7.3: Circular Motion & Torque

Vector Components continued

Uniform Circular Motion

How can an object moving at constant speed be

accelerating?

Uniform circular motion is the movement of an object or

point mass at constant speed around a circle with a fixed

radius.

Describing Circular Motion

An object’s position relative to the center of a circle

is given by the position vector, r. As the object

moves, the length of the vector remains constant,

but the direction changes.

Velocity is displacement over time. So,

displacement would be Δr. Recall that average

velocity is Δd/Δt, so for circular motion

t

rv

Circular Velocity

The velocity vector is tangent to the circular path and perpendicular to the position vector. For constant speed, as the velocity vector moves around the circle its direction changes, but the length remains constant.

Recall a = Δv/Δt

a is in the same direction as the Δv, toward the center.

r1

r2

v1

v2

Δv

v1

v2

a

Centripetal Acceleration Center-seeking acceleration for an object in

uniform circular motion is centripetal

acceleration. How is it calculated?

What about speed? During one revolution

time (T), the object covers the circumference

distance( ), so speed is v = / T.

r

vac

2

r2 r2

Reintroducing Newton’s 2nd Law

If an object is accelerating in a circle, there must be a cause (Fnet) toward the center. Identify the agent (contact or long range) and rewrite Fnet = ma.

For solving circular motion problems use coordinate system with x-axis toward acceleration (center) and called c (centripetal) and y-axis called tang (tangential) which is the direction of the velocity.

r

mvF

maF

net

cnet

2

2

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T

rmFnet

Example Problem

A 13-g rubber stopper is attached to a 0.93m string. The stopper is swung in a horizontal circle, making one revolution in 1.18s. Find the tension force exerted by the string on the stopper.

Sketch the problem with radius, direction of motion, axis with +c and tang. Show that the directions of a and FT are parallel to c.

Solution

Known: Unknown:

m = 13g FT = ?

r = 0.93m

T = 1.18s

Calculations:

ac = 4π2r/T2

ac=4(3.14)2(0.93m) / (1.18s)2= 26 m/s2

FT= ma= (0.013kg) (26 m/s2) = 0.34 N

* Note for force in N, mass must be in kg.

Can you match Centripetal force?

If forces come in pairs, is there an “outward force” causing the stopper to stay out there at the end of the string? When a car turns left sharply, is there a force causing a passenger to move toward the door? NO! Newton’s 1st law says an object in motion continues… This outward motion is simply the inertia of the object and this “centrifugal force” is FICTITIOUS!

Changing Circular Motion

In uniform circular motion, a point mass rotates around an external axis of radius, r. (Ex a person on a merry-go-round, a sock in a washing machine)

A rigid rotating object is a mass that rotates around its OWN internal axis. (Ex a merry-go-round platform, a revolving door).

To open a door, you apply a force at a distance from the hinges and perpendicular to the door.

Torque Lever arm- perpendicular distance from

the axis of rotation to the applied force.

Torque- a perpendicular force that causes an object to rotate around an axis.

Torque = force x lever arm distance

F┴ = mg x d

Torque can stop, start, or change rotation

Why is a doorknob at the edge of a door and not in the center? How can you increase Torque? Wanna seesaw?

Your turn to Practice

Do pg 166 Prac. Problems # 15 & 16

Do pg 170 #s 10, 12, 13, 26, 29, 49, 50, 53,

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