Question

6. A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 35.1 m/s2 with a beam of length 5.39 m, what rotation frequency is required?

? Hz

7. You want to design an oval racetrack such that 3200 lb racecars can round the turns of radius 1000 ft at 105 mi/h without the aid of friction. You estimate that when elements like downforce and grip in the tires are considered the cars will round the turns at a maximum of 175 mi/h. Find the banking angle θ necessary for the racecars to navigate these turns at 105 mi/h and without the aid of friction.

? degree

This banking and radius are very close to the actual turn data at Daytona International Speedway where 3200 lb stock cars travel around the turns at about 175 mi/h. What additional radial force is necessary to hold the racecar on the track at 175 mi/h?

? N

8. A 5.21-kg ball hangs from the top of a vertical pole by a 2.51-m-long string. The ball is struck, causing it to revolve around the pole at a speed of 4.83 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 m/s2.

? degree

What is the tension of the string?

? N

10. A jet airplane is in level flight. The mass of the airplane is m = 8990 kg. The plane travels at a constant speed around a circular path of radius R = 8.01 miles and makes one revolution every T = 0.106 hours. What is the magnitude of the lift force acting on the plane?

L = ? kN

Answer 1

6.

a = v²/r,

so v² = ar

v² = (35.1 m/s²)(5.39 m)

v² = 189.189 m²/s²

v = 13.75 m/s.
circumference C = 2πr = 2π(5.39 m)

C = 33.866 m.

Divide the linear speed by this to get the frequency:
(13.75 m/s) / (33.866 m) = 0.406 s^-1

7.

v = (22/15)*105 = 154 ft/sec
For no friction, q = arctan[v²/(R*g)]

q = 67.52°

Anet = √[(v²/R + g²] = 25.665 ft/sec²
Af = Anet*sin q = 23.71 ft/sec²
Ff = m*Af

Ff = (3200/25.665)*23.71 = 2956.24 lb along the slope.
Fr = Ff*cos q = 1130.35 lb

8.

Let the tension in the string be T
(mV²/R)/T = sinθ = R/2.51

T = 2.51*m*V²/R²
mg/T = cosθ = h/2.51

2.51*m*V²/R² = 2.51*mg/h

2.51² = h²+R²

R² = 2.51²-h²
V²/(2.51²-h²) = g/h

V²/g = 2.378 = (2.51²-h²)/h

h²+2.378h-2.51² = 0
Using the quadratic formula
h = 0.79 m

R = 2.382
Tanθ = R/h

θ = 71.65
T = 2.51*mg/h = 162.38N

10.

R = 12890.85 m
T = 381.6 sec → w = 0.01646537 rad/sec
Ac = R*w² = 3.4948 m/sec²
Atot = √[g²+Ac²] = 10.413925 m/sec²

Lift force = weight of plane = 8990*9.81= 88191.90 N

Force of wing on body of plane: F = m*Atot

F = 93621.1938