Further Mechanics Tutorial 4 - Examples Involving Circular Motion
The maximum speed of a vehicle around a corner can be increased by banking the track. This is used widely on high-speed railway lines, as well as motor-racing tracks.
The diagram shows a wagon going round a corner which is banked. (Sometimes this is called super-elevation or cant.)
The normal force is the hypotenuse. We can write an equation for the centripetal force:
F = N sin q
Therefore:
The normal force is related to the weight by:
mg = N cos q
This rearranges to:
We also know that:
So we can say:
The mass terms cancel out. Now, sin ÷ cos = tan, and the mass terms cancel out, so we can say:
This rearranges to give:
v^{2} = gr tan q
The radius of a railway track is 360 m. It needs to take trains that are travelling at 18 m/s. What is the angle of bank that is needed? |
These ideas are also used to explain how aeroplanes bank when turning. Flaps on the wings (ailerons) lift one wing and push the other wing down. The aeroplane turns in a circle of radius that depends on the speed and the angle of bank.
The angle of bank is shown on the attitude indicator, a vital instrument for the pilot. It is shown in the picture below.
An aeroplane is flying at constant height and constant speed of 50 m/s. The pilot banks to 30^{o} to make a turn. Use g = 9.8 m/s^{2}. (a) What is the radius of the turn? (b) What is the angular velocity of the turn? (c) The pilot has a mass of 75 kg. What is the force acting on the pilot? |
Vertical circular motion in a gravity field
Centripetal force is the main principle of many fairground and theme park rides. People pay good money to experience the changing forces. They apparently enjoy it rather like a baby in a baby-bouncer.
In the wall of death machine, people are spun around a drum which starts in the horizontal plane, and then goes vertical – a big version of a washing machine.
The centripetal force, F, and the velocity vector, cause the wall to push on the participants with a reaction force R, who feel that they are being thrown outwards against the wall of the drum. (We could describe R as the “felt” force)
When the machine is horizontal, the weight is acting at 90^{o} to the force and is balanced by the floor, so can be disregarded:
When the machine goes vertical, there are two important principles at work:
the weight mg is acting downwards all the time.
Both the reaction force and the weight vectors sum up to the centripetal force.
At the maximum height the weight and the reaction force are acting in the same direction, so the resulting force F is given by:
F = R + mg
We can rewrite this for any angle made by the radius and the centre line:
F = R + mg sin q
Rearranging:
This explains why you feel "lighter" when you go over the top of one of these machines.
At a certain speed, R becomes 0, so you feel weightless. As the m terms cancel out, we get:
v^{2} = gr
Aerobatic pilots know this well. As they go over the top of a circular manoeuvre, they feel weightless. Weightlessness training for astronauts is carried out in converted airliners that follow a path like this. These aeroplanes are known as "vomit comets" for obvious reasons.
Photo from NASA (Wikimedia Commons)
If they go even faster, they can experience "negative g". If this is excessive, it can do damage to their aeroplanes that is just as serious as positive g.
What would happen to the cosmonauts above if the aeroplane went into negative g? |
||
An aerobatic pilot makes a vertical circular loop of radius 1000 m. What speed does his machine need to travel at in order for him to feel zero g? |
Flying a perfect vertical circle is not easy. The expression "going pear-shaped" has its origins in aerobatics. Many student stunt pilots' first attempts are decidedly pear-shaped.
Back to the Wall of Death:
At the bottom of the wheel the weight is acting in the opposite direction to the reaction force, so that:
F = R - mg
Therefore:
This is why you feel heavier going past the bottom of the contraption.
A wall of death machine has a radius of 12 m, and rotates once every 6 seconds. Calculate: a. The speed of rotation of the wheel b. The centripetal acceleration of a person on the perimeter when the machine is horizontal; c. The machine is now vertical. Calculate the support force, R, when the person of mass 72 kg is at the highest point. d. …and at the lowest point. |
||
A
stone of mass 0.50 kg is attached to an inextensible string and
whirled in a vertical circle of radius 0.98 m at a constant speed of 7.0
m/s. Gravity
g = 9.8 m/s^{2}
(a)
calculate the angular speed of the stone.
(b) calculate the centripetal acceleration of the stone; (c) the least tension in the string. (Hint: think which way gravity acts) |