Problem 88 - University Physics with Modern Physics

Question

A \(20.00-\) -kg lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00 -kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Step-by-Step Solution

Answer

The dart must have an initial speed of 97.15 m/s.

Step 1: Determine the condition for completing a circular loop

For the combined object (lead sphere + dart) to complete a circular loop, at the top of the loop, the gravitational force must be equal to or less than the centripetal force needed to maintain circular motion. At the top, the tension is zero, so gravitational force is the only force providing necessary centripetal force. Thus, the velocity at the top of the loop must satisfy:\[ mg = \frac{mv^2}{r} \] which simplifies to:\[ v = \sqrt{gr} \] where \( r = 3.5 \ m \) is the radius of the loop.

Step 2: Calculate the minimum velocity at the top of the loop

Using the equation found before, calculate the velocity:\[ v = \sqrt{9.8 \, \text{m/s}^2 \times 3.5 \, \text{m}} = \sqrt{34.3} = 5.86 \, \text{m/s} \]. This is the speed needed at the top of the loop for the combination.

Step 3: Apply conservation of energy from bottom to top of loop

The system's energy is conserved from the bottom of the loop to the top. At the bottom, all energy is kinetic, but at the top, some energy is potential:\[\frac{1}{2}m_2v^2 = \frac{1}{2}m_2v_{top}^2 + m_2gh \]where \( h = 2r = 7 \ m \) is the height gain. Substitute for \( v_{top} = 5.86 \, \text{m/s} \):\[\frac{1}{2}m_2v^2 = \frac{1}{2}m_2(5.86)^2 + m_2 \times 9.8 \times 7 \]where \( m_2 = 25 \text{kg} \). Solve for \( v \).

Step 4: Solve for the velocity at the bottom of the loop

Expanding: \[\frac{1}{2}(25)v^2 = \frac{1}{2}(25)(5.86)^2 + 25 \times 9.8 \times 7 \] \[\frac{1}{2}v^2 = 0.5 \times 34.3 + 171.5 \] \[\frac{1}{2}v^2 = 17.15 + 171.5 \Rightarrow v^2 = 2(188.65) = 377.3 \Rightarrow v = \sqrt{377.3} = 19.43 \, \text{m/s} \]

Step 5: Apply conservation of momentum for the collision

Before the collision, only the dart is moving. After the collision, both the dart and the lead sphere move together. Using the conservation of momentum:\[m_1u = (m_1 + m_2) v \]where \(m_1 = 5 \, \text{kg} \), \( m_2 = 20 \, \text{kg} \), and \( v = 19.43 \, \text{m/s} \) (from Step 4):\[5u = 25 \times 19.43 \] Solve for \( u \).

Step 6: Calculate the dart's initial velocity

\[ 5u = 25 \times 19.43 \implies u = \frac{25 \times 19.43}{5} \] \[ u = 97.15 \, \text{m/s} \]. This is the initial velocity needed for the dart to ensure that the combined object can complete the loop.

Key Concepts

Circular Motion

When an object moves along a circular path, it's undergoing circular motion. This type of motion is characterized by a constant change in direction and often speed. For an object to travel in a circle, it must constantly change its velocity, even if its speed remains unchanged. This is because velocity is a vector, which means both its speed and direction matter.
The continuous change in direction requires a force directed towards the center of the circle, which is called centripetal force. Circular motion can be uniform, meaning the speed is constant, or non-uniform where the speed changes as well.
In the given exercise, we deal with an object (the lead sphere + dart) that must maintain circular motion after a collision. This requires careful calculations to ensure the necessary conditions for circular motion are met, allowing the object to complete the loop without falling out of its path.

Centripetal Force

Centripetal force is the necessary force that keeps an object moving in a circular path. Without it, the object would fly off straight instead of following the curve. This force acts at a right angle to the object's velocity, pulling the object towards the center of its circular path.

• For example, when an object swings in a circle, tension in the string, gravitational pull, or friction can serve as the centripetal force.
• In our problem, at the top of the loop, the only force providing the necessary centripetal acceleration is gravity, as tension is zero at that point.

The formula for centripetal force is given by:
\[ F = rac{mv^2}{r} \]
where \( F \) is the centripetal force, \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circle.
This equation helps determine the velocity an object needs at the top of the loop to ensure it stays on its path, as seen in Step 1 and 2 of the solution.

Energy Conservation

The principle of energy conservation states that energy in a closed system remains constant, though it may change from one form to another. In physics problems like our exercise, it is crucial to apply this principle to solve for unknown variables.
In the circular loop scenario, we consider both potential and kinetic energy.

• At the bottom, all energy is kinetic, given by \( \frac{1}{2}mv^2 \).
• At the top, the system has both potential energy, \( mgh \), due to elevation change, and kinetic energy, which must be sufficient for continued circular motion.

Energy conservation helps calculate the necessary initial speed at the bottom of the loop to achieve the required speed at the top, combining both energy forms across the journey from bottom to top.

Physics Problem Solving

Physics problem solving often involves a methodical approach to breaking down complex scenarios into manageable calculations. The problem at hand involves using two main physics principles: momentum conservation and energy conservation.

• Conservation of Momentum: This principle is used to find the initial velocity needed by the dart. Before the collision, only the dart has momentum. After the collision, its momentum is shared with the sphere, requiring calculations to find the exact starting speed of the dart.
• Energy Conservation: Applied to analyze the motion after the collision, helping determine the needed speed at the bottom to achieve circular motion.

Physics problem solving requires several steps:
1. Understanding the problem and identifying the physics principles involved.
2. Applying formulas to express these principles in mathematical forms.
3. Solving the equations to find unknown variables, step by step, ensuring logical reasoning throughout.
Such strategies teach not only how to approach complex questions but also deepen the understanding of physics concepts through practical application.