Uniform Circular MotionOctober 2014

Circular Motion• A is towards the center.

• V is tangential to the motion

• Speed is constant, V changes

• A force directed towards the center is what causes the acceleration (e.g. gravity, a string)

• If the force is removed, the ball will continue in a straight line at the speed it had.

Check your understanding …

• When a wheel rotates about a fixed axis, do all the points on the wheel have the same tangential speed?

Yes!

Do they all have the same velocity? No!

Check your understanding …

Circular Motion Equations

• Vt= 2πr/T

• ac = vt2/r

Where Vt = tangental velocity

R = radius T = period (time required to make one complete circle)Ac = centripetal acceleration

Circular Motion– We do

• The radius of a spacecraft orbiting earth is 6.67 x 106 m. If it orbits earth in 5292 seconds, what is the velocity of the spacecraft?

Vt= 2πr/T = 2*π*6.67X106m / 5292 sec = 7919 m/s

Circular Motion– We do

Jimmie Johnson is driving his #48 Lowe’s NASCAR around a bend that has a radius of 70 meters. It takes him 30 seconds to travel the track. What was the centripetal acceleration of Jimmie John’s #48 Lowe’s NASCAR?

Strategy: find vt, then find ac.

Vt= 2πr/T = 2π*70m/30sec = 14.66m/s

ac = vt2/r = (14.66 m/s)2 / 70 m = 3 m/s2

Centripetal Acceleration – You do

a. A girl sits on a tire that is attached to an overhanging tree limb by a rope. The girl’s father pushes her so that her centripetal acceleration is 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl’s tangential speed? b. A boy swings a yo-yo horizontally above his head so that the yo-yo has a centripetal acceleration of 1.5m/s2. If the yo-yo’s tangential speed is 11m/s, what is the length of the yo-yo?

c. Correct the following statement: The racing car rounds the turn at a constant velocity of 145 km/h.

Quadratic Equation

Quadratic equations, such as x = vit + ½ a t2

can be tricky to solve.

3 strategies1. Factoring (doesn’t usually work well with physics)2. Quadratic Equation solution (always works, but you need to

memorize)3. Trial & Error / Plugging in values (can be useful for multiple

choice)

We will talk about #2 and #3.

Quadratic Equation Solution• First, put your equation into this form: 0 = at2 + bt + c

for example, x = vit + ½ at2 becomes

0 = ½ at2 +vit – x

so , a = ½ a b = vi

c = -x

Then, use this equation to solve for t

Quadratic Equation Solution – We doA ball is launched upward with a speed of 15 m/s from an intial height of 5 m. What are the two approximate times that the object will be located at the height of 10 m abovethe ground?

Quadratic Equation Solution – We doA ball is launched upward with a speed of 15 m/s from an intial height of 5 m. What are the two approximate times that the object will be located at the height of 10 m abovethe ground?

0 = ½ at2 +vit – x

so , a = ½ a = ½ * - 9.8 m/s2 = -4.9 m/s b = vi

= 15 m/s

c = -x = -10 m

tt = 0.5 sec and 1 sec

Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to

find the solution.• Works best with multiple choice.

Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to

find the solution.• Works best with multiple choice.

Example – WE DO

• A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car?

A. 6.6 sB. 10.2 sC. 13.4 sD. 15.1 S

Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the

solution.• Works best with multiple choice.

Example – WE DO

• A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car?

A. 6.6 sB. 10.2 sC. 13.4 sD. 15.1 S

Use x = vit + ½ at2 for each car and compare x. At which

answer choice does the truck final reach the car?

Plugging in Numbers • Sometimes, you can just plug in numbers into the equation to find the

solution.• Works best with multiple choice.

Example – WE DO

• A car and truck start from the same position. The car has a constant velocity of 20 m/s. The truck has an initial velocity of zero, but accelerates 3 m/s. At approximately what time does the truck overtake the car?

A. 6.6 sB. 10.2 sC. 13.4 sD. 15.1 S

Use x = vit + ½ at2 for each car and compare x. At which

answer choice does the truck final reach the car?

Quadratic Equation – You do• A ball is dropped from 15 ft. At what time will the ball be 5 ft

from the ground? • Car 1 and Car 2 start at the same position. Car 1 has an initial

velocity of 10 m/s and an acceleration of 5 m/s2. Car 2 has an initial velocity of 15 m/s and an acceleration of 2 m/s2. At one time will car 2 overtake car one?

A. 1.2 secB. 2.5s

C. 3.4 s D. 4.2 s

Exit ticket & Homework

Exit ticket: write facts about circular motion

Homework:Circular acceleration problemsReview problemsExtra credit: Write 3 challenging physics problems and solve them on another sheet of paper.

Upcoming: CFA Wed / Thur next week. Exam right after break.