Download Mid Term Test 2012 Answers File

Newton to Einstein
Mid-Term Test. 9am., Thursday 15th Nov. 2012
Duration: 50 minutes.
Use g = 10 m s-2 for numerical calculations.
You may use the following expressions for the dot and cross products:
u.v  u x v x  u y v y  u z v z

u  v  u yvz  u z v y , u z vx  u xvz , u xv y  u y vx

Section A: Answer ALL Questions
A1.
Given the vectors u = (1, 0, 0) and v = (1, 1, 0), find
a) u + v
(1+1, 1+0, 0+0) = (2, 1, 0)
b) u.v
11 + 01 + 00 = 1
c) u × v
(00 – 01, 01 – 10, 11 – 01) = (0, 0, 1)
[3]
A2.
State Newton’s Second Law of Motion, and explain what it says about
momentum. F = ma (OR F = dp/dt) – it defines momentum as that quantity
whose rate of change is force.
[3]
A3.
A ramp is at 30º to the horizontal, and a mass m sits on it without sliding.
a) Give the force exerted by the ramp on the mass, and resolve it into vertical
and horizontal components. F = mg upwards, Fvert = mg, Fhoriz = 0
b) Resolve the force exerted by the ramp on the mass into components
parallel and perpendicular to the ramp. Fpar = mg sin 30° = ½mg. Fperp =
mg cos 30° = Sqrt[3]/2 mg.
c) What is the minimum coefficient of friction? min = Fpar / Fperp = 1/Sqrt[3]
= 0.58.
[6]
A4.
A steel ball weighs 2kg.
a) The ball is lifted through 20 metres vertically. Calculate its gravitational
potential energy.
mgh = 21020 = 400J
b) The ball is then dropped. Calculate its kinetic energy and velocity when it
has fallen 20 metres. mgh = ½mv2 so v2 = 2gh = 400; v = 20ms–1 [3]
A5.
A dumbbell consists of two 1kg point masses, connected by a light rod one
metre long. Calculate its moment of inertia about
a) an axis perpendicular to the rod and passing through the centre, I =
Sum[mr2] = 1½2 + 1½2 = ½ kgm2
b) an axis perpendicular to the rod and passing through one of the masses.
I = Sum[mr2] = 102 + 112 = 1 kgm2
[4]
A6.
State Kepler’s Third Law. Derive it for a planet in a circular orbit in an
inverse-square law gravitational field. T2 ∂ r3. Derive by equating centripetal
and gravitational forces:
mr2 = mr(2/T) 2 = GMm/r2, so r/T2 ∂ 1/r2, so T2 ∂ r3.
[3]
A7.
A comet is travelling at 12 km s-1 as it crosses Jupiter’s orbit on its way to the
Sun. What is its speed later in its orbit when it crosses Jupiter’s orbit again?
Explain your reasoning.
[3]
12 kms–1, because KE + PE = const; PE is the same both times, so KE and
hence speed must be the same.
Section B. Answer ONE Question
B1.
a) An empty cylindrical can (with no ends) weighing 100g rolls along the
floor at 4 ms-1. Calculate its total kinetic energy (translational plus
rotational).
[5]
Translational KE = ½mv2. Rotational KE = ½I2 = ½ (mr2) (v/r)2 = ½mv2.
So total KE = mv2 = 0.1  16 = 1.6J.
b) A pendulum consists of a 10 kg point bob hanging on a light string of
length 10m. It swings through an angle 2, i.e.  each side of vertical.
(i) Find the tension T in the string at the end-points of the motion, i.e.
when the pendulum is at the angle .
T
mg
The diagram shows the forces acting on the bob when it is stationary at the
angle . The acceleration and hence the resultant force has to be at right
angles to T. Using vectors, we thus have (T + mg).T = 0 or T2 + mgTsin(–)
= 0, i.e. T = mgsin.
Give the value of the tension in the string at the end-points if the
angle  is 90.
T = mgsin0 = 0.
(ii)
(iii)
Give the value of the acceleration of the bob at the end-points if the
angle  is 90.
T = 0, so F = mg, so acceleration is 10ms–2 vertically downwards.
[6]
c) The pendulum of part (b) is swinging with the angle  = 90. By
considering potential and kinetic energy, find
(i) the velocity of the bob at the lowest point of the motion
PE at  = 90° is mgh = 1000J. PE =0 at the bottom, so KE = ½mv2 = 1000J, so
v = Sqrt[2gh] = Sqrt[200] = 14.14ms–1.
(ii)
the tension in the string at this point.
T = mg + ma = mg + mv2/r = 100 + 200 = 300N.
[6]
d) Given that G = 6.67  10-11 N m2 kg-2, the mass of the Earth M = 5.97 ×
1024 kg and the radius of the earth is R = 6380 km,
(i) Find the escape velocity for a rocket at 200km altitude.
PE per unit mass is –GM/r, and PE + KE = 0 for escape velocity. So v = Sqrt[2
GM/r] = 11.2 kms–1.
(ii)
Find the orbital speed of a satellite in a circular orbit at 200 km
altitude.
So total KE = mv2 = 0.1  16 = 1.6J.
B2.
[7]
a) State Kepler’s Second Law and explain how it relates to angular
momentum. A planet sweeps out equal areas in equal times. The area is
½v.r per unit time. This is proportional to the angular momentum mv.r, so
the constancy of the area is equivalent to the angular momentum being
conserved.
[5]
b) A flywheel with a moment of inertia I = 5 kg m2 is mounted on a light
axle. It is initially stationary. A torque of 100 N m is applied to the axle
for 100 s.
(i) Calculate the final angular velocity of the flywheel.
v = u + at = u + F/m t, so f = I + t = 0 + /I t = 2000 rads–1.
(ii) Calculate the final angular momentum of the flywheel.
L = If = 5  2000 = 10000 kgm2s–1.
(iii) Calculate the final rotational energy of the flywheel.
E = ½I2 = ½  5  4  106 = 107J.
(iv) If a torque of 10 N m is now applied about an axis perpendicular to
the axle of the flywheel, what will the motion be? Explain your
answer.
[9]
This is a gyroscope, so it will precess around the third axis. The rate of
addition of angular momentum around this axis is the torque, so L moves to L
+  dt in time dt, i.e. with precessional angular velocity /L = 10/1000 = 10–3
rads–1.
c) The potential energy of a body varies with position as U  2x 2 .
(i) Give an expression for the force F on the body and sketch F(x).
F = –dU/dx = –4x.
F
1
x
–4
(ii)
At what position will the body be in stable equilibrium? Explain
[5]
your answer.
At x = 0, since here the force is zero and U is minimum. .
F
d) Given two vectors u and v, with magnitudes u = 1 and v = 2, and given
that the magnitude of the cross product u  v = 1, find
(i) The angle between u and v.
u  v = uv sin , so 1 = 2 sin , so sin  = ½ and  = 30°
(ii)
F
The value of the dot product u.v.
[6]
u  v = uv cos  = 2  cos30° = Sqrt[3].
End of paper
Prof. D.J. Dunstan