Motion in a Circle

Uniform Circular Motion

Motion of an object in a circle with uniform or constant speed.

Examples???

Angular Velocity

t

The angular velocity is the rate of change in angular displacement.

angular displacementangular velocitytime

minrev (Common

everyday language)

rad s-1 (physics language)

A record is rotating with an angular velocity of 45 rpm. If it rotates for 35 seconds, how many rotations does it make?

Linear Velocity

Assume it takes Sammy Speedster T seconds to drive one complete revolution around the circle with a radius of R. Draw his velocity vector at 2 different locations and label the vector with the speed he is traveling.

(T = Period = the amount of time it takes for an object to complete one revolution.)

Velocity

Does Sammy’s speed change with time?Does Sammy’s velocity change with time?

What does that mean for Sammy’s acceleration?Are there any forces acting on Sammy?

In which direction are they acting?

(A tangent line is a line which touches a circle at one point but does not intersect it.)

To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed. While the speed of the object is constant, its velocity is changing. Velocity, being a vector, has a constant magnitude but a changing direction. The direction is always directed tangent to the circle and as the object turns the circle, the tangent line is always pointing in a new direction.

A B C

The velocity and acceleration vectors are perpendicular to each other.This causes the velocity vector to change direction, but not magnitude.

An object is moving in a clockwise direction around a circle at constant speed.

A

B

C

1. Draw and label the velocity vector at each point.

2. Draw and label the acceleration vector at each point

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the

acceleration and the net force acting upon the car.

a = v2 / R a = (10.0 m/s)2 / (25.0 m)a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

Fnet = m • a Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The

halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

v = d / t v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)v = 8.97 m/s

a = v2 / R a = (8.97 m/s)2 / (12.0 m)a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

Fnet = m*a Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

945-kg car makes a 180-degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction acting upon the car.

F = mv2/R

F = (945kg)(10.0m/s)2/25.0mF = 3780 N

The roller coaster

Draw a free body diagram for the person at the top and the bottom of the loop

Suggested Method of Solving Circular Motion Problems1. construct a free-body diagram. Represent each force by a vector arrow

and label the forces according to type. 2. Identify the given and the unknown information (express in terms of

variables such as m= , a= , v= , etc.). 3. If any of the individual forces are directed at angles to the horizontal

and vertical, then resolve such forces into horizontal and vertical components.

3. Determine the magnitude of any known forces and label on the free-body diagram.

(For example, if the mass is given, then the Fgrav can be determined)4. Use circular motion equations to determine any unknown information.

(For example, if the speed and the radius are known, then the acceleration can be determined. And as another example, if the period

and radius are known, then the acceleration can be determined.) 5. Use the remaining information to solve for the requested information.

If the problem requests the value of an individual force, then use the kinematic information (R, T and v) to determine the acceleration

and the Fnet ; then use the free-body diagram to solve for the individual force value.