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
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
Centripetal acceleration also exists for non-circular curved motion
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.
Centripetal acceleration also exists for non-circular curved motion What is the radius of the kissing circle here?
The centripetal acceleration must be caused by some force. Electromagnetic force Gravitational force
Relative motion
Gympie Melbourne
y x Ground
x y Car
x y Car
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.
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!
x y Car
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
Example: Plane in a cross-wind y 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.