Physics Ch. 7 Rotational Motion Rotational motion-involves objects that spin around a

fixed point (axis) Radius-distance from the center to the edge of the circle or

object Arc length (s) is the distance the object moves, measured

along the circumference of the circle Angle (θ) is the angle through which the object moves Arc length and angle are related- if the angle doubles, then

the arc length or distance also doubles

Angles can be measured in radians as well as degrees. All angles for this chapter will need to be in radians.

When the arc length is equal to the length of the radius, then the angle swept by r is equal to 1 rad, which is equal to about 57.3°.

θ = s/r = 2π rad = 2 (3.14) = 6.28 rad = 360°

Converting degrees to radians

Multiply the angle in degrees by 2π/360°

OR

θ (rad) = π/180° θ (deg)

Angular Displacement The change in the arc length, Δs, divided by the distance

of the object from the axis.

Δθ = Δs/r Angular displacement (radians) = change in arc

length/distance from axis S is considered positive when the point rotates

counterclockwise and negative when it rotates clockwise

Angular Displacement

Sample Problem 7A pg. 246

You Try It

Earth has an equatorial radius of approximately 6380 km and rotates 360° every 24 h.

A. What is the angular displacement (in degrees) of a person standing at the equator for 1.0 h?

B. Convert this angular displacement to radians. C. What is the arc length traveled by this person?

Angular Speed

Average angular speed ωavg is the ratio of the angular displacement, Δt, that the object takes to undergo that displacement. Angular speed describes how quickly the rotation occurs.

ωavg = Δ θ/ Δ t

Average angular speed = angular displacement/time

Angular speed is given in units of radians per seconds. Sometimes they are given in revolutions per unit time. 1 rev = 2 π rad

Angular Speed

Sample Problem 7B on pg. 248

You Try It-Angular Speed

An Indy car can complete 120 laps in 1.5 h. Even though the track is an oval rather than a circle, you can still find the average angular speed. Calculate the average angular speed of the Indy car.

Angular Acceleration

Angular acceleration has the units radians per second per second

Average angular acceleration = change in angular speed/ time interval

α avg = ω2 – ω1

t2 – t1

Angular Acceleration

Sample Problem 7C pg. 249

You Try it-Angular Acceleration

A top that is spinning at 15 rev/s spins for 55 s before coming to a stop. What is the average angular acceleration of the top while it is slowing?

Do two points on a wheel with two different radi have the same angular speed and angular acceleration?

Demo 1

Compare the variables of rotational motion to linear motion from ch. 2

Linear Angularx θv ωa α

Kinematics Formulas

Formulas in Table 7-2 on pg. 251

ω in these equations represents the instantaneous angular speed of the rotating object rather than the average angular speed.

Angular Kinematics

Sample Problem 7D on pg. 251

You Try it-Angular Kinematics A barrel is given a downhill rolling start of 1.5

rad/s at the top of a hill. Assume a constant angular acceleration of 2.9 rad/s2

A. If the barrel takes 11.5 s to get to the bottom of the hill, what is the final angular speed of the barrel?

B. What angular displacement does the barrel experience during the 11.5 s ride?

Ch. 7-2 Tangential and Centripetal Acceleration

How does angular speed and acceleration relate to golfing?

Angular speed and acceleration are related to the linear speed and acceleration.

For example, the golfer wants the club to travel an approximate circle around the body. A large angular acceleration will also produce a large linear acceleration of the club on the ball.

Tangential Speed

Two objects with different distances from the axis of rotation will have the same angular speed and acceleration.

However, two objects with different distances from the axis of rotation will have different tangential speeds.

Tangential speed=instantaneous linear speed

The object on the outside of a circle must travel the same angular displacement during the same amount of time as the object on the inside.

So, the object on the outside must travel a greater distance, Δs, than the object on the inside.

An object farther from the axis must travel at a higher tangential speed to achieve the same angular displacement as an object closer to the axis.

Tangential Speed Formula Tangential speed = distance from axis x angular speed

Vt = rω

ω is equal to instantaneous angular speed and is measured in radians per unit of time only

Tangential Speed

Sample problem 7E on pg. 254

You try it! A golfer has a maximum angular speed of 6.3 rad/s for her

swing. She can choose between two drivers, one placing the club head 1.9 m from her axis of rotation and the other placing it 1.7 m from the axis.

A. Find the tangential speed of the club head for each driver.

All other factors being equal, which driver is likely to hit the ball farther?

Tangential Acceleration The instantaneous linear acceleration of an object

directed along the tangent to the object’s circular path.

If an object traveling along a circular path speeds up, it experiences an angular acceleration. The linear acceleration related to this angular acceleration is tangent to the circular path.

