A particle moving in a circular path experiences different types of acceleration, particularly tangential and centripetal acceleration. To determine the angle through which the particle turns before it begins to slip, we can use the relationship between these accelerations and the overall acceleration of the particle. Let's break it down step by step.
Understanding the Components of Acceleration
When a particle moves in a circular path, it has two components of acceleration:
- Tangential Acceleration (a_t): This is the acceleration along the path of the circle, which in this case is constant at 0.6 m/s².
- Centripetal Acceleration (a_c): This is the acceleration directed towards the center of the circular path, necessary for maintaining circular motion. It depends on the speed of the particle and the radius of the circular path.
The total acceleration of the particle (a_total) is the vector sum of tangential and centripetal accelerations. The formula for total acceleration can be expressed as:
a_total = √(a_t² + a_c²)
Finding the Point of Slipping
The problem states that the particle slips when the total acceleration reaches 1 m/s². We can set up the equation:
1 = √(0.6² + a_c²)
Squaring both sides gives us:
1 = 0.36 + a_c²
From this, we can isolate the centripetal acceleration:
a_c² = 1 - 0.36 = 0.64
a_c = √0.64 = 0.8 m/s²
Relating Centripetal Acceleration to Speed
Centripetal acceleration is also defined as:
a_c = v²/r
Where v is the linear speed of the particle and r is the radius of the circular path. From the equation we derived, we have:
0.8 = v²/r
Finding the Angle of Rotation
Next, we need to find the angle through which the particle has turned before it starts to slip. We can use the formula for angular displacement (θ) in terms of tangential acceleration:
θ = (v²)/(2 * a_t)
First, we need to express the speed (v) in terms of tangential acceleration and time. The speed increases due to tangential acceleration:
v = a_t * t = 0.6 * t
Now substituting this expression for v into the angular displacement formula gives:
θ = ((0.6 * t)²)/(2 * 0.6) = (0.36 * t²)/(1.2) = 0.3 * t²
To find the time when the total acceleration is 1 m/s², we can relate it back to the tangential acceleration:
a_total = a_t = 0.6 * t
Setting this equal to the total acceleration at the point of slipping:
0.6 * t = 1
t = 1.67 seconds
Calculating Angular Displacement
Now we substitute the value of t back into the equation for θ:
θ = 0.3 * (1.67)² = 0.3 * 2.78 ≈ 0.834 radians
To convert this into degrees, we can use the conversion factor (180/π):
θ ≈ 0.834 * (180/π) ≈ 47.8 degrees
Therefore, the angle through which the particle would have turned before it starts to slip is approximately 47.8 degrees.
