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UNIVERSITY OF TORONTO
Faculty of Applied Science and Engineering
FINAL EXAMINATION, December 2009
First Year Engineering Science
PHY180H1 F: Classical Mechanics
SOLUTIONS
Useful constants and formulae:
1. Gravitational potential energy of a system of two particles is UG= -Gm1m2/ R
2. a) Moment of inertia of a uniform solid disc of mass m and radius R about the
axis through its center of mass perpendicular to the two plane surfaces of the
disc is I = ½ mR2 ;
b) Moment of inertia of a very thin solid rod of mass m and length L about its axis
of symmetry through the center of mass is I = mL2/12.
Problem 1.
(4 marks)
Give a definition for the center of mass of a system of particles. Make a sketch to show the quantities you
are using in the definition.
Solution
The center of mass of a system of N particles is a point in space with the radius vector (=position vector)
r, given by:
m1
y
r1
N
CM
m3
r
m rv
rv =
∑
i =1
N
i i
∑m
i =1
m2
r3
r2
i
0
x
where mi is the mass of a particle number i; and ri is the radius vector (= position vector) of the particle
number i .
For correct introductory wording 1 mark;
for correct formula 1 mark;
for correct diagram 1.5 marks (a diagram may differ from the shown above);
for correct and complete list of all variables in the formula and diagram 0.5 marks.
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Problem 2. (5 marks)
Derive a formula for the escape speed of an object launched from the planet with radius R and mass M.
Explain your steps.
Solution
•
•
Escape speed is the minimum value of the initial speed of an object, launched from a planet,
needed to allow the object to move infinitely far away from the planet. (1 mark)
The total mechanical energy E of an object of mass m, launched from the surface of a planet of
mass M with initial speed vi is:
E=
•
•
•
•
mvi2
mM
−G
2
R
(1 mark)
This energy is conserved in an ideal case while the object is moving away from the planet; as the
potential energy, which is negative, is increasing and approaching 0 at infinity, the kinetic energy
is decreasing.
(1 mark)
To escape means to appear at a distance rf → ∞ , where the potential energy is zero. (0.5 mark)
Minimum initial speed corresponds to minimum final speed, which is vf = 0. (0.5 mark)
Two above statements give:
E=
mv 2f
2
−G
Possible and also correct answer is vi =
mM
= 0 → vi =
rf
2GM
R
(1 mark)
2 g R R , where gR is acceleration due to gravity on the
surface of the planet.
Problem 3. (6 marks)
Two identical particles are moving in one direction with different speeds v1 > v2 which permits them to
undergo a glancing collision. Find the angle between the velocity vectors of two particles after the
collision if the particles are moving in free space.
Solution
In this problem, the glancing collision takes place because the initial velocity vectors are parallel, but not
aligned. The problem statement does not specify whether the collision is elastic or inelastic, therefore all
cases must be studied.
(a) Perfectly elastic collision. The solution is based on a correct and complete vector diagram in two
dimensions. [Linear momentum in the diagram may be replaced by velocity because the particles have
identical masses.] The direction of initial velocities can be chosen as x-direction. The simplest solution is
obtained in the frame of reference associated with the particle with the initial speed v2.
In this frame of reference, the vector diagram is
y
as shown below:
vi = (v1 – v2) î (1 mark)
v2f
The law of conservation of energy for perfectly
α
elastic collision
vi
2
2
2
mv1 f mv 2 f
0
mvi
m
x
=
+
→ After canceling ,
2
2
2
Pythagorean Theorem gives α = 90o
2
(2 marks) for the correct diagram and (1 mark)
for the result
v1f
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(b) Perfectly inelastic collision. the vector diagram is not required, as well as the change of the frame of
reference. Two particles stick together. The law of conservation of linear momentum claims conservation
of the direction of motion → α = 0o. (1 mark)
(c) Inelastic collision. This case is an intermediate relative to the previous two ones; therefore, the
angle is 0o < α < 90o. (1 mark)
Problem 4. (7 marks)
A projectile of mass m moves to
the right with a speed vi (a). The
projectile strikes and sticks to the
end of a stationary rod of mass M
and length d, pivoted about a
frictionless horizontal axle through
its center (b).
(1) Find the angular speed ω of
the system right after the collision.
(3 marks)
(2) Determine the fractional loss in
mechanical energy due to the
collision. (4 marks)
Solution
(a) Angular momentum is conserved. We choose calculation of the angular momentum and the
moment of inertia about the axis through 0 and perpendicular to the plane of the diagram
(0.5 marks)
Lip = L f = I r + I p ω
(0.5 marks)
(
)
where Lip is the initial angular momentum of a projectile;
Lf is the angular momentum of a rod and a particle stuck to the rod;
Ir is the moment of inertia of the rod about the pivot;
Ip is the moment of inertia of a point mass (a projectile) about the axis through the pivot;
ω is the angular speed of the rod with the projectile just after they stuck together.