Tangential Acceleration Tangential acceleration = distance from axis x angular acceleration

at = rα

Tangential Acceleration

Sample problem 7F on pg. 256

You Try It!

A yo-yo has a tangential acceleration of 0.98 m/s2

when it is released. The string is wound around a central shaft of radius 0.35 cm. What is the angular acceleration of the yo-yo?

Centripetal Acceleration

As an object moves in a circular path at constant speed, it still has an acceleration.

Remember that acceleration can be produced by a change in magnitude or a change in direction.

For an object traveling at a constant speed in a circular path, the acceleration is due to a change in direction, centripetal acceleration (center seeking).

Fig. 7-8 Δv = vf – vi

Notice that V initial is negative so the vector is reversed. Also, if the change in time is very small, these two vectors will almost be perpendicular to each other, making the resultant pointing toward the center of the circle.

Centripetal Acceleration

Centripetal acceleration = (tangential speed)2

distance from axis

ac = vt2

r

Centripetal acceleration = distance from axis x (angular speed)2

ac = rω2

Centripetal Acceleration

Sample problem 7C on pg. 258

You Try It!

A cylindrical space station with a 115 m radius rotates around its longitudinal axis at an angular speed of 0.292 rad/s. Calculate the centripetal acceleration:

A. Halfway to the rim of the station B. At the rim of the station

Tangential vs. Centripetal Acceleration

Tangential acceleration is due to changing speed. Centripetal acceleration is due to changing

direction. Fig. 7-9 When both types exist simultaneously, the total

acceleration can be found using the Pythagorean theorem.

The direction of the total acceleration depends on the magnitude of each component of acceleration and can be found using the inverse of the tangent function.

θ = tan-1 ac

at

Causes of Circular Motion

A ball connected to a string that is whirled in a circular path has a centripetal acceleration because the direction of the ball is constantly changing.

The inertia of the ball tends to maintain the ball’s motion in a straight path (tangent).

However, the string is forcing the ball to follow a circular path, centripetal force.

This is the force that maintains the circular motion of the object.

Force that Maintains the Circular Motion

Fc = mvt 2

r

Force that maintains circular motion = mass X (tangential speed)2

distance to radius

Fc = mrω2

Force that maintains circular motion = mass X distance to axis X (angular speed)2

Demo 3

Any object with a constant speed will still have a centripetal acceleration due to the change in direction.

The force that keeps the object moving circular is the centripetal force.

Try this with a ball attached to string. Try this with an object that has a constant velocity.

Sample Problem 7H

Force that maintains the circular motion on pg. 261

Challenge Problem

The moon (mass = 7.36 X 1022 kg) orbits Earth at a range of 3.84 X 105 km with a period of approximately 28 days. Determine the force that maintains the circular motion of the moon.

The force that maintains circular motion is necessary for circular motion.

If the force vanishes, then the object will travel in a straight line path tangent to the circle.

Notice that the force is always at a right angle to the direction of motion.

Fig. 7-11 shows the direction the ball follows if the force vanishes.

Consider a car making a sharp left turn with a passenger inside the car. The passenger slides right and hits the door. Is there a force pushing an object outward?

Centrifugal force is often used to explain this but not used in this book.

Could it be inertia?

Inertia can explain

The passenger is traveling straight. When the car turns left, the passenger wants to continue straight because of inertia (Newton’s first Law).

However, if a larger force, that maintains the circular motion, acted on the passenger, it would force the person to move in a circular motion with the car. Friction is the source of the force (between the person and the seat).

If the force is not sufficient, the person slides across the seat.

Eventually, the door will provide a large enough force to allow the passenger to follow the circular motion with the car.

Newton’s Law of Universal Gravitation

Planets move in a circular orbit around the sun. The force keeping these planets from moving off

in a straight path is called the gravitational force. The gravitational force is the force between any

two objects. This force is a force of attraction and equal but

opposite in magnitude between two objects.

Newton’s Law of Universal Gravitation

Fg = G m1 m2

r2

Graviational Force = constant X mass1 X mass 2

(distance b/t center of masses)2

G is the constant of universal gravitaton and equal to 6.673 X 10-11

The force between two masses decreases as the masses move farther apart.

By substituting the actual values of the mass of Earth and radius of Earth, we can find the free-fall acceleration of 9.83 m/s2

Gravitational Force

Sample Problem 7I on pg. 264

You Try It!

Find the gravitational force exerted on the moon (mass = 7.36 X 1022 kg) by Earth (mass = 5.98 X 1024 kg) when the distance between them is 3.84 X 108 m).