Question

The block (0.045 kg) with the bullet (0.005 kg) then moved into a circular loop in the manner of a roller coaster at the given speed of 20.0 m/s.

If the block is barely able to pass through the top of the loop in a circular motion, what is the loop radius?

Answer 1

Answer 2

To find the radius of the circular loop, we need to consider the forces acting on the block and bullet when they are at the top of the loop.

At the top of the loop, the block and bullet experience two main forces:

1. Gravitational force (weight): This force acts downwards and is given by the formula F_gravity = (m_block + m_bullet) * g, where m_block is the mass of the block, m_bullet is the mass of the bullet, and g is the acceleration due to gravity (approximately 9.8 m/s²).

2. Centripetal force: This force acts inward toward the center of the circular loop and is responsible for keeping the block and bullet moving in a circular path. It is given by the formula F_centripetal = (m_block + m_bullet) * v² / r, where v is the speed of the block and bullet and r is the radius of the circular loop.

At the top of the loop, the centripetal force must be greater than or equal to the gravitational force to ensure that the block and bullet remain in contact with the loop and do not fall off.

Therefore, we have the following inequality:

F_centripetal ≥ F_gravity

Substitute the expressions for F_centripetal and F_gravity:

(m_block + m_bullet) * v² / r ≥ (m_block + m_bullet) * g

Now, solve for the radius (r):

v² / r ≥ g

r ≤ v² / g

Given that v = 20.0 m/s and g ≈ 9.8 m/s², we can calculate the maximum radius:

r ≤ (20.0 m/s)² / 9.8 m/s² ≈ 40.82 m

Therefore, the radius of the circular loop must be less than or equal to approximately 40.82 meters to ensure that the block and bullet can pass through the top of the loop in a circular motion without falling off.