HW09 - angular momentum conservation and statics: in which we observe the quantity L r p r p sin rp nˆrp to be conserved over the course of a reaction (Ch. 11: P42, P50, P61, P69). Then, in
chapter 12, we find force-sums on complicated structures (Ch. 12: P10, P21, P25, P28, P37, P49). •• chapter 11, problem 42: the indicated Figure
gives the torque that acts on an initially stationary disk that can rotate about its center like a merry-go-round. The scale on the t axis is set by S 4.0N m . What is the angular momentum of of
the disk about the rotation axis at times (a) t = 7.0 s and (b) t = 20 s?
The results may be found by integrating inte grating Eq. 11-29 with respect to time, keeping k eeping in mind that L that Li = 0 and that the integration may be thought of as “adding the areas” under the line-segments line-segments (in the plot of the torque versus time, with “areas” under the time axis con tributing negatively). It negatively). It is helpful to keep in mind, also, that the area 1
of a triangle is 2 (base)(height),
L(7.0 s)
7.0 s
0
dt; L(20s)
dL(t )
L L(t ) dL(t )
dt
20 s
0
dt
dt
dt; L(0s)
0
dt 0
(1.1)
0 Li ;
2 (a) We find that L = 24 k g m / s at t = t = 7.0 s.
2 (b) Similarly, L = 1.5 k g m / s at t = = 20 s.
3 2 2.0 10 kg m about ••• chapter 11, problem 50: The rotor of an electric motor has rotational inertia I m 2.0
its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia I p 12kg m2 about this axis. Calculate the number of revolutions of the rotor required to turn the probe through 30° about its central axis. We use conservation of angular momentum: I m m = I p p. The respective angles m and p by which the motor and probe rotate are therefore related by
z
I m mdt I m m
which gives m
I p p I m
z
c12 kg m
The number of revolutions for the rotor is is then
3
I p pdt I p p
2
g 180000 .
30
2.0 10 kg m
2
5
2
N = (1.8 10 )º/(360º/rev) = 5.0 10 rev.
Chapter 11, problem 61: The uniform rod (length 0.60 m, mass 1.0 kg) in Fig. 11-54 rotates in the plane of the figure about an axis through one end, 2 with a rotational inertia of 0.12 kg m . As the rod swings through its lowest position, it collides with a 0.20 kg putty wad that sticks to the end of the rod. If the rod’s angular speed just before collision is 2.4 rad/s, what is the angular speed of the rod – putty system immediately after collision?
Angular momentum is conserved in this process: Li L f I ii I f f . Solving for the desired f , and noting that I i
I rod and I f I rod I putty , f
Ii I f
i
I rod i I rod I putty
I i I mL
2
(0.12kg m2 )(2.4 rad ) s 0.12kg m (0.20kg )(0.60m ) 2
2
1.5
rad s
;
(1.2)
The in the final answer (1.2) indicates that it really isn’t too much trouble to define “ clockwise” as + or – (conventionally, it is negative). Also, notice we did not need the information about the uniform rod’s 1.0 kg mass (only its length). Chapter 11, problem 69: A certain gyroscope consists of a uniform disk with a 50 cm radius mounted at the center of an axle that is 11 cm long and of negligible mass. The axle is horizontal and supported at one end. If the disk is spinning around the axle at 1000 rev/min, what is the precession rate? 2
The precession rate can be obtained by using Eq. 11-46 with r = (11/2) cm = 0.055 m. Noting that I disk = MR /2 and its angular speed is 1000 rev/min
2 (1000) 60
rad/s 1.0 102 rad/s,
we have
Mgr 2
( MR / 2)
2 gr 2
R
2
2(9.8 m/s )(0.055 m) (0.50 m) (1.0 10 rad/s)
chapter 12, problem 10: The system in Fig. 12-26 is in equilibrium, with the string in the center exactly horizontal. Block A weighs 40 N, block B weighs 50 N, and angle is 35°. Find (a) tension
T 1 , (b) tension T 2 , (c) tension T 3 , and (d) angle .
2
2
0.041 rad/s.
