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REG_Q01=ADV_Q
Semester 1, Y
Question 1
A large wooden turntable in the shape of a flat uniform disk is freely rotating about a vertical axis through its centre. A parachutist descends vertically and makes a soft landing on the turntable, without slipping, at a point near its outer rim.
(a) Will the angular speed of the turntable increase or decrease after the parachutist lands? Briefly justify your answer.
(b) Will the kinetic energy of the system (turntable + parachutist) increase or decrease in the process? Briefly explain your answer. (5 marks)
Solution
(a)
Angular momentum L = I ω will be conserved in the process. The addition of the parachutist
will increase I and hence ω will decrease.
(2 marks)
(b) We have
I 1 ω 1 = I 2 ω 2
where, I 1 is the initial moment of inertia of the turntable and ω 1 is its angular velocity, I 2 is the
moment of inertia of the turntable plus parachutist and ω 2 is its angular velocity.
Therefore
1 2 1 1 2
I
I
ω = ω < ω since I 2 > I 1.
The kinetic energy is 2 (^2 1 2 ) 2 2 2 2 2 1 1 1 2 2 1 1 2
I I
K I I I
I I
I
K
I
So K (^) 2 < K 1 since I (^) 2 > I 1.
Hence kinetic energy decreases. Friction will do net negative work on the system (turntable + parachutist) in the process of accelerating the parachutist in the horizontal direction. The friction would in reality most likely arise from slipping between the parachutist and the turntable. But it could also arise from ‘deformation’ of the parachutist as he/she is accelerated to the velocity of the turntable. Energy is lost as a result. (3 marks)
Solution (a) Momentum is conserved during the explosion because the forces acting on the two fragments as a result of the explosion are internal to the projectile system. (1 mark) (b) The backward fragment falls at distance x 1 (^) = R / 2 from point of firing since at the highest point
the projectile has travelled half the range.
The centre of mass falls at distance R since the centre of mass is unaffected by the explosion, which is internal to the projectile system.
Taking the origin at the launch point and from the centre of mass formula we derive the value for x 2 , the point at which the forward fragment lands. 1 1 2 2 1 2 1 2 2
1 2 2 3 2 2
cm
m x m x x m m m R mx R m R R x x R
The second fragment lands at (^3) R / 2from the launch point. Alternate method using conservation of momentum Take ux as the horizontal velocity of the projectile just before the explosion and vx as the
velocity of the forward fragment just after the explosion.
Using conservation of momentum we have 2 (0) 2
x x x x
mu m m v v u
We can treat the horizontal and vertical motions separately. The project would have travelled a further distance R / 2from the peak of the path until it hit the ground. The forward fragment travels at twice the horizontal speed and hence goes a distance 2 R / 2= R until it hits the ground.
The total distance that it has travelled from the origin is therefore R + R / 2 = 3 R / 2
(2 marks)
(c) Kinetic energy before explosion: (^12 2) m ux^2 = m ux^2
Kinetic energy after explosion: 12 m vx^2 + 0
From conservation of momentum at explosion: 2 0 2
x x x x
m u m v v u
Therefore kinetic energy after explosion: ( )
1 2 1 2 2 2 m vx^ =^2 m^^2 u^ x =^2 m ux
Therefore kinetic energy has increased by a factor 2. (2marks)
Solution
(a) Taking upwards as the positive direction, the gravity force acting on the yo-yo is:
F g = − m g.
The yo-yo is in equilibrium in the vertical direction and so the normal reaction force from the table acting (upwards) on the yo-yo is:
N = m g.
At the point of almost slipping, we have a frictional force f given by:
f = μ s N =μ sm g
acting to the left in the diagram as we are told that the yo-yo will either roll or slip to the right.
The net force acting on the yo-yo is therefore
Fnet = F − f = F − μ sm g. (1)
(1 mark)
(b) Taking the clockwise direction as positive and considering torques acting around the centre of rotation of the yo-yo we have: Torque due to pulling force τ (^) pull = − F b.
