Problem#1
A 4.00-kg object is attached to a vertical rod by two strings, as in Figure 1. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string. |
| Fig.1 |
sin θ = 1.5 m/2.0 m = 0.75
θ = sin-1(0.75) = 48.60
then
r = (2 m) cos(48.60) = 1.32 m
(a) ∑F = max = mv2/r
Ta cos48.60 + Tb cos48.60 = (4 kg)(6 m/s)2/1.32 m
Ta + Tb = 109 N/cos48.60 = 165 N (1)
∑Fy = may
Ta sin48.60 – Tb sin48.60 – mg = 0
Ta – Tb = mg/sin48.60 = (4 kg)(9.80 m/s2)/sin48.60
Ta – Tb = 52.3 N
To solve simultaneously, we add the equations in Ta and Tb:
2Ta = 217
Ta = 108 N
(b) Tb = 165 N – Ta = 165 N – 108 N = 56.2 N
Problem#2
Casting of molten metal is important in many industrial processes. Centrifugal casting is used for manufacturing pipes, bearings and many other structures. A variety of sophisticated techniques have been invented, but the basic idea is as illustrated in Figure 3. A cylindrical enclosure is rotated rapidly and steadily about a horizontal axis. Molten metal is poured into the rotating cylinder and then cooled, forming the finished product. Turning the cylin- der at a high rotation rate forces the solidifying metal strongly to the outside. Any bubbles are displaced toward the axis, so unwanted voids will not be present in the casting. Sometimes it is desirable to form a composite casting, such as for a bearing. Here a strong steel outer surface is poured, followed by an inner lining of special low-friction metal. In some applications a very strong metal is given a coating of corrosion-resistant metal. Centrifugal casting results in strong bonding between the layers. Suppose that a copper sleeve of inner radius 2.10 cm and outer radius 2.20 cm is to be cast. To eliminate bubbles and give high structural integrity, the centripetal acceleration of each bit of metal should be 100g. What rate of rotation is required? State the answer in revolutions per minute.
 |
| Fig.3 |
We use ac = v2/r,
Let f represent the rotation rate. Each revolution carries each bit of metal through distance 2πr ,
so
v = 2πrf and
ac = 4π2rf2 = 100g
A smaller radius implies smaller acceleration. To meet the criterion for each bit of metal we consider
the minimum radius:
f = [100g/4π2r]1/2
f = [100g/4π2(0.021 m)]1/2 = 34.4 Hz
f = (34.4/s)(60 s/1 rev) = 2.06 x 103 rev/min
Problem#3
A 40.0-kg child swings in a swing supported by two chains, each 3.00 m long. If the tension in each chain at the lowest point is 350 N, find (a) the child’s speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Neglect the mass of the seat.)
Answer:
Given: M = 40.0 kg, R = 3.00 m, and T = 350 N
(a) ∑F = Mv2/R
2T – Mg = Mv2/R
v2 = (2T – Mg)(R/M)
v2 = [2(350 N) – (40.0 kg)(9.80 m/s2)](3.00/4.00) = 23.1 m2/s2
v = 4.81 m/s
(b) n – Mg = Mv2/R
n – (40.0 kg)(9.80 m/s2) = (40.0 kg)(23.1 m2/s2)/(3.00 m)
n = 700 N
Problem#4
A child of mass m swings in a swing supported by two chains, each of length R. If the tension in each chain at the lowest point is T, find (a) the child’s speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Neglect the mass of the seat.)
Answer:
(a) Consider the forces acting on the system consisting of the child and the seat:
∑Fy = may
2T – mg = mv2/R
v2 = [2T/m – g]R
v = {(2T/m – g)R}1/2
(b) Consider the forces acting on the child alone:
∑Fy = may
n – mg = mv2/R
n = m(g + v2/R)
and from above, v2 = [2T/m – g]R, so
n = 2T
Problem#5
Tarzan (m = 85.0 kg) tries to cross a river by swinging from a vine. The vine is 10.0 m long, and his speed at the bottom of the swing (as he just clears the water) will be 8.00 m/s. Tarzan doesn’t know that the vine has a breaking strength of 1 000 N. Does he make it safely across the river?
Answer:
Let the tension at the lowest point be T.
∑F = ma
T – mg = mv2/R
T = m(g + v2/R)
T = (85.0 kg)[9.80 m/s2 + (8.00 m/s)2/(10.0 m)]
T = 1380 N > 1000 N
He doesn’t make it across the river because the vine breaks.
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