1- In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall
ID: 1772187 • Letter: 1
Question
1- In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.6-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick"to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.7 to 1.0 and a kinetic coefficient in the range 0.2 to 0.70. What is the minimum rotational frequency, in rpm, for which the tide is safe?
2- An earth satellite moves in a circular orbit at a speed of 6,600 m/s. What is its orbital period in minutes?
3-The passengers in a roller coaster car feel 68% heavier than their true weight as the car goes through a dip with a 16 m radius of curvature. What is the car's speed at the bottom of the dip?
4- A 156-g block on a 24-cm-long string swings in a circle on a horizontal, frictionless table at 96-rpm. Whal is the tension in the string?
5- A CD-ROM drive in a computer spins the 12-cm-diameter disks at 11,600 rpm. What would be the speed of a speck of dust on the outside edge of this disk?
Explanation / Answer
1- In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.6-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick"to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.7 to 1.0 and a kinetic coefficient in the range 0.2 to 0.70. What is the minimum rotational frequency, in rpm, for which the tide is safe?
Given
Diameter of the cylinder D = 5.6 m
Solution
Radius of the cylinder
r = D/2
r = 5.6/2
r = 2.8 m
If enough friction is provided by the centripetal force then the ride is safe.
Centripetal force
FC = mv2/r
Fractional force
Fr = µFC
Fr = µmv2/r
for safe condition
Fr = W
µmv2/r = mg
µv2/r = g
Since V = r
µ2r = g
= (g/µr)
= (9.8/1.0 x 2.8)
= 1.87 rad/s
Angular frequency
n = /2
n = 0.2978 Hz
n = 0.2978 x 60 rpm
n = 17.86 rpm
2- An earth satellite moves in a circular orbit at a speed of 6,600 m/s. What is its orbital period in minutes?
Given
Orbital velocity v = 6600 m/s
Solution
Radius of the orbit
r = GM/v2
period
T = 2r/v
T = 2GM/v3
T = 2 x 3.14 x 6.67 x 10-11 x 5.98 x 1024 / 66003
T = 8712.74 s
T = 8712.74 /60 minutes
T = 145.21 minutes
3-The passengers in a roller coaster car feel 68% heavier than their true weight as the car goes through a dip with a 16 m radius of curvature. What is the car's speed at the bottom of the dip?
if the actual weight is w
Apparent weight w’ = w + 68% of w
w’ = w + 0.68w
w’ = 1.68w
This increase in weight is due to the centripetal force
0.68 w = mv2/r
0.68mg = mv2/r
v2 = 0.68gr
v = (0.68gr)
v = (0.68 x 9.8 x 16)
v = 10.32 m/s
4- A 156-g block on a 24-cm-long string swings in a circle on a horizontal, frictionless table at 96-rpm. Whal is the tension in the string?
mass
m = 156 g
m = 0.156 kg
Radius
r = 24 cm
r = 0.24 m
Frequency
n = 96 rpm
n = 96 / 60 Hz
n = 1.6 Hz
Angular velocity
= 2n
= 2 x 3.14 x 1.6
= 10.048 rad/s
Tension
T = mv2/r
T = m (r)2/r
T = mr
T = 0.156 x 0.24 x 10.048
T = 0.376 N
5- A CD-ROM drive in a computer spins the 12-cm-diameter disks at 11,600 rpm. What would be the speed of a speck of dust on the outside edge of this disk?
frequency
n = 11600 rpm
n = 11600 / 60 Hz
n = 193.33 Hz
angular speed
= 2n
= 2 x 3.14 x 193.33
= 1214.1124 rad/s
linear speed
v = r
v = 1214.1124 x 0.12
v = 145.69 m/s
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