Circular Motion © David Hoult 2009

Acceleration of a Body Moving in a Circular Path at Constant Speed The magnitude of the velocity of the body is constant but the direction is constantly changing, therefore, the body is © David Hoult 2009

Acceleration of a Body Moving in a Circular Path at Constant Speed The magnitude of the velocity of the body is constant but the direction is constantly changing, therefore, the body is accelerating At any instant, the direction of the velocity is a © David Hoult 2009

Acceleration of a Body Moving in a Circular Path at Constant Speed The magnitude of the velocity of the body is constant but the direction is constantly changing, therefore, the body is accelerating At any instant, the direction of the velocity is a tangent to the circular path © David Hoult 2009

The magnitude of the acceleration depends on © David Hoult 2009

The magnitude of the acceleration depends on i) the speed of the body © David Hoult 2009

The magnitude of the acceleration depends on i) the speed of the body ii) the radius of the circular path © David Hoult 2009

We might suggest that a  © David Hoult 2009

We might suggest that a  v © David Hoult 2009

We might suggest that a  v and that a  © David Hoult 2009

We might suggest that a  v and that a  1 r and therefore © David Hoult 2009

We might suggest that a  v and that a  1 r Consideration of the units suggests that and therefore a  v r © David Hoult 2009

We might suggest that a  v and that a  1 r Consideration of the units suggests that and therefore a  v r v2v2 r © David Hoult 2009

It can be shown that the magnitude of the acceleration is given by a = v2v2 r © David Hoult 2009

It can be shown that the magnitude of the acceleration is given by a = v2v2 r or in terms of angular speed © David Hoult 2009

It can be shown that the magnitude of the acceleration is given by a = v2v2 r or in terms of angular speed a = r  2 or in terms time period © David Hoult 2009

It can be shown that the magnitude of the acceleration is given by a = v2v2 r or in terms of angular speed a = r  2 or in terms time period a = 4 2 r4 2 r T2T2 © David Hoult 2009

The direction of this acceleration is © David Hoult 2009

The direction of this acceleration is towards the centre of the circle For this reason it is called a © David Hoult 2009

The direction of this acceleration is towards the centre of the circle For this reason it is called a centripetal acceleration and is said to be caused by a centripetal force © David Hoult 2009

The direction of this acceleration is towards the centre of the circle For this reason it is called a centripetal acceleration and is said to be caused by a centripetal force F c = m v2v2 r F c = mr  2 © David Hoult 2009

A centripetal force does not change the kinetic energy of the body on which it acts because it acts © David Hoult 2009

A centripetal force does not change the kinetic energy of the body on which it acts because it acts at 90° to the direction of the motion of the body © David Hoult 2009

Estimate the magnitude of the force needed to cause a car to move around a curve in a road at 50 km h -1. What force causes the centripetal acceleration in this situation ? © David Hoult 2009

Estimate the magnitude of the force needed to cause a car to move around a curve in a road at 50 km h -1. What force causes the centripetal acceleration in this situation ? Friction © David Hoult 2009

Estimate the magnitude of the force needed to cause a car to move around a curve in a road at 50 km h -1. What force causes the centripetal acceleration in this situation ? Estimates needed: Friction © David Hoult 2009

Estimate the magnitude of the force needed to cause a car to move around a curve in a road at 50 km h -1. What force causes the centripetal acceleration in this situation ? Estimates needed: mass of car, m Friction © David Hoult 2009

Estimate the magnitude of the force needed to cause a car to move around a curve in a road at 50 km h -1. What force causes the centripetal acceleration in this situation ? Estimates needed: mass of car, m radius of path,r Friction © David Hoult 2009

F c = m v2v2 r © David Hoult 2009

A small mass hangs on a string inside the car. It is observed by a passenger. © David Hoult 2009

If the car turns to the left: A small mass hangs on a string inside the car. It is observed by a passenger*. * the mass is in front of the passenger. © David Hoult 2009

If the car turns to the left: A small mass hangs on a string inside the car. It is observed by a passenger. © David Hoult 2009

Find the angle  during the time the car is moving round the curved path. A small mass hangs on a string inside the car. If the car turns to the left:  © David Hoult 2009

mg © David Hoult 2009

Te cos  mg © David Hoult 2009

Te cos  mg Te sin  © David Hoult 2009

Te cos  mg Te sin  The vertical forces acting on the mass are in equilibrium, therefore © David Hoult 2009

Te cos  mg Te sin  The vertical forces acting on the mass are in equilibrium, therefore Te cos  must have the same magnitude as mg © David Hoult 2009

Te cos  mg Te sin  The mass is accelerating to the left, horizontally © David Hoult 2009

Te cos  mg Te sin  The mass is accelerating to the left, horizontally The horizontal component of the tension provides the centripetal force needed for this acceleration. © David Hoult 2009

Te cos  mg Te sin  Therefore Te sin  = m v2v2 r © David Hoult 2009

...need I say more ? © David Hoult 2009