Problem#1
An air puck of mass m1 is tied to a string and allowed to revolve in a circle of radius R on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a counterweight of mass m 2 is tied to it (Fig. 1). The suspended object remains in equilibrium while the puck on the tabletop revolves. What is (a) the tension in the string? (b) the radial force acting on the puck? (c) the speed of the puck? |
| Fig.1 |
(a) Since the object of mass m2 is in equilibrium,
∑Fy = may
T – m2g = 0
T = m2g
(b) The tension in the string provides the required centripetal acceleration of the puck.
Thus,
FC = T = m2g
(c) Centripetal force is also given by
FC = m1v2/r
We have
m2g = m1v2/r
v = [m2rg/m1]1/2
Problem#2
The pilot of an airplane executes a constant-speed loop-theloop maneuver in a vertical circle. The speed of the airplane is 300 mi/h, and the radius of the circle is 1,200 ft. (a) What is the pilot’s apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)
Answer:
Given: v = 300 mi/h = 300 x 1609.34 m/3600 s = 134 m/s and r = 1,200 x 0.305 m = 366 m
(a) At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase.
His apparent weight (160 lb = 0.454 x 160 kg = 72.6 kg) is
mg' = mg + mv2/r
mg’ = 72.6 kg[9.80 m/s2 + (134 m/s)2/366 m] = 4.27 x 103 N
(b) At the highest point, the force of the seat on the pilot is directed down and
mg' = mg – mv2/r
mg’ = 72.6 kg[9.80 m/s2 – (134 m/s)2/366 m] = –2.85 x 103 N
Since the plane is upside down, the seat exerts this downward force.
(c) When mg′ = 0 , then mg = mv2/R. If we vary the aircraft’s R and v such that the above is true,
then the pilot feels weightless.
Problem#3
A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk (Fig. 2). The coefficients of friction between block and disk are 0.750 (static) and 0.640 (kinetic) while those for the penny and block are 0.520 (static) and 0.450 (kinetic). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk?
 |
| Fig.2 |
The forces acting on penny and beams are shown in the figure below,
For the block to remain stationary,
∑Fy = may = 0
n1 – mbg – mpg = 0
n1 = g(mb + mp)
At the point of slipping, the required centripetal force equals the maximum friction force:
f = (mb + mp)vmax2/r
with f ≤ µs1n = µs1(mb + mp)g, then
µs1g = vmax2/r
vmax2 = µs1gr = 0.750(0.120 m)(9.80 m/s2)
vmax = 0.939 m/s
For the penny to remain stationary on the block:
∑Fy = 0
n2 – mpg = 0
n2 = mpg
and ∑Fx = ma
fp = mpv2/r
When the penny is about to slip on the block,
fp = fp,max = µs2n2
so that
µs2mpg = mpvmax2/r
vmax2 = µs2gr = 0.520(0.120 m)(9.80 m/s2)
vmax = 0.782 m/s
This is less than the maximum speed for the block, so the penny slips before the block starts to slip.
The maximum rotation frequency is
Max rpm = vmax/2πr
Max rpm = (0.782 m/s)/(2π x 0.120 m)(60 rev.s/min) = 62.2 rev/min
Problem#4
Figure 4 shows a Ferris wheel that rotates four times each minute. It carries each car around a circle of diameter 18.0 m. (a) What is the centripetal acceleration of a rider? What force does the seat exert on a 40.0-kg rider (b) at the lowest point of the ride and (c) at the highest point of the ride? (d) What force (magnitude and direction) does the seat exert on a rider when the rider is
halfway between top and bottom?
 |
| Fig.4 |
Given: T = ¼ min = 15.0 s, m = 40.0 kg, and d = 18.0 m
Then v = 2πr/T = 2π(9.00 m)/(15.0 s) = 3.77 m/s
(a) the centripetal acceleration of a rider is
ac = v2/r = (3.77 m/s)2/(9.00 m) = 1.58 m/s2
(b) Flaw = m(g + ac) = 40.0kg(9.80 m/s2 + 1.58 m/s2) = 455 N
(c) Fhight = m(g – ac) = 40.0kg(9.80 m/s2 – 1.58 m/s2) = 328 N
(d) Fmind = m(g2 + ac2)1/2
Fmind = (40.0 kg)[(9.80 m/s2)2 + (1.58 m/s2)2]1/2
Fmind = 397 N upward
and
tan θ = ac/g
θ = tan-1(1.58/9.80) = 9.150 inward
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