To determine the radius of curvature of a particle fired at an angle, we first need to analyze its motion. The particle is launched with an initial speed of 40 m/s at an angle of 53° to the horizontal. We want to find the radius of curvature at the moment when the particle's velocity becomes perpendicular to its initial velocity.
Understanding the Motion
The initial velocity can be broken down into its horizontal and vertical components using trigonometric functions:
- Horizontal component, \( u_x = u \cdot \cos(\theta) = 40 \cdot \cos(53°) \)
- Vertical component, \( u_y = u \cdot \sin(\theta) = 40 \cdot \sin(53°) \)
Calculating Components
Using the cosine and sine values for 53°:
- \( \cos(53°) \approx 0.6018 \)
- \( \sin(53°) \approx 0.7986 \)
Now, substituting these values:
- Horizontal component: \( u_x \approx 40 \cdot 0.6018 \approx 24.072 \, \text{m/s} \)
- Vertical component: \( u_y \approx 40 \cdot 0.7986 \approx 31.944 \, \text{m/s} \)
Finding Velocity Perpendicular to Initial Velocity
The velocity of the particle becomes perpendicular to the initial velocity when the angle between them is 90°. At this point, we can use the concept of centripetal acceleration to find the radius of curvature.
The centripetal acceleration \( a_c \) is given by the formula:
Radius of curvature (R) = v² / a_c
Where \( v \) is the speed of the particle at that instant, and \( a_c \) is the centripetal acceleration. The centripetal acceleration can be derived from the vertical motion, which is influenced by gravity.
Velocity at Perpendicular Condition
At the moment the velocity is perpendicular, the vertical component of the velocity will be equal to the horizontal component of the initial velocity:
Let \( v \) be the speed at that moment. Thus, \( v_y = u_x \) or \( v_y = 24.072 \, \text{m/s} \).
Using the vertical motion equation, we can find \( v \) using the following relationship:
v_y = u_y - g \cdot t
Where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Rearranging gives us:
t = (u_y - v_y) / g
Substituting the values:
t = (31.944 - 24.072) / 9.81 \approx 0.79 \, \text{s}
Calculating Radius of Curvature
Now we can find the speed \( v \) at this time:
v = u_y - g \cdot t = 31.944 - 9.81 \cdot 0.79 \approx 24.072 \, \text{m/s}
Now substituting \( v \) into the radius of curvature formula:
R = v² / a_c
Where \( a_c = g \) (since the only force acting in the vertical direction is gravity):
R = (24.072)² / 9.81 \approx 56.25 \, \text{m}
Final Result
Thus, the radius of curvature of the particle at the instant its velocity becomes perpendicular to the initial velocity is approximately 56.25 m. Therefore, the correct answer is option (1) 56.25 m.
