1. y
An object is moving along a circular path at constant speed. At point A, its velocity is < 0, -2, 0 > m/s. A B x
What will the velocity be when it reaches point B?
(1) < 0, 2, 0 > m/s (3) < -2, 0, 0 > m/s
(2) < 0, -2, 0 > m/s (4) < 0, 4, 0 > m/s

2. What will the acceleration be when it reaches point B?y
An object is moving along a circular path at constant speed. At point A, its velocity is < 0, -2, 0 > m/s.
R = 2 m A B x
What will the acceleration be when it reaches point B?
(1) < 0, 4, 0 > m/s (3) < 2, 0, 0 > m/s
(2) < 0, 2, 0 > m/s (4) < -2, 0, 0 > m/s

3. Centripetal acceleration also exists for non-circular curved motion

4. Centripetal acceleration also exists for non-circular curved motion
Use the radius of the ‘kissing circle’ at any point of the curve to calculate the centripetal acceleration there.

5. Centripetal acceleration also exists for non-circular curved motion
What is the radius of the kissing circle here?

6. The centripetal acceleration must be caused by some force.
Electromagnetic force
Gravitational force

7. Relative motion

8. Gympie
Melbourne

9. y x
Ground

10. x y
Car

11. x y
Car

12. In general:
This is called a Galilean transformation, after Galileo.
It is not exactly correct:
But it works if v << c, the speed of light.

13. In general:
For speeds close to the speed of light, Einstein showed that we must use a Lorentz transformation instead.
You’ll see it next semester!

14. x y
Car

15. From a car travelling at a velocity of < 120, 0, 0> km/hr with respect to the ground, I see a kangaroo, moving backwards with respect to the car with a velocity
<-60, 0, 0> km/hr.
What is the velocity of the kangaroo with respect to the ground? x y
Car
<20, 0, 0> km/hr
<180, 0, 0> km/hr
<-180, 0, 0> km/hr
<60, 0, 0> km/hr
<-60, 0, 0> km/hr

16. Example: Plane in a cross-windy
20° θ
The speed of the plane relative to the wind is 215 km/hr.
The speed of the wind relative to the ground is 65 km/hr.
Find θ, and the speed of the plane relative to the ground.