7) A 0.50-kg mass is attached to the end of a 1.0-m string. The system is whirle

Question

7) A 0.50-kg mass is attached to the end of a 1.0-m string. The system is whirled in a horizontal
circular path. If the maximum tension that the string can withstand is 350 N. What is the
maximum speed of the mass if the string is not to break?
A) 700 m/s
B) 26 m/s
C) 19 m/s
D) 13 m/s
Answer: B

8) A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At
the exact top of the path the tension in the string is 3 times the stone's weight. The stone's
speed at this point is given by
A) 2(gr)1/2.
B) (2gr)1/2.
C) (gr)1/2.
D) 2gr.
Answer: A

9) A stone, of mass m, is attached to a strong string and whirled in a vertical circle of radius r. At
the exact bottom of the path the tension in the string is 3 times the stone's weight. The stone's
speed at this point is given by
A) 2(gr)1/2.
B) (2gr)1/2.
C) (gr)1/2.
D) 2gr.
Answer: B

10) A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive.
What is the radius of curvature of the loop in which the plane is flying?
A) 640 m
B) 1200 m
C) 7100 m
D) 9200 m
Answer: D

11) A pilot makes an outside vertical loop (in which the center of the loop is beneath him) of
radius 3200 m. At the top of his loop he is pushing down on his seat with only one-half of his
normal weight. How fast is he going?
A) 5.0 m/s
B) 25 m/s
C) 125 m/s
D) 625 m/s
Answer: C

Explanation / Answer

7) Max tension = 350 N = centripetal force
Therefore, mv2/r2 = 350

=> Max Velocity = Sqrt ( 350 * 12 / 0.5) = 26.4 m/s

8) When the stone is at the top of the string, forces acting on the stone are the weight of the stone (downwards), tension in the stirng (downwards) and the centrifugal force (upwards)

Therefore, T + mg = mv2/r2

Given that T = 3mg

Therefore, 4mg = mv2/r2

=> V = 2 (gr)1/2

9) Using the same concept as above,

When the stone is at the bottom of the string, forces acting on the stone are the weight of the stone (downwards), tension in the stirng (upwards) and the centrifugal force (downwards)

Therefore, T = mv2/r2 +mg

Given that T = 3mg

Therefore, 2mg = mv2/r2

=> V = (2gr)1/2

10) In a ciruclar motion, acceleration = v2/r = 4g

=> r = V2 / 4g = 360000/4*9.8 = 9200 m (approx)

11) When he is at the top of the loop, the forces acting on the pilot are the centrifugal force (upwards), normal force from the seat (upwards) and his weight (downwards)

Given that normal force = 1/2 * mg

Therefore, normal force + centrifugal force = weight

=> 1/2*mg + mv2/r2 = mg

=> mv2/r2 = 1/2*mg => v = Sqrt (0.5*g*r2) = 125 m/s

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