(a) The tension T 1 can be determined by the equation
F
1 y
1 y
solve for T 0 T1, y wA T1 cos wA T 1
w A
1
(b) The tension T 2 can be determined by the equation
F
F 0 at the point where string-1 terminates,
F
2 x
cos
40N cos 35
49N ;
(1.3)
0 at the point where string-2 meets string-1,
solve for T 0 T2,x T1,x T2 cos 0 T1 sin T2 T 1 sin (49N)sin 35 28 N ;
2 x
2
(1.4)
(c) the tension T 3 is obviously the Pythagorean resultant of the forces T 2 and W B , as can obviously be seen by writing the x and y force-balances at the point where string-2 meets string-3,
F F
x 3
y 3
0 T2 T 3, x T3 T3, x 2 T3, y 2 T2 2 WB 2 (28N )2 (50N )2 57N ; 0 W B T 3, y
(1.5)
(d) The angle of string 3 (measured from vertical) is
1
T x 1 28 tan 29. 50 T y
tan
•• chapter 12, problem 21: The system in Fig. 1236 is in equilibrium. A concrete block of mass 225 kg hangs from the end of the uniform strut of mass 45.0 kg. For angles 30.0 and 45.0 , find
(a) the tension T in the cable and the (b) horizontal and (c) vertical components of the force on the strut from the hinge. (a) Two angles for later convenience: (1) the an gle between the cable and the strut, , and (2) the angle between the strut and any vertical force, , – 45º – 30º 15º; 90º – 45º 45º ;
(1.6)
Denoting M = 225 kg and m = 45.0 kg, and as the length of the strut, we compute torques about the hinge. Let “into the page” be negative, and “out of the page” be positive; using the right hand rule. The strut’s CMcoordinate is at / 2 , so the massive strut’s torque is as if the whole mass of the strut is concentrated there, 0
r F sin i
i
solve for T T
i
(T )sin (Mg )sin 2 mg sin T Mg strut CM
Mg sin mg 2 sin sin
M 12 m sin
g sin 6.63 10 N ; 3
(1.7)
(b) Despite the complications (i.e., three torques) of the above -analysis, the only two elicitors of reaction-forces at the hinge are T, Mg, and mg; of these, only T has an x-component (specifically, a negative-x component),
F 0 T F T cos F F = T cos30 = 5.74 10 N ; 3
x
x
x
x
x
(1.8)
(c) As advertised in part-b, the forces -Mg, -mg, and -Ty show up in the y-direction’s force-balance,
F
y
solve for F y
0 Fy Mg mg Ty Fy Mg mg T sin 30 5.96 103 N ;
(1.9)
••chapter 12, problem 25: In Fig. 12-40, what magnitude of (constant) force applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height h = 3.00 cm? The wheel’s radius is r = 6.00 cm, and its mass is m = 0.800 kg.
We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the step corner, shown as the two components f h and f v. If the minimum force is applied the wheel does not accelerate, so both the total force and the total torque acting on it are zero.
We calculate the torque around the step corner. The second diagram indicates that the distance from the line of F to the corner is r – h, where r is the radius of the wheel and h is the height of the step. The distance from the line of mg to the corner is
b g
r r h 2
2
2rh h . Thus, 2
g
F r h mg 2rh h2 0 . The solution for F is
F =
2rh h r h
2
2
mg
2
2
2(6.00 10 m)(3.00 10 m) (3.00 10 m) 2
2
(6.00 10 m) (3.00 10 m)
2 2
(0.800 kg)(9.80 m/s )
13.6 N. Note: The applied force here is about 1.73 times the weight of the wheel. If the height is increased, the force that must be applied also goes up. Next we plot F /mg as a function of the ratio h / r . The required force increases rapidly as h / r 1 .
chapter 12, problem 28: In Fig. 12-43, suppose the length L of the uniform bar is 3.00 m and its weight is 200 N. Also, let the block’s weight W = 300 N and the angle 30.0 . The wire can withstand a maximum tension of 500 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x (pretending the string doesn’t break) , what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
(a) the torque-balance about point-A determines the max distance the block can stand on the beam, 0
i
(W ) sin 90 12 L(Wbar ) sin 90 wire block rod ,CM ri Fi sin i L(Tmax )sin xmax
(1.10) (500 N)sin30.0 (200 N) / 2 T sin W b / 2 solve for x L 3.00 m 1.50 m ; xmax max W 300 N max
(b) The x-reaction-force Fx is due to
F
x ,i
0 , in which there are two forces: the (negative) x-component of
tension, and the (positive) reaction-force,
F
x,i
solve for F 0 Tx Fx Tmax cos Fx Fx = T max cos = 433N ;
(c) the y-reaction force Fy is due to
x
F
y ,i
(1.11)
0 ; unlike the x-direction (1.11), there are four forces in this
balance,
F
y ,i
solve for F 0 Fy Ty Wblock Wbar Fy Tmax sin W Wb Fy = W + Wb T max sin = 250 N ; (1.12) x
chapter 12, problem 37: In Fig. 12-49, a uniform plank, with a length L of 6.10 m and a weight of 445 N, rests on the ground and against a frictionless roller at the top of a wall of height h = 3.05 m. The plank remains in equilibrium for any value of u > 70° but slips if u < 70°. Find the coefficient of static friction between the plank and the ground.
