At some instant, a particle traveling in a horizontal circular path of radius 3.30 m has a total acceleration with a magnitude of 18.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant and (1/8) revolution later.
initial speed: _____m/s
final speed: ______m/s
Why did the particle join a circus? Because it wanted to experience some "revolutions"! Okay, let's calculate the speed, shall we?
We know that the total acceleration is the combination of radial acceleration and tangential acceleration. Since the particle is moving in a horizontal circular path, the radial acceleration is zero. Therefore, the total acceleration is equal to the tangential acceleration.
Given that the tangential acceleration is 12.0 m/s², we can now calculate the speed at this instant using the formula:
acceleration = change in velocity / time
Rearranging the formula, we get:
change in velocity = acceleration × time
Since the particle is experiencing a constant tangential acceleration, we can assume the time interval is 1 second.
Hence, the change in velocity is equal to the acceleration, which is 12.0 m/s². So, the initial speed is also 12.0 m/s.
Now, let's move to the second part of the question - 1/8 revolution later.
Since the particle is moving in a circular path, it completes one revolution in 2π seconds. Therefore, 1/8 revolution corresponds to (1/8) × 2π seconds.
Using the same formula as before, change in velocity = acceleration × time, we can calculate the change in velocity as:
change in velocity = 12.0 m/s² × (1/8) × 2π seconds
Simplifying further, we get:
change in velocity = 6π m/s
So, the final speed would be the initial speed of 12.0 m/s plus the change in velocity of 6π m/s.
final speed = 12.0 m/s + 6π m/s
And thus, the final speed is 12.0 m/s + 6π m/s, or roughly 12.0 m/s if we're feeling lazy.
To determine the initial speed of the particle, we need to use the formula for total acceleration in circular motion:
a = sqrt(ar^2 + at^2)
Where:
a = magnitude of total acceleration
ar = radial acceleration
at = tangential acceleration
r = radius of the circular path
Given:
a = 18.0 m/s^2
at = 12.0 m/s^2
r = 3.30 m
Substituting the values into the formula, we get:
18.0 = sqrt(ar^2 + 12.0^2)
Squaring both sides of the equation, we have:
324 = ar^2 + 144
Rearranging the equation:
ar^2 = 324 - 144
ar^2 = 180
Taking the square root of both sides, we get:
ar = sqrt(180)
ar ≈ 13.42 m/s^2
Now, we can calculate the initial speed using the equation:
v_initial = sqrt(ar * r)
v_initial = sqrt(13.42 * 3.30)
v_initial ≈ 14.28 m/s
Therefore, the initial speed of the particle is approximately 14.28 m/s.
To determine the final speed of the particle (1/8) of a revolution later, we can use the equation for tangential acceleration:
at = (v_final - v_initial) / t
Where:
v_final = final speed
t = time interval
Given:
v_initial = 14.28 m/s
at = 12.0 m/s^2
t = (1/8) revolution = (1/8) * 2π radians
Substituting the values into the equation and solving for v_final:
12.0 = (v_final - 14.28) / ((1/8) * 2π)
Multiplying both sides by ((1/8) * 2π), we have:
24π = v_final - 14.28
v_final = 24π + 14.28
v_final ≈ 14.28 + 24 * 3.14
v_final ≈ 14.28 + 75.36
v_final ≈ 89.64 m/s
Therefore, the final speed of the particle (1/8) of a revolution later is approximately 89.64 m/s.
To determine the speed of the particle at the given instant, we need to break down the total acceleration into its components - centripetal acceleration and tangential acceleration.
1. Centripetal acceleration (ac):
The centripetal acceleration is given by the equation ac = v^2 / r, where v is the speed of the particle and r is the radius of the circular path. Rearranging the equation, we have v = sqrt(ac * r).
Given:
Radius (r) = 3.30 m
Total acceleration (at) = 18.0 m/s^2
Substituting the given values into the equation, we can solve for the speed (v):
v = sqrt(ac * r)
v = sqrt(18.0 * 3.30)
v = sqrt(59.4)
v ≈ 7.71 m/s (rounded to two decimal places)
Therefore, the initial speed of the particle is approximately 7.71 m/s.
2. Final speed (v') after (1/8) revolution later:
To determine the change in speed after (1/8) revolution, we need to consider the tangential acceleration (atangential) and the time it takes to complete the (1/8) revolution.
Given:
Tangential acceleration (atangential) = 12.0 m/s^2
Fraction of revolution (1/8)
Since the tangential acceleration is constant, it can be used to calculate the change in speed using the kinematic equation:
Δv = atangential * t
where Δv is the change in velocity, atangential is the tangential acceleration, and t is the time.
To calculate the time taken to complete (1/8) revolution, we need to use the formula:
t = (fraction of revolution) * (time period for a complete revolution)
A complete revolution takes 1/f = 1/(2πr / v) seconds, where f is the frequency (number of revolutions per second).
So the time for (1/8) revolution is t = (1/8) * (2πr / v).
Substituting the known values, we can calculate the time (t):
t = (1/8) * (2π * 3.30 / 7.71)
t = 0.805 seconds (rounded to three decimal places)
Using the calculated time, we can find the change in velocity (Δv):
Δv = atangential * t
Δv = 12.0 * 0.805
Δv ≈ 9.66 m/s (rounded to two decimal places)
To find the final speed (v'), we add the change in velocity (Δv) to the initial speed (v):
v' = v + Δv
v' = 7.71 + 9.66
v' ≈ 17.37 m/s (rounded to two decimal places)
Therefore, the final speed of the particle after (1/8) revolution later is approximately 17.37 m/s.
(total acc)^2 = tangential^2 + centripetal^2
centripetal = V1^2 / r
1/8 2 π r = 1/2 (tngtl) t^2 + V1 t
V2 = V1 (tngtl) t