An amusement park ride consists of airplane-shaped cars attached to steel rods (

Question

An amusement park ride consists of airplane-shaped cars attached to steel rods (Figure 1). Each rod has a length of 15.8 m and a cross-sectional area of 8.00 cm2 How much is the rod stretched when it is vertical and the ride is at rest? (Assume that each car plus two people seated in it has a total weight of 1870 N Express your answer with the appropriate units. When operating, the ride has a maximum angular speed of 11.0 rev/min How much is the rod stretched then? Express your answer with the appropriate units.

Explanation / Answer

dL = T*L/(A*E) where T = tension in rod = weight supported by rod, and E is Young's modulus for steel (2.00*10^11 Pa)

dL = 1870*15.8/(8.0E-4*2.00*10^11) = 0.0184 cm

When rotating, the problem is to find T and substitute into the above equation:
a = centripetal acceleration
w = angular velocity, rad/sec = 11.0 rpm*2/60 = 1.15
g = gravitational acceleration = 9.8 m/s²
r = radius of revolution
L = length of rod = 15.8 m
= angle of rod wrt the axis of rotation
The upper end of the rod is assumed to be attached at the axis of rotation.

a = r*w².....r = L*sin.....
1) a = w²*L*sin
A vector diagram will show that

2) T = m*[g² + w^4*L²*sin²]
also,
m*g = T*cos

g² = cos²*[g² + w^4L²sin²], which can be manipulated to

sin = [1 - g²/(w^4L²)]

sin= sqrt(1-(9.8*9.8/(1.15^4*15.8^2))

sin= 0.7800

= sin-1 (0.7800)

= 51.26
substituting from (2) and squaring,

m*g = T*cos

T= m*g/ cos= 1870/cos(51.26)

T= 2988.23 N

dL = T*L/(A*E)= 2988.23*15.8/(8.0E-4*2.00*10^11)

dL= 0.0295 cm

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