The free-body diagram below shows the forces a cting on the plank. Since the roller is frictionless, the force it exerts is normal to the plank and makes the angle with the vertical.
Its magnitude is designated F . W is the force of gravity; this force acts at the center of the plank, a distance L/2 from the point where the plank touches the floor. F N is the normal force of the floor and f is the force of friction. The distance from the foot of the plank to the wall is denoted b y d . This quantity is not given directly but it can be computed using d = h/tan . The equations of equilibrium are: horizontal force components:
F sin f 0
vertical force components:
F cos W F N 0
torques:
F N d fh W d L2 cos 0.
The point of contact between the plank and the roller was used as the origin for writing the torque equation. When = 70º the plank just begins to slip and f = s F N , where s is the coefficient of static friction. We want to use the equations of equilibrium to compute F N and f for = 70º, then use s = f / F N to compute the coefficient of friction. The second equation gives F = (W – F N )/cos and this is substituted into the first to obtain f = (W – F N ) sin /cos = (W – F N ) tan . This is substituted into the third equation and the result is solved for F N :
F N =
d L/2 cos + h tan d + h tan
W
h(1 tan2 ) (L / 2) sin h(1 tan2 )
W ,
where we have used d = h/tan and multiplied both numerator and denominator by tan . We use the 2
2
2
trigonometric identity 1+ tan = 1/cos and multiply both numerator and denominator by cos to obtain
F N = W 1
L 2h
2
cos sin .
Now we use this expression for F N in f = (W – F N ) tan to find the friction:
f =
WL 2h
2
sin cos .
Substituting these expressions for f and F N into s = f / F N leads to
s =
L sin 2 cos 2h L sin cos2
.
Evaluating this expression for = 70º, L = 6.10 m and h = 3.05 m gives
6.1msin270cos70 s = = 0.34. 2 2 3.05m 6.1m sin70cos 70
chapter 12, problem 49: In Fig. 12-58, a 103 kg uniform log hangs by two steel wires, A and B, both of radius 1.20 mm. Initially, wire A was 2.50 m long and 2.00 mm shorter than wire B. The log is now horizontal. What are the magnitudes of the forces on it from (a) wire A and (b) wire B? (c) What is the ratio d A /d B?
(a) Let F A and F B be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium, F A + F B – mg = 0. Information given about the stretching of the wires allows us to find a relationship between F A and F B. If wire A originally had a length L A and stretches by L A , then
L A FA L A / AE , where A is the cross-sectional area of the wire and E is Young’s modulus for steel (200 × 109 2 N/m ). Similarly, L B FB LB / AE . If is the amount by which B was originally longer than A then, since they have the same length after the log is attached, L A LB . This means F A LA AE
FB LB AE
.
We solve for F B: F B
F A LA L B
AE LB
.
We substitute into F A + F B – mg = 0 and obtain mgL B AE F A . L A LB
The cross-sectional area of a wire is
c
h 4.52 10
A r 2 120 . 10 3 m
2
6
m2 .
Both L A and L B may be taken to be 2.50 m without loss of significance. Thus (103kg) (9.8m/s ) (2.50 m) (4.52 10 m ) (200 10 N/m ) (2.0 10 m) 2
F A
6
2
9
2
3
2.50m 2.50 m
866 N. (b) From the condition F A + F B – mg = 0, we obtain F B mg F A (103kg) (9.8m/s2 ) 866 N 143 N.
(c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes F Ad A – F Bd B = 0, which leads to d A d B
F B F A
143N 866 N
0.165.
