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Rotational Motion and The Law of Gravity

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Pure Rotational Motion

A rigid body moves in pure rotation if every point of the body moves in a circular path, with all such paths laying on a common straight line called the

axis of rotation.

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Angular Displacement

Angular Displacement of a body is the angle through which a point or line has been rotated

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Angular Displacement

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Radians

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Radians-Degree Conversions to Internalize

Degrees Radians Degrees Radians0˚ 0 210˚ 7π/6

30˚ π/6 225˚ 5π/445˚ π/4 240˚ 4π/360˚ π/3 270˚ 3π/290˚ π/2 300˚ 5π/3

120˚ 2π/3 315˚ 7π/4135˚ 3π/4 330˚ 11π/12150˚ 5π/6 360˚ 2π180˚ π

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Converting Between Units of Angular Displacement

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Angular Speed

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Angular Acceleration

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Rotational and Linear Kinematic Equations

Notice that the set of new equations on the left have the same form as the set of familiar kinematics equations on the right, with

the substitution of old (linear) quantities with new (angular) quantities as follows.

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Rotational and Linear Kinematic Equations

Note: The linear kinematics equations on the left were derived under the assumption of constant acceleration and so we could only use them in that case. If acceleration was not constant we had to split the problem up into portions in which acceleration

was constant. A similar case applies for the rotational kinematic equations on the right.

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Tangential (Linear) Velocity

• Same velocity we’ve been discussing all along (m/s)• Has the following magnitude:

• Directed tangent to circular path.• Two points that are located at different

distances from axis or rotation haveequal angular speeds but differenttangential speeds.

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Tangential Speed

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Tangential Acceleration

If a rotating object speeds up, each point on that object experiences an (equal) angular acceleration. The linear acceleration related to this angular acceleration is called the tangential acceleration and can be found as follows:

If the object’s angular speed changes by an amount ∆ω in a time interval ∆t, the linear speed of a point a distance r away from the axis of rotation changes by an amount ∆vt as follows:

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Tangential Acceleration

Dividing by ∆t gives to examine the rate at which the speed changes, we obtain:

For short time intervals (∆t small), the quantity on the left is the tangential acceleration and the ratio on the right is the angular acceleration so that we have the tangential acceleration of a point on a rotating body in terms of the angular acceleration of the body:

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Tangential Acceleration

Note that only the magnitude, not direction, of the velocity vector changed so that the tangential acceleration vector points parallel to the linear velocity vector , that is, tangent to the circular path.

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Centripetal Acceleration

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Centripetal AccelerationCentripetal Acceleration: acceleration directed towards the

center of a circular path.

Note that even though an object moves at a constant speed, it still has an acceleration

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Acceleration Vector in Circular Motion

When an object has both angular and centripetal acceleration, the total acceleration vector can be found using the usual techniques:

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Centripetal Force

Centripetal Force: The force required to maintain centripetal motion; directed towards the center of the circular path

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Force Diagrams Revisited

A car travels at a constant speed of 30.0 mi/h (13.4 m/s) on a flat circular turn of radius 50.0 m. What minimum coefficient of static friction, μs, between the tires and the roadway will allow the car to make the circular turn without sliding?

As the car is moving at a constant speed, there is no tangential acceleration.

There is, however, a centripetal (or radial) acceleration as a result of the force of static friction.

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Force Diagrams Revisited

The maximum static friction force, fs,max provides the radial acceleration:

In the vertical direction, the force of gravity and the normal force are in equilibrium:

Substituting this expression in for FN,

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Banked Curve Problem

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Starship 3000

Find the rate which the Starship 3000 rotates if Alec descends at a rate of 0.5 m/s2.

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Newton’s Law of Gravitation

Magnitude of the Gravitational Force exerted by a spherical mass on a particle

Gravitational Constant

“Inverse-Square Law”

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Newton’s Law of Gravitation

Gravitational Force exerted by a spherical mass on a particle

Gravitational Constant

“Inverse-Square Law”

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Newton’s Law of Gravitation

Find the gravitational force between two billiards balls each of mass 0.15 kg, the centers of which are spaced a distance of 50 cm apart.

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Net Gravitational Force

Find the net gravitational force on the red billiard ball due to the two blue billiards balls as shown, where each ball is of mass 0.15 kg as before.

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Newton’s Law of Gravitation

The magnitude of the gravitational force on an object of mass m at the Earth’s surface is:

Gravitational Force due to the Earth

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Newton’s Law of Gravitation

If you were to dig some depth d into a spherically symmetric mass of constant density, the gravitational force exerted on you by the entire sphere would be equivalent to the g-force exerted by the smaller sphere of reduced mass “underneath your feet” – the donut exerts no net gravitational force!

One amazing consequence of inverse-square nature of gravitational force is the following:

d

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Gravitational Potential Energy Revisited

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Gravitational Potential Energy Revisited

Find the Potential Energy (PE) of an object of mass m an altitude h above the surface of the Earth (h<<RE).

Let’s take a look at the function :

I claim that we can approximate this with another function:

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Gravitational Potential Energy Revisited

Let’s ensure that this is actually a good approximation:

On your graphing calculator, plot the following functions:y1(x)=1/(1+x)y2(x)=1-xy3(x)=y1(x)-y2(x)

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Gravitational Potential Energy Revisited

It appears we can reach an x-value of 0.2 before we see more than even a 5% error.

So long as our altitude is less than one-fifth the radius of the Earth, or, approximately 1275 km, we are well within 5% error.

Recall that our x is h/RE so h/RE < 0.2 is within a tolerance of 5%, or h<0.2 RE.

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Gravitational Potential Energy Revisited

Back to finding the PE of an object of mass m an altitude h above the surface of the Earth (h<<RE).

(up to an adjustable constant, a reference level)

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Escape Velocity

Find the minimum speed necessary to escape the Earth’s gravitational pull. This is known as the escape velocity.

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Kepler’s Laws of Planetary Motion

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Kepler’s 1st Law

All planets move in an elliptical orbit with the Sun at one of the focal points.

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Kepler’s 2nd Law

A line drawn from the Sun to any planet sweeps out equal areas in equal time intervals.

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Kepler’s 3rd Law

The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun

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