A wheel starts rotating from rest with uniform angular acceleration.After 3.0,its completes one revolution.Calculate a) angular velocity after 3.0s. b)angular acceleration c)angular velocity after completing d)time taken for second revolution
To calculate the angular velocity and angular acceleration of a rotating wheel, as well as the time taken for the second revolution, we can use the equations of rotational motion.
a) We can use the equation for angular velocity:
ω = ω_0 + αt
Since the wheel starts from rest, the initial angular velocity (ω_0) is zero. Given that the time (t) is 3.0 s, we can substitute the values to find the angular velocity (ω) after 3.0 s.
b) To find the angular acceleration (α), we can use the equation:
θ = ω_0t + 0.5αt^2
Since the wheel completes one revolution (2π radians) after 3.0 s, we can substitute the values to solve for the angular acceleration (α).
c) After completing one revolution, the angular displacement (θ) is 2π radians. The final angular velocity (ω) can be obtained using the equation:
ω^2 = ω_0^2 + 2αθ
Substituting the values, we can solve for the final angular velocity (ω).
d) To find the time taken for the second revolution, we can rearrange the equation for angular displacement:
θ = ω_0t + 0.5αt^2
Since one revolution corresponds to an angular displacement of 2π radians, we can substitute the values and rearrange the equation to solve for the time (t).
Let's calculate each of these step-by-step:
a) Angular velocity after 3.0 s:
Given:
ω_0 = 0 (initial angular velocity)
t = 3.0 s
Using the equation:
ω = ω_0 + αt
Substituting the values:
ω = 0 + α * 3.0
Therefore, angular velocity after 3.0 s is ω = 3α rad/s.
b) Angular acceleration:
Given:
θ = 2π (angular displacement after 3.0 s)
ω_0 = 0 (initial angular velocity)
t = 3.0 s
Using the equation:
θ = ω_0t + 0.5αt^2
Substituting the values:
2π = 0 * 3.0 + 0.5α * (3.0)^2
Simplifying the equation:
2π = 0 + 4.5α
Therefore, the angular acceleration (α) is α = (2π) / 4.5 rad/s^2.
c) Angular velocity after completing one revolution:
Given:
θ = 2π (angular displacement after completing one revolution)
ω_0 = 0 (initial angular velocity)
α = (2π) / 4.5 (angular acceleration)
Using the equation:
ω^2 = ω_0^2 + 2αθ
Substituting the values:
ω^2 = 0^2 + 2 * (2π / 4.5) * 2π
Simplifying the equation:
ω^2 = (2π^2) / (2.25)
Taking the square root on both sides:
ω = √((2π^2) / (2.25))
Therefore, the angular velocity after completing one revolution is ω = √((2π^2) / 2.25) rad/s.
d) Time taken for the second revolution:
Given:
θ = 2π (angular displacement for one revolution)
ω_0 = 0 (initial angular velocity)
α = (2π) / 4.5 (angular acceleration)
Using the equation:
θ = ω_0t + 0.5αt^2
Substituting the values:
2π = 0 + 0.5 * (2π / 4.5) * t^2
Simplifying the equation:
2π = 0.222t^2
Solving for t:
t^2 = 2π / 0.222
t = √(2π / 0.222)
Therefore, the time taken for the second revolution is approximately t = √(2π / 0.222) s.
To solve this problem, we'll first need to understand the basic equations for rotational motion.
1. Angular velocity (ω) is defined as the rate at which an object rotates. It is measured in radians per second (rad/s).
2. Angular acceleration (α) is defined as the rate at which the angular velocity changes. It is measured in radians per second squared (rad/s^2).
3. The relationship between angular velocity, angular acceleration, and time can be expressed using the equation: ω = ω0 + α*t, where ω0 is the initial angular velocity, t is the time, and α is the angular acceleration.
4. The relationship between angular displacement (θ), angular velocity (ω), and time can be expressed using the equation: θ = ω0*t + (1/2)*α*t^2. This equation is similar to the equation for linear motion: s = v0*t + (1/2)*a*t^2, where s is the displacement, v0 is the initial velocity, a is the acceleration, and t is the time.
Let's solve the problem step by step:
a) Angular velocity after 3.0 s:
We are given that the wheel completes one revolution in 3.0 s. We know that one revolution is equal to 2π radians. Therefore, the angular displacement (θ) after 3.0 s is 2π radians.
Using the equation θ = ω0*t + (1/2)*α*t^2, we can rearrange it to solve for the angular velocity (ω):
2π = ω0*3.0 + (1/2)*α*(3.0)^2
We also know that the wheel starts from rest, so the initial angular velocity (ω0) is 0.
Substituting these values into the equation, we have:
2π = 0 + (1/2)*α*(3.0)^2
Simplifying the equation yields:
2π = (1/2)*9.0*α
4π = 9.0*α
α = 4π/9.0
Plugging this value of α into the equation ω = ω0 + α*t, where t = 3.0 s, we get:
ω = 0 + (4π/9.0)*3.0
ω = 4π/3.0 rad/s (final answer)
Therefore, the angular velocity after 3.0 s is 4π/3.0 rad/s.
b) Angular acceleration:
From the result in part (a), we found that α = 4π/9.0 rad/s^2.
c) Angular velocity after completing one revolution:
In one revolution, the angular displacement (θ) is 2π radians. We know that ω = ω0 + α*t, and that the wheel starts from rest initially, so ω0 = 0.
Plugging these values into the equation, we get:
2π = 0 + (4π/9.0)*t
Simplifying the equation yields:
2π = (4π/9.0)*t
t = (9.0*2π)/(4π)
t = 4.5 s
Plugging the value of t into the equation ω = ω0 + α*t, where t = 4.5 s, we get:
ω = 0 + (4π/9.0)*4.5
ω = 2π rad/s (final answer)
Therefore, the angular velocity after completing one revolution is 2π rad/s.
d) Time taken for the second revolution:
From the result in part (c), we found that t = 4.5 s for one revolution. To find the time taken for the second revolution, we simply multiply this time by 2:
Time for second revolution = 4.5 s * 2 = 9.0 s (final answer)
Therefore, the time taken for the second revolution is 9.0 s.
2•π•N = ε•t^2/2,
ε = 4•π•N/t^2 = 4•π•1/3^2 =1.39 rad/s^2
ω = ε•t =1.39•3 = 4.19 rad/s.
2•π•N1 = ε•t1^2/2,
t1 =sqrt(4•π•N1/ ε) = sqrt(4•π•2/ 1.39) = 4.25 s.