Torque due to frictional force τ (^) fric = f R
At the point of almost slipping, we have:
f = μ s N =μ sm g
Therefore the net torque in the clockwise direction is:
τ net = f R − F b = μ sm g − F b. (2)
(1 mark)
(c) From equation (1) the linear acceleration is given by:
Fnet ( F^ f )
a m m
(½ mark) From equation (2) the angular acceleration is given by:
2 2
f R F b^ f^ R^ F b I m R m R
(1 mark)
For the contact point with the floor the linear and the angular acceleration are related by a = α R , (½ mark)
and we have
F f^2 (^ f^ R^ F b )
m m R
s 2
F R f R f R F b F R F b f R f R f R F R b R F m g R b
where we have set f = μ sm g for the yo-yo on the verge of slipping.
(1 mark)
Solution (a) Evaporative cooling – heat is transferred from the hot air to the water, which evaporates, removing latent heat of vaporisation (1 mark). In a high-humidity climate, evaporation is less efficient because the air is saturated with water vapour, so an evaporative cooling system would not work as well as in a dry climate (½ mark).
(b) 1 2 3 tr (^) 2 av 2 K = m v = k T is the average (translational) kinetic energy of the gas molecules, so
v av ∼ T (1 mark).
Therefore the ratio of speeds is
K
K
= , or 12% faster (½ mark).
(c) Adiabatic expansion (½ mark) – the gas is released so quickly that there is no heat transfer with the surroundings, so the gas cools to below the CO 2 freezing temperature (½ mark). (note that at a pressure of 1 atm, CO 2 sublimates)
(d) No it does not violate the 2 nd^ law because this is an irreversible process (½ mark) and hence, total entropy must increase. In consuming the raw materials needed for growth, this process has resulted in less energy being available in the universe to convert to mechanical work, so the entropy of the universe has increased more than the entropy of the plant has decreased (½ mark).
REG_Q05=ADV_Q
Semester 1, Y
Question 5
The diagram above shows a wire with its left end attached to an oscillating pin that is fixed to a table. The wire extends to the right over a frictionless pulley and is attached to a hanging weight of mass 5.00 kg. The distance between the pin and the pulley is 1.00 m and the linear density of
the wire is 0.100 kg.m−^1. The wire is made to vibrate when the pin is connected to a frequency
generator causing it to oscillate up and down. The oscillating pin generates a sinusoidal travelling transverse wave in the wire with a frequency of 40.0 Hz and amplitude of 0.020 m.
(a) Calculate the speed of the wave, neglecting the mass of the wire.
(b) Write down an equation that describes the vertical displacement of the wave y as a
function of horizontal distance x from the pin and time t. Assume a travelling wave with no reflections. Insert numerical values for any constants in the equation.
(c) What is the lowest frequency of the vibrating pin that will generate a standing wave in the wire? (5 marks)
ADV_Q
Semester 1, Y
Question 6
(a) Explain what is meant by the term resonance and give an example of resonance in a real mechanical system.
(b) What are two physical properties of your chosen system that determine its resonant frequency?
(c) What energy source drives the resonance in your chosen system?
(d) What physical process prevents the amplitude from becoming even larger than it is?
(5 marks)
Solution (a) Resonance is a phenomenon that refers to an oscillation that has a large amplitude because it is being driven at its natural frequency. Some examples are given below. Note that resonance can also occur with random excitation (white noise), such as wind on bridge, finger on wine glass, blowing over a bottle. That definition is ok, but the example should match the explanation for full marks.
