Problem 12 - Fundamentals of Physics

Question

A solid brass ball of mass 0.280 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius \(R=14.0 \mathrm{~cm},\) and the ball has radius \(r

Step-by-Step Solution

Answer

The height for leaving the track is 35 cm; force direction at Q concerns horizontal balance.

Step 1: Determine the height for looping

To find the minimum height \(h\) from which the ball must be released to be on the verge of leaving the track at the top of the loop, we use energy conservation. The gravitational potential energy at height \(h\) converts into kinetic energy and potential energy at the top of the loop. At the top of the loop, the critical condition to just stay in contact requires the centripetal force to equal the gravitational force. Therefore,\[ mg(h - 2R) = \frac{1}{2}mv^2 + mg(2R), \]where \(v^2 = gR\) for this condition. Solving gives:\[ gh = \frac{5}{2}gR, \]so \( h = \frac{5}{2}R.\)Hence, for the minimal case, \(h = 35\, \mathrm{cm}.\)

Step 2: Solve for actual height release (h=6R)

The ball is actually released from a height of \(h = 6R\). At height \(6R\), its potential energy is \(mg(6R)\). At point \(Q\), the energies are redistributed:\[mg(6R) = \frac{1}{2}mv^2 + mg(2R),\]where 6R is the total height and 2R is the top of the loop's height.

Step 3: Calculate the velocity at the top of the loop

Rearranging to solve for the velocity at the top of the loop, we have:\[ mg(6R - 2R) = \frac{1}{2}mv^2 \]\[ 4gR = \frac{1}{2}v^2 \]Solving for \(v\), we find:\[ v^2 = 8gR. \]

Step 4: Calculate force component at point Q

At the top of the loop, the ball maintains contact maintaining minimum velocity needed for centripetal force balance: \(mg = \frac{mv^2}{R}\). At the point \(Q\), assume horizontal motion without external vertical component, consider force by track providing centripetal acceleration compensating for insufficient gravitational component:\[ F = \,?\]

Step 5: Calculate normal force at the top of the loop

The normal force at the top of the loop (point Q), when the ball is released from \(6R\) is higher than equilibrium, and it will have been subjected to greater centrifugal force.Solve: normal force, horizontal component was central to solving, but formula needed modifying in intuition assessment intimating gravity influences continue reduced by force override balancing scope: finding forces momentum fluxes anticipated.

Key Concepts

Centripetal Force

When an object moves in a circular path, it constantly changes direction. This change implies the object is undergoing acceleration, which requires a force. The force responsible for this is the centripetal force. In the context of the brass ball looping the track, the centripetal force is what keeps it moving in a circle. To stay on the track, particularly at the top of the loop, the centripetal force must equal the gravitational force pulling the ball downward. Formulaically, this force is given by

• \( F_c = \frac{mv^2}{R} \), where
  • \( F_c \) is the centripetal force,
  • \( m \) is the mass of the ball,
  • \( v \) is the velocity of the ball, and
  • \( R \) is the radius of the loop.
  When balancing forces, at the critical point where the ball may leave the track, the necessary centripetal force must match the gravitational force:
• \( mg = \frac{mv^2}{R} \)

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed in an isolated system. It can only be transferred from one form to another. For the brass ball, energy transitions between potential energy, due to height, and kinetic energy, due to motion.Starting from rest at a height \( h \), the ball possesses gravitational potential energy, \( U = mgh \). As it descends, potential energy converts to kinetic, % and the sum remains constant, as indicated by the equation below:

• \( mg(h - 2R) = \frac{1}{2}mv^2 + mg(2R) \)

At the top, kinetic and potential energies balance to maintain the necessary speed ensuring centripetal balance. By applying conservation of energy from a higher release at \( 6R \), more energy transitions into kinetic, affecting velocities along the loop.

Gravitational Potential Energy

Gravitational potential energy (\( U_g \) ) is the energy an object possesses due to its position in a gravitational field. For the ball, this is entirely determined by its height above the base level of the loop.The potential energy at a height \( h \) is given by:

• \( U_g = mgh \)

At the start from height \( 6R \), it's the initial reserve energy. As the ball moves downwards and reaches height \( 2R \) (top looping point), gravitational potential energy is used up in converting to kinetic energy and maintaining necessary motion through the loop.At the critical point where the ball is about to potentially lose contact with the track, the potential energy complement at \( 2R \) will ensure minimal velocity that aligns with centripetal force requirements.

Circular Motion

Objects moving in a circle have to continuously change direction, which involves acceleration called centripetal acceleration. This motion is evident in the circular path of the brass ball. The velocity of the ball is always tangent to the circle, while acceleration, hence force required, is directed towards the center of the circle. This force manifests itself as the centripetal force explained as:

• \( a_c = \frac{v^2}{R} \)

Here, it is essential not just for the ball to move in a circle but also to address how circular motion affects forces on the ball.As the ball travels through the loop, velocities dictated through energy conservation determine the centripetal force, keeping it moving smoothly or risking it flying off if too slow. This interplay explains why particular speeds at certain points need maintaining for consistent track contact.

Kinetic Energy

Kinetic energy (\( KE \) ) is the energy of motion. When the brass ball moves down from its height, its initial gravitational potential energy transforms into kinetic energy, making the ball accelerate along the loop.The relationship between height change and kinetic energy can be expressed as:

• \( KE = \frac{1}{2}mv^2 \)

At the loop's bottom, the ball possesses maximum kinetic energy due to higher velocity. As it climbs back up, some kinetic energy transforms back into potential energy. Maintaining minimum kinetic energy at the loop's top ensures continuous motion needed for centripetal force and equilibrium to prevent it from losing track contact—an essential balance taught by conservation principles disallowing total kinetic loss as potential energy climbs.