Question

An electromagnetic doorbell has been constructed by wrapping 70 turns of wire around a long, thin rod, as shown in the figure. The rod has a mass of \(30.0 \mathrm{~g}\), a length of \(8.00 \mathrm{~cm},\) and a cross-sectional area of \(0.200 \mathrm{~cm}^{2}\). The rod is free to pivot about an axis through its center, which is also the center of the coil. Initially, the rod makes an angle of \(\theta=25.0^{\circ}\) with the horizontal. When \(\theta=0.00^{\circ}\), the rod strikes a bell. A uniform magnetic field of 900. G is directed at an angle \(\theta=0.00^{\circ}\). a) If a current of \(2.00 \mathrm{~A}\) is flowing in the coil, what is the torque on the rod when \(\theta=25.0^{\circ} ?\) b) What is the angular velocity of the rod when it strikes the bell?

Short answer

Answer: The torque on the rod when it is at an angle of 25 degrees is approximately 0.0516 N⋅m, and its angular velocity when it strikes the bell is approximately 58.4 rad/s.

Step-by-step solution

Determine the area of the coil

We are given the cross-sectional area (0.200 cm^2) and length (8.00 cm) of the rod. We can treat this coil as a rectangular loop, so the area of the coil is just the cross-sectional area of the rod. Therefore, \(A = 0.200 \,\mathrm{cm^2}\). Note that we should convert this area to SI units (square meters). To do this, we have: $$A = 0.200 \times (10^{-2})^2 \,\mathrm{m^2} = 0.000200 \,\mathrm{m^2}.$$

Calculate the torque

We can now determine the torque on the rod by using the formula for the torque on a coil in a magnetic field: $$\tau = NIAB \sin \theta.$$ We are given the number of turns (\(N = 70\)), current (\(I = 2.00 \,\mathrm{A}\)), and magnetic field strength (\(B = 900. \,\mathrm{G} = 900. \times 10^{-4} \,\mathrm{T}\)). With the area \(A\) and angle \(\theta = 25.0^{\circ}\), we can substitute these values into the equation and calculate the torque: $$\tau = (70)(2.00\,\mathrm{A})(0.000200\,\mathrm{m^2})(900.\times 10^{-4}\,\mathrm{T}) \sin 25.0^{\circ}.$$ $$\tau \approx 0.0516 \,\mathrm{N\cdot m}.$$ b) Find the angular velocity of the rod when it strikes the bell

Determine the moment of inertia of the rod

The moment of inertia of a rod pivoting about its center can be determined using the formula: $$I = \frac{1}{12}mL^2,$$ where \(m\) is the mass of the rod (in kg) and \(L\) is its length (in meters). We are given the mass as \(30.0\,\mathrm{g}\), which we should convert to kilograms: \(m = 0.030\,\mathrm{kg}\). Also, the length is given as \(8.00\,\mathrm{cm}\), so we convert it to meters: \(L = 0.080\,\mathrm{m}\). Now we can calculate the moment of inertia: $$I = \frac{1}{12}(0.030\,\mathrm{kg})(0.080\,\mathrm{m})^2 = 1.6 \times 10^{-5} \,\mathrm{kg\cdot m^2}.$$

Apply the conservation of mechanical energy

When the rod is at an initial angle of \(25.0^{\circ}\) and reaches \(\theta=0^{\circ}\), mechanical energy is conserved in the system: $$\frac{1}{2}I\omega^2 = \tau \,\Delta \theta,$$ where \(\omega\) is the unknown angular velocity and \(\Delta \theta = 25.0^{\circ}\). We want to find the angular velocity \(\omega\), so we can rearrange the formula: $$\omega = \sqrt{\frac{2\tau \Delta \theta}{I}}.$$ Now we can substitute the known values: $$\omega = \sqrt{\frac{2(0.0516\,\mathrm{N\cdot m})(25.0^{\circ} \times \frac{\pi}{180}\,\mathrm{rad})}{1.6\times10^{-5}\,\mathrm{kg\cdot m^2}}}.$$

Calculate the angular velocity

Computing the expression, we get: $$\omega \approx 58.4 \,\mathrm{rad/s}.$$ So, the angular velocity of the rod when it strikes the bell is approximately \(58.4 \,\mathrm{rad/s}\).

Key concepts

Torque Calculation

Understanding the concept of torque is vital when analyzing the rotational motion of objects, such as the rod in an electromagnetic doorbell system. Torque can be envisioned as the rotational equivalent of force; it's a measure of how much a force acting on an object causes that object to rotate around an axis.

The formula \(\tau = NIAB \sin \theta\) provides the torque experienced by a coil in a magnetic field, where \(\tau\) is the torque, \(N\) the number of turns in the coil, \(I\) the current running through the coil, \(A\) the area of the coil, \(B\) the magnetic field strength, and \(\theta\) the angle between the plane of the coil and the magnetic field.

The key aspect to note here is the sine function, which indicates how the angle of the coil relative to the field affects the torque. When the angle is zero, sin(0) is zero, and there's no torque. Conversely, maximum torque occurs at \(\theta = 90^\circ\), where sin(90) is one. The problem provides all the necessary values to plug into the equation, leading to the successful computation of the torque.

Moment of Inertia

The moment of inertia is a property of any object which can be rotated; it quantifies how much torque is needed for a given angular acceleration around a pivot point. It depends on the mass of the object, the shape of the object, and the distribution of mass relative to the axis of rotation.

For a rod of uniform density pivoting around its center, like the one in our doorbell, the moment of inertia \(I\) is given by \(I = \frac{1}{12}mL^2\), where \(m\) is the mass and \(L\) is the length of the rod.

This formula highlights the importance of geometry and mass distribution in rotational dynamics. Heavier rods, or those with mass distributed further away from the pivot, have higher moments of inertia, requiring more torque to achieve the same angular acceleration.

Angular Velocity

The concept of angular velocity is another cornerstone of rotational dynamics, representing the rate of change of the angular displacement over time, or how fast an object spins around an axis. The angular velocity, denoted by \(\omega\), is crucial for describing the motion of the rod in our problem as it strikes the bell.

It can be related to torque and moment of inertia through the work-energy principle or, in this case, the conservation of mechanical energy. This relationship allows for the calculation of the angular velocity of the rod just before impact, assuming no other external forces do work on the system.

Magnetic Field

The magnetic field (\(B\)) is a fundamental concept in electromagnetism, representing the area where magnetic forces are exerted. It's significant in the context of an electromagnetic doorbell because it directly affects the torque on the current-carrying rod. A uniform magnetic field implies a constant force on a moving charge, enabling straightforward calculations of electromagnetic effects.

In this scenario, the magnetic field's strength and direction play a critical role. The magnetic field vector's alignment with the coil’s plane affects the magnitude of torque, as seen in the sinusoidal relationship in the torque equation.

Conservation of Mechanical Energy

Finally, the principle of conservation of mechanical energy states that in an isolated system free from non-conservative forces (like friction), the total mechanical energy (the sum of kinetic and potential energies) remains constant over time.

Applying this principle to our doorbell rod, we can state that the gravitational potential energy lost by the rod as it falls to the horizontal is converted into rotational kinetic energy. This conversion allows us to link the change in orientation (\(\Delta \theta\)) to the final angular velocity (\(\omega\)) of the rod, assuming energy is conserved. The computation of the angular velocity leverages this conservation, yielding the final solution to the problem.