment of inertia is the rotational equivalent of mass. It is more difficult to spin the merry-go-round when people sit far from the center because it has a higher inertia than when people sit close to the center.
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Interpreting the Moment of Inertia
Text: p. 208
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Example 7.15 Calculating the moment of inertia
Your friend is creating an abstract sculpture that consists of three small, heavy spheres attached by very lightweight 10-cm-long rods as shown in FIGURE 7.36. The spheres have masses m1 = 1.0 kg, m2 = 1.5 kg, and m3 = 1.0 kg. What is the object’s moment of inertia if it is rotated about axis A? About axis B? prepare We’ll use Equation 7.21 for the moment of inertia: I = m1r12 + m2r22 + m3r32 In this expression, r1, r2, and r3 are the distances of each particle from the axis of rotation, so they depend on the axis chosen.
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Example 7.15 Calculating the moment of inertia (cont.)
Particle 1 lies on both axes, so r1 = 0 cm in both cases. Particle 2 lies 10 cm (0.10 m) from both axes. Particle 3 is 10 cm from axis A but farther from axis B. We can find r3 for axis B by using the Pythagorean theorem, which gives r3 = 14.1 cm. These distances are indicated in the figure.
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Example 7.15 Calculating the moment of inertia (cont.)
solve For each axis, we can prepare a table of the values of r, m, and mr2 for each particle, then add the values of mr2. For axis A we have
[Insert Figure 7.36 (repeated)]
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Example 7.15 Calculating the moment of inertia (cont.)
For axis B we have
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Example 7.15 Calculating the moment of inertia (cont.)
assess We’ve already noted that the moment of inertia of an object is higher when its mass is distributed farther from the axis of rotation. Here, m3 is farther from axis B than from axis A, leading to a higher moment of inertia about that axis.
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The Moments of Inertia of Common Shapes
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Try It Yourself: Hammering Home Inertia
Most of the mass of a hammer is in its head, so the hammer’s moment of inertia is large when calculated about an axis passing through the end of the handle (far from the head), but small when calculated about an axis passing through the head itself. You can feel this difference by attempting to wave a hammer back and forth about the handle end and the head end. It’s much harder to do about the handle end because the large moment of inertia keeps the angular acceleration small.
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Using Newton’s Second Law for Rotation
Text: p. 211
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Example 7.16 Angular acceleration of a falling pole
In the caber toss, a contest of strength and skill that is part of Scottish games, contestants toss a heavy uniform pole, landing it on its end. A 5.9-m-tall pole with a mass of 79 kg has just landed on its end. It is tipped by 25° from the vertical and is starting to rotate about the end that touches the ground. Estimate the angular acceleration.
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Example 7.16 Angular acceleration of a falling pole (cont.)
prepare The situation is shown in FIGURE 7.37, where we define our symbols and list the known information. Two forces are acting on the pole: the pole’s weight which acts at the center of gravity, and the force of the ground on the pole (not shown). This second force exerts no torque because it acts at the axis of rotation. The torque on the pole is thus due only to gravity. From the figure we see that this torque tends to rotate the pole in a counterclockwise direction, so the torque is positive.
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Example 7.16 Angular acceleration of a falling pole (cont.)
solve We’ll model the pole as a uniform thin rod rotating about one end. Its center of gravity is at its center, a distance L/2 from the axis. You can see from the figure that the perpendicular component of is w⊥ = w sin θ. Thus the torque due to gravity is
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Example 7.16 Angular acceleration of a falling pole (cont.)
From Table 7.1, the moment of inertia of a thin rod rotated about its end is Thus, from Newton’s second law for rotational motion, the angular acceleration is
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Example 7.16 Angular acceleration of a falling pole (cont.)
assess The final result for the angular acceleration did not depend on the mass, as we might expect given the analogy with free-fall problems. And the final value for the angular acceleration is quite modest. This is reasonable: You can see that the angular acceleration is inversely proportional to the length of the pole, and it’s a long pole. The modest value of angular acceleration is fortunate—the caber is pretty heavy, and folks need some time to get out of the way when it topples!
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Example 7.18 Starting an airplane engine
The engine in a small air-plane is specified to have a torque of 500 N  m. This engine drives a 2.0-m-long, 40 kg single-blade propeller. On start-up, how long does it take the propeller to reach 2000 rpm?
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Example 7.18 Starting an airplane engine (cont.)
prepare The propeller can be modeled as a rod that rotates about its center. The engine exerts a torque on the propeller. FIGURE 7.38 shows the propeller and the rotation axis.
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Example 7.18 Starting an airplane engine (cont.)
solve The moment of inertia of a rod rotating about its center is found in Table 7.1: The 500 N  m torque of the engine causes an angular acceleration of
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Example 7.18 Starting an airplane engine (cont.)