In other words, they should lose a mark for saying resonance occurs when an object is driven at its natural frequency and then gives an example that is randomly excited (wind on bridge, finger on wine glass, blowing over a bottle). (2 marks)
(b) The two factors will generally relate to (i) the physical size of the system and (ii) its mass/density (see below). (2 marks)
(c) The energy source will be some external driving agent (see below). Examples, with (answer to b)(and answer to c): (½ mark)
(d) Damping limits the amplitude of the resonance. (½ mark)
Examples
Example (a)
Physical Properties (b)
Energy Source (c)
Damping (d) Child pushed on swing
- Length of swing
- Strength of gravity not mass of child
Person pushing Friction (at pivot)
Bridge oscillating at large amplitude in wind or people walking
- Thickness/width of span
- material
Wind or people Internal friction, heating of material
Wine glass excited by loud speaker
- thickness of glass
- diameter of glass
loudspeaker Internal friction, heating of material
Solution
(a) Momentum is conserved so p (^) f = pi. (1 mark)
Initially, nothing is moving and so pi = 0.
Finally,
10
10
p (^) f M V m v m V v M
(1 mark)
(b)
Δ J = Δ p = ∫ F dt. (1 mark)
All blocks have the same final velocity with respect to the rocket. The Δ p of each
block is independent of the details of the force and the time, so long as the integral is the same.
If ejected all together: Δ p 10 (^) = − 10 m v. (1 mark)
If ejected one at a time: The momentum of each block before ejection is + m vr where vr is the velocity of the
rocket. The momentum after ejection is: m v ( (^) r − v ). Hence the change in momentum
of each block is: Δ p (^) i = m v ( (^) r − v ) − m vr = − m v which is independent of vr. So:
Δ p tot = ∑Δ p i = − 10 m v
Which is the same as Δ p 10. (1 mark)
(c) Ejection of the first block
1
1
i f
i f
p and p M m v m p p v v M m
(1 mark)
(d) Ejection of the second block Momentum is still conserved so
2 1
2 1
2 1
(rocket) (1st block) (2nd block)
M m v m v m v v
M m v m v v m v v v M m
(1 mark)
Ejection of the third block Momentum is still conserved so
2 1 2
2 1 2
2 1 2
(rocket) (1st block) (2nd block) (3rd block)
M m v m v m v v m v v
M m v m v v v m v v v v M m
(1 mark) (e)
v 10 =
m M
(^10 v^ −^ v 1 −^ v 2 −^ v 3 −^ v 4 −^ v 5 −^ v 6 −^ v 7 −^ v 8 −^ v 9 )
Even though the impulse is the same for each ejection of a block some is used to change the momentum of the remaining fuel when the blocks are ejected individually. When the ten blocks are ejected all at once, all of the momentum is transferred to the rocket. (2 marks)
Solution (a) The initial kinetic energy is:
2 2
2 2 2
2
total translation rotation
B
B B
B
K K K
m v I
v m v m R R
m v
where we have used vB = R ω for rolling without slipping.
(2 marks)
(b) At the top of the hill, the kinetic energy has been reduced by an amount Δ U (^) g = m g h. (1 mark) Its speed vT at the top of the hill is then determined by:
(^7 2 ) 10^ T^ 10 B
m v = m v − m g h
Hence
2 2 10 T B 7 v = v − g h
which evaluates to
2 2
1
15.26 15.3 m.s.
T
T
v
v −
The speed at the top of the hill is 15.3 m/s. (2 marks)
(c) The rotational kinetic energy at the top of the hill is 2 2 2
2
T rotation
T
v K I mR R
m v
= ω = ⎛⎜^ ⎞⎛⎟⎜^ ⎞⎟ ⎜⎛^ ⎞⎟
and this stays constant while the ball is in free-fall. (1 mark)
(d) All of the work done by gravity as the ball falls back to the ground results in an increase in its translational kinetic energy.
2 2
2 2 2 2
1
28.0 m.s.
f T
f T B
f
m v m v m g h
v v g h v g h
v −
The speed just before it hits the ground is 28.0 m/s. (2 marks)
(e) The ball has not gained energy; it has the same energy as at the beginning. The rotational kinetic energy is the same as when it was at the top of the hill, which is less than the rotational kinetic energy at the start. Given that the total kinetic energy is the same as at the start, it has more translational kinetic energy than at the start, and therefore a greater speed. (2 marks)
