Question 1183038: A particle accelerates uniformly from rest at 0.15pi rad/s^2 from rest. The particle's initial position is at 45 degrees from the negative x-axis. If the particle moves a distance 15m along the arc of radius 2m for a time of 1.15 seconds, determine (a) position of the particle after 1.15 seconds from the +x-axis, (b) the initial and final angular velocity, (c) the tangential velocity and acceleration of the particle at 2 seconds.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this rotational motion problem:
**(a) Position after 1.15 seconds:**
1. **Angular displacement (θ):** The particle travels a distance (s) of 15 m along an arc of radius (r) 2 m. The angular displacement is given by:
θ = s / r = 15 m / 2 m = 7.5 radians
2. **Initial position:** The initial position is 45 degrees from the negative x-axis. We need to convert this to radians measured counterclockwise from the positive x-axis:
* 45 degrees from the negative x-axis is equivalent to 180° + 45° = 225°
* 225° * (π/180°) = 5π/4 radians
3. **Final position:** The final position is the initial position plus the angular displacement:
* Final position = 5π/4 + 7.5 radians
* Final position ≈ 3.927 + 7.5 ≈ 11.427 radians
4. **Convert to degrees (optional):** If you want the final position in degrees:
* 11.427 radians * (180°/π) ≈ 654.7°
* To find the angle from the +x-axis, you may need to reduce this value to its principal value, by subtracting multiples of 360 until the value is between 0 and 360 degrees.
**(b) Initial and final angular velocity:**
1. **Initial angular velocity (ω₀):** The particle starts from rest, so ω₀ = 0 rad/s.
2. **Final angular velocity (ω):** We can use the following equation of rotational motion:
ω = ω₀ + αt
where α is the angular acceleration and t is the time.
ω = 0 + (0.15π rad/s²)(1.15 s)
ω ≈ 0.542 rad/s
**(c) Tangential velocity and acceleration at t = 2 s:**
1. **Angular velocity at t = 2 s:**
ω(t=2) = ω₀ + αt = 0 + (0.15π rad/s²)(2 s) ≈ 0.942 rad/s
2. **Tangential velocity (v):**
v = ωr = (0.942 rad/s)(2 m) ≈ 1.884 m/s
3. **Tangential acceleration (a_t):**
a_t = αr = (0.15π rad/s²)(2 m) ≈ 0.942 m/s²
4. **Radial (centripetal) acceleration (a_r):**
a_r = ω²r = (0.942 rad/s)²(2 m) ≈ 1.776 m/s²
5. **Total acceleration (a):** The total acceleration is the vector sum of the tangential and radial accelerations.
a = sqrt(a_t² + a_r²) = sqrt((0.942 m/s²)² + (1.776 m/s²)²) ≈ 2.00 m/s²
The direction of the total acceleration can be found using trigonometry:
angle = arctan(a_t / a_r) = arctan(0.942/1.776) ≈ 28 degrees from the radial direction.
**Summary of Results:**
* (a) Position after 1.15 seconds: ≈ 11.427 radians (or ≈ 654.7 degrees) from the +x-axis
* (b) Initial angular velocity: 0 rad/s; Final angular velocity (at 1.15 s): ≈ 0.542 rad/s
* (c) Tangential velocity at 2 s: ≈ 1.884 m/s
* (c) Tangential acceleration at 2 s: ≈ 0.942 m/s²
* (c) Total acceleration at 2 s: ≈ 2.00 m/s² at an angle of approximately 28 degrees from the radial direction.
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