Circular Motion Flashcards
- A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion?
Here we have the following values :
R = 5m , T = 8sec weight =670N
V = ω × R
V = (2π/(8))×R
V = 3.928m/s
M = 670/(9.8) = 68.367kg
Wapp = W + Mv2/R
Wapp = 670 + ((68.367)(3.928)2/(5))
Wapp = 880N (approx.)
18) A car moving at a steady 10 m/s on a level highway encounters a bump that has a circular cross-section with a radius of 30 m. The car maintains its speed over the bump. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?
A) 200 N
B) 390 N
C) 790 N
D) 490 N
E) 590 N
B) 390 N
r = 30 m
v = 10 m/s
m = 60 kg
(a) Let N be the normal reaction.
At the bump
N = mg - mv^2 / r
N = 60 x 9.8 - 60 x 10 x 10 / 30 = 588 - 200 = 388 newton
19) Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70-kg pilot at that point?
A) 3300 N
B) 690 N
C) 2600 N
D) 490 N
E) 1400 N
A) 3300 N
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
W= Fv
Fv= mr^2/r
Fv= (70x150^2)/600
Fv= 2625 N
Real weight of the pilot
WR= mg
WR= 70 x 9.8
WR= 686 N
Apparent weight - Real weight of the pilot= weight due to centripetal force
FN- mg= mv^2/r
FN= mv^2/r + mg
FN= 2625N + 686 N
FN= 3311 N
20) In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. If the passengers are to experience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.5 km?
A) 79 m/s
B) 310 m/s
C) 260 m/s
D) 160 m/s
E) 510 m/s
D) 160 m/s
-N + mg = ma = mv²/r
For “weightlessness” N = 0, so
0 = mg - mv²/r
g - v²/r = 0
v =√( gr)
g = 9.8 and r = 2.5km = 2500 m
v = √(9.8 x 2500)
= 156.6 m/s
21) A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s.
(a) What are the magnitude and direction of the acceleration of the ball?
(b) Find the tension in the string.
a = v^2 / r
v = 2 * pie * r / T
T is the time of one revolution = 0.5 sec
v = 2 * 3.14 * 0.5 / 0.5 = 6.28 m/s
then
a = v^2 / r
a = 6.28^2 / 0.5 = 78.87 m/s^2
b)
the tension in the string is
T = m * a
T = 0.175 * 78.87
T = 13.8 N
22) A car traveling at a steady 20 m/s rounds an 80-m radius horizontal unbanked curve with the tires on the verge of slipping. What is the maximum speed with which this car can round a second unbanked curve of radius 320 m if the coefficient of static friction between the car’s tires and the road surface is the same in both cases?
A) 160 m/s
B) 80 m/s
C) 70 m/s
D) 40 m/s
E) 30 m/s
D) 40 m/s
a=v^2/r
umg=m v^2/r
u=v^2/r
u= (20m/s)^2/ (80m) x (9.8 m/s^2) = 0.51
(v m/s)^2/(320m) x (9.8 m/s^2) = 0.51
v^2= 1.6 x 10^3
v = 40 m/s
23) A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned person on Earth to lie in bed. In a space station, the effect of gravity can be reduced or even eliminated. How long should each rotation take for a doughnut-shaped hospital of 200-m radius so that persons on the outer perimeter would experience 1/10 the normal gravity of Earth?
A) 91 min
B) 8.7 min
C) 4.6 min
D) 1.5 min
E) 0.011 min
D) 1.5 min
centrip acc = (1/10) g = 0.980m/s2
.
centrip acc = ω2 r
.
0.980 = ω2 *200
.
ω = 0.07 radians per second
.
time for one rotation = distance / speed = 2π radians / 0.07 rad per second = 89.76 seconds or 1.5 mins (period of rotation)
24) A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?
A) 29.1°
B) 13.5°
C) 33.8°
D) 56.2°
E) 60.9°
A) 29.1
Ncos ꬾ= mg
Nsinꬾ = mv²/r
tanꬾ = v²/rg = 24.5 m/s)²/(110 m)(9.8 m/s²)
ꬾ = 29.1°
25) The curved section of speedway is a circular arc having a radius of 190 m. This curve is properly banked for racecars moving at 34 m/s. At what angle with the horizontal is the curved part of the speedway banked?
A) 32°
B) 34°
C) 30°
D) 28°
E) 26°
C) 31.83
tanꬾ = v˳²/Rg
= (34)²/190 x 9.8
=.621
ꬾ= 31.83°
26) A highway curve of radius 80 m is banked at 45°. Suppose that an ice storm hits, and the curve is effectively frictionless. What is the speed with which to take the curve without tending to slide either up or down the surface of the road?
A) 9.4 m/s
B) 28 m/s
C) 780 m/s
D) The curve cannot be taken safely at any speed
B) 28 m/s
The formula for the banking angle is
v = √(rgtanθ)
v = √[(80)(9.8)(tan 45)]
v = 28 m/s
27) What is the gravitational force acting on a 59-kg person due to another 59-kg person standing 2.0 m away? We can model each person as a small sphere. (G = 6.67 × 10-11 N ∙ m2/kg2)
A) 5.8 × 10-8 N
B) 8.5 × 103 N
C) 1.2 × 10-7 N
D) 9.8 × 10-10 N
E) 2.0 × 10-9 N
B) 5.8 x 10^-8 N
F= G x (m1m2)/d^2
F= 6.67 x 10^-10 x (59 kg x 59 kg)/ 2^2
F= 5.804 X 10^-8 N