The time needed to reach ωf = 2000 rpm = 33.3 rev/s = 209 rad/s is
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Example 7.18 Starting an airplane engine (cont.)
assess We’ve assumed a constant angular acceleration, which is reasonable for the first few seconds while the propeller is still turning slowly. Eventually, air resistance and friction will cause opposing torques and the angular acceleration will decrease. At full speed, the negative torque due to air resistance and friction cancels the torque of the engine. Then and the propeller turns at constant angular velocity with no angular acceleration.
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Example Problem
A baseball bat has a mass of 0.82 kg and is 0.86 m long. It’s held vertically and then allowed to fall. What is the bat’s angular acceleration when it has reached 20° from the vertical? (Model the bat as a uniform cylinder).
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Rolling Motion
Rolling is a combination motion in which an object rotates about an axis that is moving along a straight-line trajectory.
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Rolling Motion
The figure above shows exactly one revolution for a wheel or sphere that rolls forward without slipping. The overall position is measured at the object’s center.
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Rolling Motion
In one revolution, the center moves forward by exactly one circumference (Δx = 2πR).
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Rolling Motion
Since 2π/T is the angular velocity, we find This is the rolling constraint, the basic link between translation and rotation for objects that roll without slipping.
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Rolling Motion
The point at the bottom of the wheel has a translational velocity and a rotational velocity in opposite directions, which cancel each other. The point on the bottom of a rolling object is instantaneously at rest. This is the idea behind “rolling without slipping.”
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Example 7.20 Rotating your tires
The diameter of your tires is 0.60 m. You take a 60 mile trip at a speed of 45 mph. During this trip, what was your tires’ angular speed? How many times did they revolve?
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Example 7.20 Rotating your tires (cont.)
prepare The angular speed is related to the speed of a wheel’s center by Equation 7.25: ν = ωR. Because the center of the wheel turns on an axle fixed to the car, the speed v of the wheel’s center is the same as that of the car. We prepare by converting the car’s speed to SI units: Once we know the angular speed, we can find the number of times the tires turned from the rotational-kinematic equation Δθ = ω Δt. We’ll need to find the time traveled Δt from ν = Δx/Δt.
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Example 7.20 Rotating your tires (cont.)
solve a. From Equation 7.25 we have b. The time of the trip is
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Example 7.20 Rotating your tires (cont.)
Thus the total angle through which the tires turn is Because each turn of the wheel is 2π rad, the number of turns is
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Example 7.20 Rotating your tires (cont.)
assess You probably know from seeing tires on passing cars that a tire rotates several times a second at 45 mph. Because there are 3600 s in an hour, and your 60 mile trip at 45 mph is going to take over an hour—say, ≈ 5000 s—you would expect the tire to make many thousands of revolutions. So 51,000 turns seems to be a reasonable answer. You can see that your tires rotate roughly a thousand times per mile. During the lifetime of a tire, about 50,000 miles, it will rotate about 50 million times!
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Rotational momentum
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Rotational momentum
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Rotational momentum
For each experiment, the rotational inertia of the spinning person decreased (the mass moved closer to the axis of rotation). Simultaneously, the rotational speed of the person increased. We propose tentatively that when the rotational inertia I of an extended body in an isolated system decreases, its rotational speed ω increases, and vice versa. We can test this idea with a testing experiment.
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Rotational momentum
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Rotational momentum is constant for an isolated system
If a system with one rotating body is isolated, then the external torque exerted on the object is zero. In such a case, the rotational momentum of the object does not change: 0 = Lf – Li Lf = Li
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Example 8.7
Attach a 100-g puck to a string, and let the puck glide in a counterclockwise circle on a horizontal, frictionless air table. The other end of the string passes through a hole at the center of the table. You pull down on the string so that the puck moves along a circular path of radius 0.40 m. It completes one revolution in 4.0 s. If you pull harder on the string so that the radius of the circle slowly decreases to 0.20 m, what is the new period of revolution?
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Rotational momentum of an isolated system is constant
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Example 8.8
Imagine that our Sun ran out of nuclear fuel and collapsed. The Sun's current period of rotation is 25 days. What would the Sun's radius have to be for its period of rotation to be the same as during the pulsar described earlier?
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Rotational kinetic energy
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Rotational kinetic energy
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Rotational kinetic energy
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Flywheels for storing and providing energy
In a car with a flywheel, instead of rubbing a brake pad against the wheel and slowing it down, the braking system converts the car's translational kinetic energy into the rotational kinetic energy of the flywheel. As the car's translational speed decreases, the flywheel's rotational speed increases. This rotational kinetic energy could then be used later to help the car start moving again.
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Example 8.9
A 1600-kg car traveling at a speed of 20 m/s approaches a stop sign. If it could transfer all of its translational kinetic energy to a 0.20-m-radius, 20-kg flywheel while stopping, what rotational speed would the flywheel acquire?
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