Week

Centripetal Acceleration

Objective

Student will: Solve problems involving centripetal acceleration and centripetal force

Cornell Notes (1/3)Centripetal Acceleration: The rate of change in the direction of a moving object.

An object has constant speed but is accelerating because of change in direction.

The units for centripetal acceleration (m/s2)

Formula:

Centripetal Accel. (m/s2)Tangential Velocity (m/s) Radius (m)

Questions1) How would you

define centripetal acceleration?

a⃗c=𝑣𝑡

2

r

Cornell Notes (2/3)Vt: Tangential velocity. Velocity’s direction is tangent to the circular motion.

The object’s speed along an imaginary line drawn tangent to the circular path.

Example

2) What is the difference between tangential velocity and tangential speed?

3) What is the direction of centripetal acceleration?

When tangential speed is constant, the motion is called uniform circular motion

Acceleration’s direction is always towards the center of the circle or curve.

Cornell Notes (3/3)Centripetal Force: The net force on an object in uniform circular motion

Friction between a car’s tires and a circular track

Gravitational force is a centripetal force that keeps the moon in its orbit.

Formula:

Centripetal Force (N)Mass (kg)Tang. Velocity (m/s)Radius (m)

4) How would you define centripetal force?

F⃗ c=m𝑣𝑡2

r

Cornell Notes (1/5)

Example: Centripetal AccelerationThe thunder God, Thor, flies around Earth with a

tangential speed of 343 m/s around Earth. If the radius of Earth is 6,371 km, what is Thor’s centripetal acceleration? Assume his mass to be 290.299 kg

Cornell Notes (2/5)Given:

Unknown:

Steps1) Define

a⃗c=?

v⃗=343m /sr=6,371km r=6,371,000mm=290.299kg

The thunder God, Thor, flies around Earth with a tangential speed of 343 m/s around Earth. If the radius of Earth is 6,371 km, what is Thor’s centripetal acceleration? Assume his mass to be 290.299 kg

Cornell Notes (3/5)Choose an equation or situation:

Rearrange the equation to isolate the unknown:

2) Plan

a⃗c=v⃗ t2

r

=

Cornell Notes (4/5)Substitute the values into the equation and solve

3) Calculate

=

= 0.018466331816

=

=

Cornell Notes (5/5)4) Evaluate Compare to the radius of the

Earth and Thor’s velocity, the centripetal acceleration should be a small number.

Thor’s centripetal acceleration is 0.0185 m/s2

What is Thor’s centripetal force then?

Extension

Fc = mac

F c=290.366 ∙0.0184F c=5.3619949041F c=5.361N