Problem #1
An object rotates horizontally on a circular road at a speed of 10 m / s and does 120 turns in one minute. (a) Determine the frequency and period of the object, and (b) Look for changes in the speed vector when rotating 600, 900 and 1800!Answer:
(a) frequency and period
n = 120 rounds, t = 1 minute = 60 seconds, then
f = n/t = 120 rounds / 60 seconds = 2 put / second
T = 1/f = ½ second
(b) change in velocity vector
If an object moves from point A and rotates through 600 at point B, then
Δv = vB - vA
Δv = 2v.cos 1200/2 = 2.10.cos 600 = 20/2 = 10 m/s
Δv = vB - vA
Δv = 2v.cos 1200/2 = 2.10.cos 600 = 20/2 = 10 m/s
If the object rotates to 900;
Δv = vC – vA
Δv = (vA2 + vC2)1/2 = [(10)2 + (10)2]1/2 = 10√2 m/s
If the object rotates 1800;
Δv = vD – vA = –10 m / s – 10 m/s = –20 m/s
Problem #2
A dish rotates 1800 turns for 6 minutes. The fingers are 8 cm long. Determine (a) the frequency and period of the disk, (b) the speed of the disk angle, (c) the linear velocity of a point at 3 cm from the edge of the dish and (d) the centripetal acceleration of the dish!
Answer:
(a) Frequency and period of disk
f = n/t = 1800 turns/(6 x 60 seconds) = 5 Hz
T = 1/f = 0.2 seconds
(b) the speed of the disk angle
ω = 2πf = 2π x 5 Hz = 10π rad/s
(c) the linear speed of the point at 3 cm from the edge of the disk is
v = ωR = (10π rad/s) (5 cm) = 50π cm/s = 0.5π m/s
(d) centripetal acceleration
as = v2/R = ω2R = (10π rad/s)2(8 cm) = 8000 cm/s2 = 80 m/s2
Problem #3
A wall clock must have seconds, minutes and clock hands. Determine the ratio of seconds to seconds, minutes and hours!
Answer:
In 1 round = 3600 = 2π radians, the second hand takes 1 minute = 60 seconds, while the minute hand takes 1 hour = 3600 seconds and the clock needs 12 hours = 12 x 3600 seconds, then
ωs (second hand) = 2π/(60 s) = π/30 rad / s
ωm (minute hand) = 2π/(3600 s) = π/1800 rad / s
ωh (hour hand) = 2π/(12 x 3600 s) = π/21600 rad / s
then ωs: ωm: ωh = (π/30): (π/1800) : (π/21600) = 720 : 12: 1
Problem #4
You design an aircraft propeller that rotates at 2400 rpm like the picture. The speed of the forward plane is 75.0 m/s and the tip speed of the air propeller blades must not exceed 270 m/s. (This is about 80% of the sound speed in the air. If the propeller speed is greater than this, they will produce a lot of noise.) (A) What is the maximum radius of the propeller? and (b) With this radius, what is the speed of the propeller?
Answer:
Changing the angular speed of the propeller in the rad / s is
ω = 2400 put/minute = 2400 (2π rad)/(60 seconds) = 80π rad/s = 251 rad/s
(a) from the picture we wrote,
vT2 = v2tan + v2pesawat = r2ω2 + v2plane
(270 m/s)2 = r2(251 rad/s)2 + (75.0 m/s)2
(251 rad/s)2r2 = (270 m/s)2 – (75.0 m/s)2 = 67275
r = 1.03 m
(b) The acceleration of centripetal particles is
as = ω2r = (251 rad/s)2(1.03 m) = 6.5 x 104 m/s2 = 6600g
Problem #5
A wheel rotates 150 revolutions in 30 seconds. What is the angular speed?
