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Text Book Problem 11.7 (with permission)
In Fig. 11.31, a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls
without slipping a distance L  6.0 m down a roof that is inclined at the angle   30 .
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roof's edge is at height H  5.0 m . How far horizontally from the roof's edge
does the cylinder hit the level ground?
Figure 11.31 Problem 11.7.
Solution
(a) We find its angular speed as it leaves the roof using conservation of energy. Its initial
kinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where
h  6.0sin 30  3.0 m (we are using the edge of the roof as our reference level for
computing U). Its final kinetic energy (as it leaves the roof) is the sum of rotational and
translational kinetic energies.
K f  21 Mv 2  21 I 2 .
Here we use v to denote the speed of its center of mass and  is its angular speed — at
the moment it leaves the roof. Since (up to that moment) the ball rolls without sliding we
can set v = R = v where R = 0.10 m. Using I  21 MR 2 (from tables), conservation of
energy leads to
1
1
1
1
3
Mgh  Mv 2  I  2  MR 2 2  MR 2 2  MR 2 2 .
2
2
2
4
4
The mass M cancels from the equation, and we obtain

c
hb
g
1 4
1
4
gh 
9.8 m s2 3.0 m  63 rad s .
R 3
010
. m 3
(b) Now this becomes a projectile motion of the type examined in Chapter 4. We put the
origin at the position of the center of mass when the ball leaves the track (the “initial”
position for this part of the problem) and take +x leftward and +y downward. The result
of part (a) implies v0 = R = 6.3 m/s, and we see from the figure that (with these positive
direction choices) its components are
v0 x  v0 cos 30  5.4 m s
v0 y  v0 sin 30  3.1 m s.
The projectile motion equations become
x  v0 x t and y  v0 y t 
1 2
gt .
2
We first find the time when y = H = 5.0 m from the second equation (using the quadratic
formula, choosing the positive root):
t
v0 y  v02y  2 gH
g
 0.74s.
b
gb
g
Then we substitute this into the x equation and obtain x  5.4 m s 0.74 s  4.0 m.
Text Book Problem 11.17 (with permission)
A yo-yo has a rotational inertia of 950 g  cm 2 and a mass of 120 g. Its axle radius is
3.2 mm, and its string is 120 cm long. The yo-yo rolls from rest down to the end of the
string.
(a) What is the magnitude of its linear acceleration?
(b) How long does it take to reach the end of the string? As it reaches the end of the
string, what are its
(c) linear speed,
(d) translational kinetic energy,
(e) rotational kinetic energy, and
(f) angular speed?
Solution:
(a) The expression for acceleration is found using either conservation of energy, or
Newton’s second law.
Energy method:
The initial kinetic and potential energy are zero: K i  0 , U i  0 , where we take the initial
position as our gravitational potential energy reference. When it falls it will convert
potential energy into kinetic energy of translation and rotation. Hence at some distance y
below the start point:
U f   Mgy (t has less potential energy now) and
K f  12 Mv 2  12 I COM  2
By conservation of energy,  K f  U f    K i  U i  and so
1
2
Mv 2  12 I COM  2  Mgy . Then
taking   v / R0 we get
2
1
2
Mv 2  12 I COM
v
 Mgy
Ro2
2Mgy
2 gy

. To get acceleration, we
2
M  I com / Ro 1  I com MRo2
differentiate both sides, and find:
dv
2g
dy
2v

and so:
2
dt 1  I com MRo dt
g
acom  
1  I com MR02
solving for v 2 gives: v 2 
where the positive direction is upward. We use I com  950 g  cm2 , M =120g, R0 = 0.320
cm, and g = 980 cm/s2 and obtain
| acom | 
980 cm/s 2
1   950 g  cm
2
 120 g  0.32 cm 
2
 12.5 cm/s 2  13 cm/s 2 .
(b) Taking the coordinate origin at the initial position leads to ycom  21 acomt 2 . Thus, we
set ycom = – 120 cm, and find
t
2  120cm 
2 ycom

 4.38 s  4.4 s.
acom
12.5 cm s 2
(c) As it reaches the end of the string, its center of mass velocity is given by the equations
above, or:
vcom  acom t   12.5 cm s 2   4.38s   54.8 cm s ,
so its linear speed then is approximately | vcom |  55 cm/s.
(d) The translational kinetic energy is
K trans 
1 2
1
2
mvcom   0.120 kg  0.548 m s   1.8 10 2 J .
2
2
(e) The angular velocity is given by  = – vcom/R0 and the rotational kinetic energy is
2
K rot
2
v  1
 0.548 m s 
1
1
 I com 2  I com  com   (9.50 10 5 kg  m 2 ) 
  1.4 J .
3
2
2
2
3.2
10
m
R



