1.To consider speed & velocity around a circle 2.To consider acceleration as a change in velocity 3.To define an equation for centripetal acceleration.

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Presentation transcript:

1.To consider speed & velocity around a circle 2.To consider acceleration as a change in velocity 3.To define an equation for centripetal acceleration 4.To define an equation for centripetal force Book Reference : Pages 24-25

If an object is moving in a circle with a constant speed, it’s velocity is constantly changing.... Because the direction is constantly changing.... If the velocity is constantly changing then by definition the object is accelerating If the object is accelerating, then an unbalanced force must exist Velocity v acceleration

Velocity v B  Velocity v A Consider an object moving in circular motion with a speed v which moves from point A to point B in  t seconds (From speed=distance / time), the distance moved along the arc AB,  s is v  t Velocity v B Velocity v A vv CA B  The vector diagram shows the change in velocity  v : (v B – v A )

Velocity v B  Velocity v A The triangles ABC & the vector diagram are similar If  is small, then  v / v =  s / r Velocity v B Velocity v A vv CA B  Substituting for  s = v  t  v / v = v  t / r  v / v = v  t / r (a = change in velocity / time) a =  v /  t = v 2 / r

We can substitute for angular velocity.... a = v 2 / r From the last lesson we saw that: v = r  (substituting for v into above) a = (r  ) 2 / r a = r  2

In exactly the same way as we can connect force f and acceleration a using Newton’s 2 nd law of motion, we can arrive at the centripetal force which is keeping the object moving in a circle f = mv 2 / r or f = mr  2 Any object moving in a circle is acted upon by a single resultant force towards the centre of the circle. We call this the centripetal force

Gravity which keeps satellites in orbit around Earth and the Earth in orbit around the sun is a classic example of a centripetal force. Planet satellite Gravity

The wheel of the London Eye has a diameter of 130m and takes 30mins for 1 revolution. Calculate: a.The speed of the capsule b.The centripetal acceleration c.The centripetal force on a person with a mass of 65kg

The speed of the capsule : Using v = r  we know that we do a full revolution (2  rad) in 30mins (1800s) v = (130/2) x (2  / 1800) v = 0.23 ms -1

The centripetal acceleration: Using a = v 2 / r a = (0.23) 2 / (130/2) a = 7.92 x ms -2 The centripetal force: Using f = ma F = 65 x 7.92 x F = N

An object of mass 0.15kg moves around a circular path which has a radius of 0.42m once every 5s at a steady rate. Calculate: a.The speed and acceleration of the object b.The centripetal force on the object [.528 ms -1, 0.663ms -2, 0.100N]