Answer:
(a) revolutions per minute (rpm)
150 revolutions /30 seconds = 150 revolutions /0.5 minute = 300 revolutions/minute = 300 rpm
(b) degrees per second (o/s)
1 revolution = 360o, 300 revolutions = 54000o
150 revolutions/30 seconds = (150)(360o)/30 seconds = 54000o/30 seconds = 1800o/second
(c) radians per second (rad/s)
1 revolution = 6.28 radians
150 revolutions/30 seconds = (150)(6.28) radians/30 seconds = 942 radians/30 seconds = 31.4 radians/second.
Δv = vC – vA
Δv = (vA2 + vC2)1/2 = [(10)2 + (10)2]1/2 = 10√2 m/s
If the object rotates 1800;
Δv = vD – vA = –10 m / s – 10 m/s = –20 m/s
Problem #2
A dish rotates 1800 turns for 6 minutes. The fingers are 8 cm long. Determine (a) the frequency and period of the disk, (b) the speed of the disk angle, (c) the linear velocity of a point at 3 cm from the edge of the dish and (d) the centripetal acceleration of the dish!
Answer:
(a) Frequency and period of disk
f = n/t = 1800 turns/(6 x 60 seconds) = 5 Hz
T = 1/f = 0.2 seconds
(b) the speed of the disk angle
ω = 2πf = 2π x 5 Hz = 10π rad/s
(c) the linear speed of the point at 3 cm from the edge of the disk is
v = ωR = (10π rad/s) (5 cm) = 50π cm/s = 0.5π m/s
(d) centripetal acceleration
as = v2/R = ω2R = (10π rad/s)2(8 cm) = 8000 cm/s2 = 80 m/s2
Problem #3
A wall clock must have seconds, minutes and clock hands. Determine the ratio of seconds to seconds, minutes and hours!
Answer:
In 1 round = 3600 = 2π radians, the second hand takes 1 minute = 60 seconds, while the minute hand takes 1 hour = 3600 seconds and the clock needs 12 hours = 12 x 3600 seconds, then
ωs (second hand) = 2π/(60 s) = π/30 rad / s
ωm (minute hand) = 2π/(3600 s) = π/1800 rad / s
ωh (hour hand) = 2π/(12 x 3600 s) = π/21600 rad / s
then ωs: ωm: ωh = (π/30): (π/1800) : (π/21600) = 720 : 12: 1
Problem #4
You design an aircraft propeller that rotates at 2400 rpm like the picture. The speed of the forward plane is 75.0 m/s and the tip speed of the air propeller blades must not exceed 270 m/s. (This is about 80% of the sound speed in the air. If the propeller speed is greater than this, they will produce a lot of noise.) (A) What is the maximum radius of the propeller? and (b) With this radius, what is the speed of the propeller?
Answer:
Changing the angular speed of the propeller in the rad / s is
ω = 2400 put/minute = 2400 (2π rad)/(60 seconds) = 80π rad/s = 251 rad/s
(a) from the picture we wrote,
vT2 = v2tan + v2pesawat = r2ω2 + v2plane
(270 m/s)2 = r2(251 rad/s)2 + (75.0 m/s)2
(251 rad/s)2r2 = (270 m/s)2 – (75.0 m/s)2 = 67275
r = 1.03 m
(b) The acceleration of centripetal particles is
as = ω2r = (251 rad/s)2(1.03 m) = 6.5 x 104 m/s2 = 6600g
Problem #5
A wheel rotates 150 revolutions in 30 seconds. What is the angular speed?
Answer:
(a) revolutions per minute (rpm)
150 revolutions /30 seconds = 150 revolutions /0.5 minute = 300 revolutions/minute = 300 rpm
(b) degrees per second (o/s)
1 revolution = 360o, 300 revolutions = 54000o
150 revolutions/30 seconds = (150)(360o)/30 seconds = 54000o/30 seconds = 1800o/second
(c) radians per second (rad/s)
1 revolution = 6.28 radians
150 revolutions/30 seconds = (150)(6.28) radians/30 seconds = 942 radians/30 seconds = 31.4 radians/second.
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