 0 
(f) The angular speed is

vcom
0.548 m/s

 1.7 102 rad/s  27 rev s .
3
3.2 10 m
R0
Note: As the yo-yo rolls down, its gravitational potential energy gets converted into both
translational kinetic energy as well as rotational kinetic energy of the wheel. To show that
the total energy remains conserved, we note that the initial energy is
U i  Mgyi  (0.120 kg)(9.80 m/s 2 )(1.20 m)  1.411 J
which is equal to the sum of K trans (= 0.018 J) and K rot (= 1.393 J).
Tutorial set week 12 2011
M80. A uniform bar 7.5 m long is held against
a rough wall by a light rope as shown in the
diagram. The pulley may be assumed to be
frictionless,
W = 140 N and the weight of the bar is w = 40
N. Find:
(a) the tension in the rope attached to the
bar (T=109 N)
(b) the minimum coefficient of friction between the bar and the wall if the bar does not
slip. (µ = 0.112)
Solution.
The bar is in static equilibrium, so the sum of forces and sum of torques must be zero. The Free
body diagram is given below.
We take the sum of torques about the left end and take positive torque clockwise. Then.
X
τ = 0 = 3.75mg + 4.5M g − 3T sin 70 − 7.5T sin 35
3.75(40) + 4.5(140) − 2.82T − 4.3T = 0
7.12T = 780
T = 109.6
From the vertical forces we get:
X
Fy = 0 = f + T sin 70 + T sin 35 − mg − M g
1
We also know that the frictional force f = µs N
0 = µs N + (109.6) sin 70 + (109.6) sin 35 − 40 − 140
µs N = 14.2
From the horizontal forces we get:
X
Fx = 0 = N − T cos 70 − T cos 35
N = (109.6) cos 70 + (109.6) cos 35 = 127.2
Hence the coefficient of friction is: µs = 14.2/127.2 = 0.112
M82. The metre rule AB has mass 2 kg, rests
against a wall at A, and is supported by cable
BC. A mass M is hung from the rule at B. What
is the largest M for which the rule will not slip at
the wall, if the coefficient of static friction is µs =
0.4? (7.0kg).
Solution.
The ruler is in static equilibrium, so the sum of forces and sum of torques must be zero. The Free
body diagram is given below.
From the horizontal forces we get:
X
Fx = 0 = N − T cos 45
2
N = 0.707T
From the vertical forces we get:
X
Fy = 0 = T sin 45 − f − mg − M g
We also know that the frictional force f = µs N and so
0 = T sin 45 − µs N − 2g − M g
0 = 0.707T − (0.4)(0.707)T − 2g − M g
0.424T = (M + 2)g
We take the sum of torques about the left end and take positive torque clockwise. Then.
X
τ =(0.5)mg sin 65 + (1.0)M g sin 65 − (1.0)T sin 20 = 0
8.882 + 8.882M = 0.342T
T = 26M + 26
This can be used to eliminate T ,
0.424(26M + 26) = (M + 2)g
1.126M + 1.126 = M + 2
M = 6.95kg
M85. A 75.0 kg window cleaner used a 10.0 kg ladder which
was 5.0 m long. He put one end down 2.5 m from a wall
and rested the upper end against a cracked window and
climbed the ladder. He had climbed 3.0 m up along the
ladder when the window broke. Neglecting friction between
the ladder and the window and assuming that the base of
the ladder had not slipped, calculate
(a) the force exerted on the window by the ladder
just before the window broke, (283 N)
(b) the magnitude and direction of the force exerted
on the ladder by the ground just before the window broke.
(880 N at 71◦ elevation)
Solution.
The ladder is in static equilibrium, so the sum of forces and sum of torques must be zero. The Free
body diagram is given below.
3
(a) We take the sum of torques about the bottom end and take positive torque counter-clockwise.
Then,
X
τ = 0 ⇒ 5Nw sin θ − 2.5mg cos θ − 3M g cos θ = 0
We also know the angle of the ladder, as cos θ = 2.5
and so θ = 60◦ . Then we can solve for the force
5
on the window:
4.33Nw − 122.5 − 1102.5 = 0
4.33Nw = 1225
Nw = 283N
(b) From the horizontal forces we get:
X
Fx = 0
⇒
Nw − f = 0
Nw = f
Hence
f = 283N
From the vertical forces we get:
X
Fy = 0
⇒
Nf − mg − M g = 0
Nf = (m + M )g
Nf = (75 + 10)9.8 = 833N
The total force at the floor will then be
q
p
F = f 2 + NF2 = 2832 + 8332 = 880N
and the angle with respect to the horizontal will be given by:
tan φ =
833
= 2.94
283
φ = 71.2◦
4
M87. A crane used on a construction site has a uniform
boom 30 m long which weighs 20 kN. The boom is supported at its base by a horizontal pivot P, and is held at
an angle of 60◦ to the horizontal by a horizontal cable C.
A 104 kg load is held stationary at a height of 3 m by a
second cable D which runs over a pulley of small diameter
at the top of the boom and down to a winch adjacent to P.
Neglecting friction in the pivot at P and the masses of the
cables calculate
(a) the tension in the horizontal cable C, (62.4 kN)
(b) the magnitude and direction of the force exerted
by the pivot P on the boom (111 kN, 203 kN)
Solution.
The beam is in static equilibrium, so the sum of forces and sum of torques must be zero. The Free
body diagram of the beam is given below.
(a) We take the sum of torques about the bottom end and take positive torque clockwise. Then,
X
τ = 0 ⇒ (30)TC sin θ − (15)20000 cos θ − (30)10000g cos θ = 0
where the tension on the cable running back to the pivot does not produce a torque as it is parallel
to the beam. Solving for the tension in the cable gives:
TC = 62, 400N
(b) From the horizontal forces we get:
X
Fx = 0
⇒
TC + TD cos 60 − Nx = 0
But TD = M g as they are connected via the pulley at the top. Hence:
Nx = TC + TD cos 60 = 62, 400 + 10000g cos 60 = 111, 400N
5
From the vertical forces we get:
X
Fy = 0
⇒
Ny − M g − TD sin 60 − mg = 0
Ny = M g + TD sin 60 = M g + M g sin 60 + mg
Ny = 10000g(1 + sin 60) + 20, 000 = 202, 900N
The total force at the pivot will then be:
q
p
FP = Nx2 + Ny2 = 111, 4002 + 202, 9002 = 232kN
and the angle with respect to the horizontal will be given by:
tan φ =
202, 900
= 1.82
111, 400
φ = 61.2◦
6