1.1. To apply what we have learnt about circular To apply what we have learnt about circular motion to rides at the fairgroundmotion to rides at the fairground

2.2. To consider three particular cases of motion:To consider three particular cases of motion:

• The big dipperThe big dipper

• The long swingThe long swing

• The “centrifuge” wall of deathThe “centrifuge” wall of death

Book Reference : Pages 28-29Book Reference : Pages 28-29

At the bottom of a big dipper you are pushed At the bottom of a big dipper you are pushed into your seat and feel “heavier”...into your seat and feel “heavier”...

Centre of curvature

S

mg

velocity

At the bottom of the dip at speed At the bottom of the dip at speed vv with radius with radius rr, , Resolving the vertical forces:Resolving the vertical forces:

S – mg = mvS – mg = mv2 2 / r/ r

S = mg + mvS = mg + mv22 / r / r

The “The “extraextra” weight you experience when feeling ” weight you experience when feeling ““heavyheavy” is given by the centripetal force” is given by the centripetal force

Consider a person of mass Consider a person of mass mm on a very long on a very long swing of length swing of length rr

Fixed point

To winch

Initial position

r

S

mgVelocity

As the swing is released we can consider the As the swing is released we can consider the conservation of energy, loss in potential energy conservation of energy, loss in potential energy is the gain in the kinetic energyis the gain in the kinetic energy

mgh = ½mvmgh = ½mv22 v v22 = 2gh = 2gh

The passenger is on a circular path with radius The passenger is on a circular path with radius rr. . At the bottom of the swing the support force At the bottom of the swing the support force SS acts against the person’s weight mg. This acts against the person’s weight mg. This provides the centripetal force:provides the centripetal force:

S – mg = mvS – mg = mv22/r/r

Substituting for vSubstituting for v22::

S – mg = mvS – mg = mv22/r/r

S – mg = 2mgh/rS – mg = 2mgh/r

S = mg + 2mgh/rS = mg + 2mgh/r

The person “feels heavier” by 2mgh/r, if the The person “feels heavier” by 2mgh/r, if the swing drops through 90°, then the extra support swing drops through 90°, then the extra support force is twice the persons weight (mg)force is twice the persons weight (mg)

Consider a fairground ride which spins fast Consider a fairground ride which spins fast enough to keep you in place even when upside enough to keep you in place even when upside down at the top of the ride....down at the top of the ride....

Velocity

Rotation

Reaction R & weight mg

The wheel turns fast enough to keep the The wheel turns fast enough to keep the passenger in position as they pass over the top.passenger in position as they pass over the top.

At the top, the reaction R acts downwards and At the top, the reaction R acts downwards and together with the weight provides the together with the weight provides the centripetal force:centripetal force:

mg + R = mvmg + R = mv22/ r/ r

R = mvR = mv22/ r – mg/ r – mg

At a certain speed vAt a certain speed v00 such that v such that v0022 = gr, then the = gr, then the

reaction from the wall will be zeroreaction from the wall will be zero

A car on a big dipper starts from rest and descends A car on a big dipper starts from rest and descends though 45m into a dip which has a radius of though 45m into a dip which has a radius of curvature of 78m. Assuming that air resistance & curvature of 78m. Assuming that air resistance & friction are negligible. Calculate:friction are negligible. Calculate:

a.a. The speed of the car at the bottom of the dipThe speed of the car at the bottom of the dip

b.b. The centripetal acceleration at the bottom of The centripetal acceleration at the bottom of the dipthe dip

c.c. The extra force on a person with a weight of The extra force on a person with a weight of 600N in the train600N in the train

A swing at a fair has a length of 32m. A passenger of A swing at a fair has a length of 32m. A passenger of mass 69kg falls from the position where the swing is mass 69kg falls from the position where the swing is horizontal. Calculate:horizontal. Calculate:

a.a. The speed of the person at the lowest pointThe speed of the person at the lowest point

b.b. The centripetal acceleration at the lowest The centripetal acceleration at the lowest pointpoint

c.c. The support force on the person at the lowest The support force on the person at the lowest pointpoint

A wall of death ride at the fairground has a radius A wall of death ride at the fairground has a radius of 12m and rotates once every 6s. Calculate:of 12m and rotates once every 6s. Calculate:

a.a. The speed of rotation at the perimeter of the The speed of rotation at the perimeter of the wheelwheel

b.b. The centripetal acceleration of a person on The centripetal acceleration of a person on the perimeterthe perimeter

c.c. The support force on a person of mass 72kg at The support force on a person of mass 72kg at the highest pointthe highest point