A small object slides without friction from height H=50cm and then loops the vertical loop of radius R=20cm from which a symmetrical section of angle 2A has been removed. find the angle A and flying through the air, the object will reach point B?

To tackle this problem, we need to analyze the motion of the object as it slides down from a height and then enters a vertical loop. The key concepts involved here are energy conservation and projectile motion. Let's break it down step by step.

Understanding the Energy Conservation

Initially, the object has gravitational potential energy when it is at height H. As it slides down, this potential energy converts into kinetic energy. The potential energy (PE) at height H is given by:

• PE = mgh

Where:

• m = mass of the object
• g = acceleration due to gravity (approximately 9.81 m/s²)
• h = height (0.5 m in this case)

At the bottom of the slide, all this potential energy will have converted into kinetic energy (KE), which is expressed as:

• KE = (1/2)mv²

Calculating the Speed at the Bottom

Setting the potential energy equal to the kinetic energy gives us:

• mgh = (1/2)mv²

We can cancel the mass (m) from both sides (assuming it’s not zero), leading to:

• gh = (1/2)v²

Now, substituting h = 0.5 m and g = 9.81 m/s²:

• 9.81 * 0.5 = (1/2)v²
• 4.905 = (1/2)v²
• v² = 9.81
• v = √9.81 ≈ 3.13 m/s

Analyzing the Vertical Loop

Next, we need to consider the vertical loop with radius R = 20 cm (0.2 m). The object will experience centripetal acceleration when it is in the loop. For the object to successfully navigate the loop, it must maintain a minimum speed at the top of the loop to counteract gravitational force. The minimum speed (v_min) at the top of the loop can be derived from the centripetal force requirement:

• mg = (mv²)/R

From this, we can derive:

• v² = gR

Substituting g = 9.81 m/s² and R = 0.2 m:

• v² = 9.81 * 0.2 = 1.962
• v = √1.962 ≈ 1.4 m/s

Since the speed at the bottom (3.13 m/s) is greater than the minimum speed required (1.4 m/s), the object can complete the loop.

Finding the Angle A

Now, we need to determine the angle A for the symmetrical section that has been removed. The object will exit the loop at a certain angle, which we can find using the conservation of energy again. As the object moves through the loop, it will lose some height and thus some potential energy. The height at which the object exits the loop can be calculated based on the geometry of the loop and the angle A.

Let’s denote the height at which the object exits the loop as h_exit. The height at the top of the loop is R (0.2 m), and the object will exit at an angle A, which means it will lose height based on the cosine of that angle:

• h_exit = R(1 - cos(A))

Using energy conservation again, the potential energy lost equals the kinetic energy gained:

• mgh_exit = (1/2)mv²

Substituting for h_exit:

• mgR(1 - cos(A)) = (1/2)mv²

Canceling m and rearranging gives:

• gR(1 - cos(A)) = (1/2)v²

Substituting the values we have:

• 9.81 * 0.2(1 - cos(A)) = (1/2)(3.13)²

Calculating the right side:

• 9.81 * 0.2(1 - cos(A)) = 4.905

Now, solving for cos(A):

• 1 - cos(A) = 4.905 / (9.81 * 0.2)
• 1 - cos(A) = 4.905 / 1.962
• 1 - cos(A) ≈ 2.5

This indicates that A is not physically possible since the cosine function cannot exceed 1. Therefore, we need to reassess the conditions under which the object exits the loop. The object will exit the loop at the angle where the centripetal force equals gravitational force, which typically occurs at the top of the loop.

Determining the Exit Point B

As the object exits the loop, it will follow a projectile motion path. The exit speed can be calculated using the speed at the top of the loop, which is the minimum speed required to maintain the circular motion. The angle of projection will determine how far it travels horizontally before hitting the ground.

Using the angle of exit and the speed, we can apply projectile motion equations to find the horizontal distance to point B. The horizontal distance can be calculated using:

• Range = (v * cos(θ) * t)

Where θ is the angle of exit and t is the time of flight, which can be calculated based on the vertical drop from the height of the loop to the ground.

In summary, the angle A is determined by the conditions of motion and energy conservation, and the object will reach point B based on its exit speed and angle. To find the exact numerical value for A and the distance to B, further calculations involving the specific geometry of the loop and the trajectory would be necessary.