2
mv i d ⎛ 1
⎛d⎞ ⎞
2
= ⎜⎜ M d + m ⎜ ⎟ ⎟⎟ ω
2
⎝2⎠ ⎠
⎝ 12
ω=
(b)
6mv i
Md + 3md
The original energy is
(1 mark)
(1 mark)
1 2
mv . (1 mark)
2 i
The final energy is:
3m 2 v i2 d
1 2 1⎛ 1
md 2 ⎞ 36m 2 v i2
2
Iω = ⎜ Md +
=
⎟
2
2 ⎝ 12
4 ⎠ ( Md + 3md )2 2 ( Md + 3md )
(1 mark)
4
The absolute loss of energy is
3m 2 v i2 d
mM v i2 d
1
2
mv i −
=
2
2 ( M d + 3md ) 2 ( M d + 3md )
(1 mark)
and the fractional loss of energy is
mMv i2 d 2
M
=
2
M + 3m
2 ( Md + 3md ) mv i
(1 mark)
L
Problem 5. (8 marks)
Four identical bricks of length L each are placed on
a solid base and shifted with respect to each other
by a different fraction of L, as shown.
Find the maximum value x of the shift of the upper
brick relative to the brick under it which still permits
the system to be in equilibrium. Explain your steps.
x
L/4
L/6
Base
L/8
Solution
Calculations of torques will be performed about the
axis that coincides with the top right edge of the
base. Solutions may be different.
0
x
Possible solution 1: set x-axis as shown. The axis of rotation penetrates the plane of the diagram at
point 0. (1 mark)
Torque for each brick is calculated as a product of the force of gravity and the arm of action of the force.
For the lowest brick (#1), the torque is τ1 = mgL ( 1/2 - 1/8) = 3/8 mgL (counterclockwise rotation);
for the brick #2, the torque is τ2 = mgL [ 1/2 - (1/8 + 1/6)] = 5/24 mgL (counterclockwise rotation);
for the brick #3, the torque is τ3 = mgL [ 1/2 - (1/8 + 1/6 + 1/4)] = - 1/24 mgL (clockwise rotation);
for the brick #4, the torque is τ4 = mgL [ 1/2 - (1/8 + 1/6 + 1/4 + x)] = (- 1/24 – x/L ) mgL (clockwise
rotation).
(1 mark x 4 = 4 marks)
Condition of equilibrium: τ1 + τ2 + τ3 + τ4 = 0
(1 mark)
The common factor mgL can be canceled, and the equation for equilibrium can be written in terms of
fractions of L.
3 5
1
1 x
+
−
−
− = 0 (1 mark)
8 24 24 24 L
x = ½ L (1 mark)
Possible solution 2: The center of mass of the object that consists of four bricks can be to the left or
over the edge of the base. The x-axis is chosen in the same way as for the previous solution. (1 mark).
All other calculations are same except the dimension of the equation of equilibrium: it is written for
position of the center of mass, and not for the torque. The marking scheme is same as for the solution #1.
The diagram is not required if the choice of axes and of the origin is clearly explained in words.
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Problem 6. (10 marks)
A uniform solid disc of mass M and radius R is rolling down an incline making angle θ with the horizontal.
What is the minimum coefficient of friction required to maintain pure rolling motion for the disc?
(8 marks)
What is the direction of the force of friction: up or down the incline? (1 mark)
Is friction static or kinetic?
(1 mark)
Solution
The diagram shows the force of friction f (static 1 mark and upward 1 mark), the normal force n exerted
on the sphere by the ramp and the force of gravity Mg.
n
For the correct and complete diagram 2 marks without
correct direction of f, which is already considered.
The coefficient of static friction μ is given by
μ=
M
R
f
Mg cos θ
(1)
(1 mark)
f
Mg
fR = I 0α
(2)
(1 mark)
θ
α is the angular acceleration of the disc; I0 is the moment of
inertia of the disc about the axis through its center of mass.
MgR sin θ = Iα
(3)
(1 mark)
I is the moment of inertia of the disc about the instantaneous axis through the point of contact of the disc
and the surface of the incline.
I0 =
1
MR 2
2
I = I 0 + MR 2 =
(4)
3
MR 2
2
(is given among the formulae provided)
(5)
(1 mark)
Combination of (1), (2), (3), (4) and (5) gives for μ:
μ min =
I 0 tan θ
tan θ
=
2
3
I 0 + MR
(1 mark)
If μ was less than obtained minimum value, the motion of the disc would be a combination of sliding down
the ramp and rolling. (1 mark).