A flywheel with a diameter of 2.45 m is rotating at an angular speed of 74.1 rev/min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 1260 rev/min in 102 s? (d) How many revolutions does the wheel make during that 102 s?
(a) To convert from revolutions per minute (rpm) to radians per second, we need to use the following conversion factor:
1 revolution = 2π radians
So, to convert from rpm to radians per second, we can use the following steps:
angular speed (in radians per second) = (angular speed in rpm) * (2π radians / 1 revolution) * (1 minute / 60 seconds)
Given the angular speed in rpm as 74.1 rev/min, we can substitute this value into the formula:
angular speed = 74.1 rev/min * (2π radians / 1 revolution) * (1 min / 60 s)
Calculating this expression, we get:
angular speed = 74.1 * (2π / 1) * (1/60) ≈ 7.78 radians/second
So, the angular speed of the flywheel is approximately 7.78 radians/second.
(b) The linear speed of a point on the rim of the flywheel can be calculated using the formula:
linear speed = (angular speed) * (radius)
Given the diameter of the flywheel as 2.45 m, the radius is half of the diameter, so:
radius = diameter / 2 = 2.45 m / 2 = 1.225 m
Substituting the angular speed of 7.78 radians/second and the radius of 1.225 m into the formula, we get:
linear speed = 7.78 radians/second * 1.225 m ≈ 9.51 m/second
So, the linear speed of a point on the rim of the flywheel is approximately 9.51 m/second.
(c) To find the constant angular acceleration required to increase the angular speed of the flywheel to 1260 rev/min in 102 s, we can use the following formula:
final angular speed = initial angular speed + (angular acceleration) * (time)
Given the initial angular speed as 74.1 rev/min and the final angular speed as 1260 rev/min, we need to convert these values to radians per second:
initial angular speed = 74.1 rev/min * (2π radians / 1 revolution) * (1 min / 60 s) ≈ 7.78 radians/second
final angular speed = 1260 rev/min * (2π radians / 1 revolution) * (1 min / 60 s) ≈ 132.75 radians/second
We can substitute these values into the formula to solve for the angular acceleration:
132.75 radians/second = 7.78 radians/second + (angular acceleration) * 102 seconds
Simplifying this equation, we get:
(angular acceleration) * 102 seconds = 132.75 radians/second - 7.78 radians/second
(angular acceleration) * 102 seconds = 124.97 radians/second
angular acceleration = 124.97 radians/second / 102 seconds ≈ 1.23 radians/second^2
So, the constant angular acceleration required to increase the wheel's angular speed to 1260 rev/min in 102 s is approximately 1.23 radians/second^2.
(d) To find the number of revolutions the wheel makes during the 102 s time period, we can use the formula:
number of revolutions = (angular speed) * (time) / (2π)
Given the angular speed as 7.78 radians/second and the time as 102 seconds, we can substitute these values into the formula:
number of revolutions = 7.78 radians/second * 102 seconds / (2π)
Calculating this expression, we get:
number of revolutions ≈ 40.08 revolutions
So, the wheel makes approximately 40.08 revolutions during the 102 s time period.
To solve these problems, we need to understand some basic concepts.
For part (a), we need to convert the angular speed from rev/min to radians/second.
To convert from rev/min to radians/second, we need to use the conversion factors:
1 revolution = 2π radians
1 minute = 60 seconds
Therefore, the conversion factor from rev/min to radians/second is:
(2π radians / 1 revolution) * (1 revolution / 1 minute) = (2π/60) radians/second
So, to find the angular speed in radians per second, we multiply the given angular speed of 74.1 rev/min by the conversion factor:
(74.1 rev/min) * (2π/60) radians/second = 7.74 radians/second
Therefore, the angular speed of the flywheel is 7.74 radians/second.
For part (b), we need to find the linear speed of a point on the rim of the flywheel.
The linear speed of a point on the rim of a rotating object is given by the equation:
v = rω
where v is the linear speed, r is the radius of the wheel, and ω is the angular speed.
Given that the diameter of the wheel is 2.45 m, the radius is half of that:
r = 2.45 m / 2 = 1.225 m
Substituting the values into the equation, we get:
v = 1.225 m * 7.74 radians/second = 9.47 m/second
Therefore, the linear speed of a point on the rim of the flywheel is 9.47 m/second.
For part (c), we need to find the constant angular acceleration required to increase the wheel's angular speed to 1260 rev/min in 102 seconds.
The formula for angular acceleration is given by:
α = (ωf - ωi) / t
where α is the angular acceleration, ωf is the final angular speed, ωi is the initial angular speed, and t is the time taken.
Given that ωi is the initial angular speed of 74.1 rev/min and ωf is the final angular speed of 1260 rev/min, we need to convert them to radians/second using the conversion factor we derived earlier:
ωi = 74.1 rev/min * (2π/60) radians/second = 7.74 radians/second
ωf = 1260 rev/min * (2π/60) radians/second = 131.95 radians/second
Substituting the values into the formula, we get:
α = (131.95 radians/second - 7.74 radians/second) / 102 seconds = 1.26 radians/second^2
Therefore, the constant angular acceleration required is 1.26 radians/second^2.
For part (d), we need to find the number of revolutions the wheel makes during the 102 seconds.
The formula for angular displacement is given by:
θ = ωi * t + 0.5 * α * t^2
Given that ωi is the initial angular speed of 74.1 rev/min (which, as we calculated earlier, is equivalent to 7.74 radians/seconds), α is the constant angular acceleration of 1.26 radians/second^2, and t is the time of 102 seconds, substituting the values into the formula, we get:
θ = 7.74 radians/second * 102 seconds + 0.5 * 1.26 radians/second^2 * (102 seconds)^2
Simplifying the equation, we get:
θ ≈ 795.11 radians
Since one revolution is equal to 2π radians, we can convert the radians to revolutions:
θ ≈ 795.11 radians * (1 revolution / 2π radians) ≈ 399.55 revolutions
Therefore, the wheel makes approximately 399.55 revolutions during the 102 seconds.
a. C = pi*D = 3.14 * 2.45 = 7.7m.
V=74.1rev/min * 6.28rad/rev *(1/60min/s
= 7.76 rad/s.
b. V = 7.76rad/s * 7.7m/6.28rad=9.51m/s
c. t = 102s / 60s/min = 1.7 min.
V = Vo + at = 1260rev/min.
74.1 + 1.7a = 1260,
1.7a = 1260 - 74.1 = 1185.9,
a = 698rev/min^2.
d. rev = 1260rev/min * 1.7min = 2142