Problem 54 - College Physics

Question

3\. An apparatus for launching a small boat consists of a 150.0 kg cart that rides down a set of tracks on four solid steel wheels, each with radius 20.0 \(\mathrm{cm}\) and mass 45.0 \(\mathrm{kg}\) . The tracks slope at an angle of \(7.50^{\circ}\) to the horizontal, and the boat's mass is 750.0 kg. If the boat is released from rest a distance of 16.0 \(\mathrm{m}\) from the water (measured along the slope), how fast will it be mov- ing when it reaches the water? Assume the wheels roll without slipping, and that there is no energy loss due to friction.

Step-by-Step Solution

Answer

The boat hits the water at approximately 6.38 m/s.

Step 1: Determine the total mass

First, calculate the total mass of the system which includes the cart, the boat, and the wheels. The cart is 150.0 kg, the boat is 750.0 kg, and each of the four wheels is 45.0 kg. Thus, the total mass \(M\) is given by: \[ M = 150.0 + 750.0 + 4 \times 45.0 = 1080.0 \text{ kg} \]

Step 2: Calculate the rotational inertia of the wheels

Each wheel has a rotational inertia \(I\) given by the formula for a solid disk: \[ I = \frac{1}{2} m_w r^2 \]Where \(m_w = 45.0\, \text{kg}\) is the mass of one wheel, and \(r = 0.2\, \text{m}\) is the radius. Calculate \(I\):\[ I = \frac{1}{2} \times 45.0 \times (0.2)^2 = 0.9\, \text{kg} \cdot \text{m}^2 \]

Step 3: Calculate the initial gravitational potential energy

The potential energy \(U\) at the top of the slope is given by:\[ U = M g h \]Where \(g = 9.8\, \text{m/s}^2\) and \(h = 16.0 \times \sin(7.5^\circ)\). The height \(h\) can be calculated as:\[ h = 16.0 \times \sin(7.5^\circ) = 2.09\,\text{m} \]So the potential energy is:\[ U = 1080.0 \times 9.8 \times 2.09 = 22,162.64\,\text{J} \]

Step 4: Set up conservation of energy

Since there is no energy loss and wheels roll without slipping, the mechanical energy is conserved. The total initial potential energy converts to translational kinetic energy \(K_{trans}\) and rotational kinetic energy \(K_{rot}\) when the boat reaches the water:\[ U = K_{trans} + K_{rot} \]\[ m v^2/2 + 4 \times I \left(\frac{v}{r}\right)^2/2 = 22,162.64 \]

Step 5: Solve for velocity

Translate the equations into a formula:\[ \frac{1}{2} M v^2 + \frac{4}{2} \frac{I}{r^2} v^2 = 22,162.64 \]This simplifies to:\[ \left(\frac{1}{2} \times 1080 + 4 \times \frac{0.9}{(0.2)^2}\right) v^2 = 22,162.64 \]\[ 545 v^2 = 22,162.64 \]\[ v^2 = \frac{22,162.64}{545} \approx 40.66 \]\[ v \approx \sqrt{40.66} \approx 6.38 \text{ m/s} \]

Step 6: Conclusion

The boat is moving at approximately \( 6.38\, \text{m/s} \) when it reaches the water. This calculation assumes ideal conditions without energy losses from friction or air resistance, and that the wheels roll perfectly without slipping.

Key Concepts

Conservation of Energy

The principle of conservation of energy is a fundamental concept in physics. It states that energy in a closed system remains constant over time. For our boat and cart system, this means that the energy present at the beginning (as gravitational potential energy) is transformed into kinetic energy by the time it reaches the water. There are no external forces, like friction or air resistance, to dissipate energy. This simplifies the calculation process significantly. In essence:

• Initial potential energy \( U \) is converted to kinetic energy \( K \).
• The equation to represent this is \( U = K_{trans} + K_{rot} \).
• Understanding this principle helps us solve for unknown factors, like velocity, using the initial conditions of the system.

Rotational Inertia

Rotational inertia, or moment of inertia, represents how mass is spread in an object as it rotates. It's a measure of an object's resistance to changes in its rotational motion. Each wheel of the cart contributes rotational inertia, which must be calculated separately from the translational movement of the entire system. In our problem:

• The rotational inertia \( I \) for each wheel is calculated using \( I = \frac{1}{2} m r^2 \).
• This tells us how the mass of the wheels contributes to the system's total energy when they are rolling.
• Understanding rotational inertia is crucial because it affects the total kinetic energy and thus the speed at which the boat enters the water.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. In our rolling cart situation, the system's kinetic energy comprises two components: the translational kinetic energy of the cart and the boat moving down the hill, and the rotational kinetic energy of the wheels turning as they roll.

• Translational kinetic energy \( K_{trans} \) is expressed as \( \frac{1}{2} m v^2 \).
• Rotational kinetic energy \( K_{rot} \) is \( 4 \times \frac{1}{2} I \left(\frac{v}{r}\right)^2 \), accounting for each wheel's rotation.
• Understanding both types of kinetic energy ensures complete use of the mechanical energy principle.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. For our system, it's defined by the height from which the boat is released down the slope.

• The formula \( U = m g h \) calculates this potential energy.
• In our example, it's based on the given height \( h = 16.0 \times \sin(7.5^\circ) \).
• This energy represents the starting point: the potential to do work as the boat moves downwards, converting potential to kinetic energy as it descends.