Ch
10.1 #1
During a certain period
of time, the angular position of a swinging door is described by θ = 2t^{2}
+ 10t + 5, where θ is in radians and t is in seconds. Determine the angular position, angular
speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t =
3.00 seconds.
q = 5 + 10t +
2t^{2} 
w = dq/dt w = 10 + 4t 
a = dw/dt a = 4 
(a) t = 0 

q = 5 rad 
w = 10 rad/s 
a = 4 rad/s^{2} 
(b) t = 3 

q = 53 rad 
w = 22 rad/s 
a = 4 rad/s^{2} 
Ch
10.2 #2
A dentists drill
starts from rest. After 3.2 s of
constant angular acceleration, it turns as a rate of 2.51 x 10^{4}
rpm. (a) Find the drill’s angular
acceleration. (b)
Determine the angle (in radians) through which the drill rotates during this
period. a = Dv/Dt a = Dw/Dt
(a) a = 2.51 x 10^{4} rpm (2prad/rev) (1min/60s)/ 3.2 sec a = 821 rad
/ s^{2} 
(b) q = ½ a t^{2} q = 4200 radians 
Ch
10.2 #8
A rotating wheel requires
3.00 s to rotate through 37.0 revolutions.
Its angular speed at the end of the 3.00 s interval is 98.0 rad/s. What is the
constant angular acceleration of the wheel?
a = (ω_{f} – ω_{0})
/ t ω_{0} = ω_{f}
 a
t 
OR 
ω_{ave}
= Δq / Δ t ½ (ω_{f} + ω_{0}) = (q_{f}  q_{0}) / t ½ (98 + ω_{0})
= 37(2 π) / 3 ω_{0}
= 57 rad 
q  q_{0} = ½ at^{2}
+ (ω_{f}  a t)t q = ω_{f}t
 ½ at^{2} 37(2π) = 98(3) – ½ a3^{2} a = 13.7 rad/s^{2} 
a = (ω_{f} – ω_{0}) / t a = (98 – 57) / 3 a = 13.7 rad/s^{2} 
Ch
10.3 #17
A disk 8 cm in
radius rotates at a constant rate of 1200 rpm about its central axis. Determine (a) its angular speed, (b) the
tangential speed at a point 3 cm from its center, (c) the radial acceleration
of a point on the rim, and (d) the total distance a point on the rim moves in 2
sec.
(a) w = 1200 rpm
(2prad/rev) (1min/60s) w = 126 rad / sec 
(b) v = w r v =
126rad/sec (0.03m) v = 3.77
m/s 
(c) a = v^{2} / r = w^{2}r^{2} / r a = w^{2}r a = 1260 m/s^{2} 
Ch
10.4 #21
The 4 particles
are connected by rigid rods of negligible mass. The origin is at the center. If the system rotates in the xy plane
about the z axis at 6 rad/sec, calc (a) moment of
inertia of the system about the zaxis (b) the
rotational kinetic energy of the system 


r_{1} =
r_{2} = r_{3} = r_{4} r = ((6/2)^{2}
+ (4/2)^{2})^{½} r = Ö13 meters (a) I = S_{j}m_{j}r_{j}^{2} I = (3 + 2 + 2 + 4) 13 I = 143 kg m^{2} 
(b) K_{R} =
½ Iw^{2} K_{R} =
½ 143 (6 rad/s)^{2} K_{R}
= 2.57 x 10^{3} J 

Ch
10.5 #23
Three identical
this rods, each of length L and mass m are welded perpendicular to one
another. The assembly is rotated about
an axis that passes through the end of one rod and is parallel to
another. Determine the moment of
inertia of this structure. Ans: 11 mL^{2} / 12 Hint: We
need to add the Inertia for all three rods I = I_{x}
+ I_{y} + I_{z} 

For I_{y} all of the mass is located
at L/2 from the pivot point I_{y} = m(L/2)^{2} I_{y} = 1/4 mL^{2} For I_{z} the rod on the zaxis
is being pivoted about the end (formula from chart) I_{z} = 1/3 mL^{2} For I_{x} the mass varies from L/2 to L/√2 The Right Half (also can be done all at once) I_{x}
= r^{2}dm … l = m/L and l = dm/dx I_{x}
= r^{2} l dx where
r^{2} = x^{2} + (L/2)^{2} And x varies from 0 to L/2 Work to the
right à I = I_{x}
+ I_{y}
+ I_{z} I = mL^{2}/3 +
mL^{2}/4 + mL^{2}/3 I = 11mL^{2}/12 
Right Half of x I_{x} =
r^{2} l dx where r^{2}
= x^{2} + (L/2)^{2} I_{x} =l (x^{2} + (L/2)^{2}) dx I_{x} = l x^{2}dx + l (L/2)^{2}
dx (x à 0 to L/2) I_{x} = l/3
x^{3} + l/4 L^{2} x (x à 0 to L/2) I_{x} = l/3 (L^{3}/8) + l/4 L^{2} (L/2) I_{x} =
(m/L)(L^{3}/24) + (m/L) (L^{3}/8) I_{x} =
mL^{2}/24 + mL^{2}/8 I_{x}
= mL^{2}/6 (For right
half) I_{x}
= mL^{2}/3 for all of the rod on the xaxis 
Ch
10.6 #31 Find the net
torque on the wheel about the axle through O if a = 10 cm and b = 25 cm. 

S t = t_{ccw} 
t _{cw} S t = 0.1m*12N – 0.25m*(9N + 10N) S t = 3.55 Nm 
Ch
10.7 #35
A model airplane with
mass 0.75 kg is tethered by a wire so that it flies in a circle 30 m in
radius. The airplane engine provides a
net thrust of 0.8 N perpendicular to the tethering
wire. (a) Find the torque the net thrust produces about the center of the
circle. (b) Find the angular acceleration of the airplane when it is in level
flight. (c) Find the linear acceleration of the airplane tangent to its flight
path.
(a) t =
r x F t = 30 m * 0.8 N t = 24 Nm 
(b) t =
I a 24 = (0.75 * 30^{2}) a a = 0.0356 rad / s^{2}

(c) a_{T} = a * r a_{T} = 0.0356 * 30 a_{T} = 1.07 m/s^{2} 
Ch
10.8 #47 This
problem describes an experimental method for determining the moment of
inertia of an irregular shaped object such as the payload for a
satellite. A counter weight of mass m, is suspended by a cord wound around a spool of radius
r, forming part of a turntable supporting the object. The turntable can rotate w/o friction. 

When the
counterweight is released from rest it descends through a distance h,
acquiring a speed v. Show that the
moment of inertia I of the rotating apparatus is mr^{2}
(2gh/v^{2} – 1). 

K_{R} + K = U_{grav} ½ Iw^{2} + ½ mv^{2} = mgh ½ Iw^{2} = mgh 
½ mv^{2} I = 2(mgh  ½
mv^{2}) / w^{2} I = (2mgh
 mv^{2}) / (v/r)^{2} I = mr^{2} (2gh  v^{2}) / v^{2} I = mr^{2} (2gh/v^{2}  1) 
Ch
10.9 #52
A bowling ball has
mass M, radius R, and a moment of inertia of 2/5 MR^{2}. If it starts from rest, how much work must be
done on it to set it rolling without slipping at a linear speed, v? Express the work in terms of M and v.
Work
= DK Work
= K + K_{R} Work = ½Mv^{2}
+ ½ Iw^{2} 
Work = ½Mv^{2}
+ ½ 2/5MR^{2} (v/R)^{
2} Work = ½Mv^{2}
+ 2/10 Mv^{2} Work =
7/10 Mv^{2} 