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} 