An insect of mass \(100 \mathrm{mg}\) is resting on the edge of a flat uniform disc of mass \(3 \mathrm{~g}\) and radius \(50 \mathrm{~mm}\), which is rotating at 60 r.p.m. about a smooth pivot. The insect crawls in towards the centre of the disc. Find the angular velocity when it reaches it, and the gain in kinetic energy. Where does this kinetic energy come from, and what happens to it when the insect crawls back out to the edge?

Short Answer

Expert verified
Answer: The gain in kinetic energy can be calculated using the formula ∆KE = KE₂ - KE₁, where KE₁ is the initial kinetic energy and KE₂ is the final kinetic energy of the system. It is important to note that this gain in kinetic energy comes from the work done by the insect as it crawls from the edge of the disc to its center. When the insect crawls back to the edge, the gained kinetic energy will be converted back into potential energy.

Step by step solution

01

Calculate initial angular momentum

Firstly, we need to calculate the initial angular momentum of the system (disc and insect) when the insect is at the edge of the disc. The initial angular velocity \(ω_1\) is given in rev/min, so let's convert it to rad/s: $$ω_1 = \frac{60}{60} × 2π$$ $$ω_1 = 2π \mathrm{~rad/s}$$ The angular momentum \(L\) is given by the formula: $$L = Iω$$ where \(I\) is the moment of inertia. For a flat uniform disc, the moment of inertia is given by: $$I_{disc} = \frac{1}{2}MR^2$$ For a point mass (insect) at the edge of the disc, the moment of inertia is given by: $$I_{insect} = mR^2$$ Therefore, the initial angular momentum is: $$L_1 = I_{disc}ω_1 + I_{insect}ω_1$$
02

Calculate final angular momentum

Now, let's calculate the angular momentum of the disc and insect when the insect reaches the center of the disc (where \(I_{insect}\) becomes zero). The moment of inertia of the disc will remain unchanged: $$I_{disc} = \frac{1}{2}MR^2$$ We know that the angular momentum is conserved, so: $$L_1 = L_2$$ $$I_{disc}ω_1 + I_{insect}ω_1 = I_{disc}ω_2$$ Where \(ω_2\) is the final angular velocity of the system. Now, let's find the angular velocity \(ω_2\): $$ω_2 = \frac{I_{disc}ω_1 + I_{insect}ω_1}{I_{disc}}$$
03

Calculate the gain in kinetic energy

To find the gain in kinetic energy, we must find the initial and final kinetic energy of the system and then subtract the initial value from the final value. The formula for kinetic energy is: $$KE = \frac{1}{2}Iω^2$$ Initial kinetic energy is: $$KE_1 = \frac{1}{2}(I_{disc} + I_{insect})ω_1^2$$ Final kinetic energy is: $$KE_2 = \frac{1}{2}I_{disc}ω_2^2$$ Gain in kinetic energy: $$∆KE = KE_2 - KE_1$$
04

Determine where the kinetic energy comes from

The gain in kinetic energy comes from the work done by the insect as it crawls from the edge of the disc to its center. When the insect crawls back to the edge, the gained kinetic energy will be converted back into potential energy. Now, you can substitute the given values (masses and radius) into the formulas above to calculate the angular velocity \(ω_2\) and the gain in kinetic energy ΔKE.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Think of the moment of inertia as an object’s resistance to change in its rotation. It is the rotational equivalent to mass in linear motion. The more an object’s mass is distributed far from the axis of rotation, the higher its moment of inertia will be, making it harder to spin or stop once it starts spinning.

For our disc and insect system, we have two different moments of inertia to consider. For the uniform disc that spins around a pivot at its center, we use the formula: \[I_{disc} = \frac{1}{2}MR^2\], where \(M\) is the mass of the disc and \(R\) is its radius. The moment of inertia of the insect, when at the edge of the disc, is calculated like a point mass at a distance \(R\) from the pivot: \[I_{insect} = mR^2\]. When the insect moves, the total moment of inertia of the system changes because the distribution of mass relative to the axis of rotation changes. This plays a crucial role in the conservation of angular momentum, which we’ll explore in depth later in the article.
Kinetic Energy in Rotational Systems
In a rotational system, kinetic energy is associated with the rotation of an object. The formula to determine kinetic energy for a rotating object is \[KE = \frac{1}{2}I\omega^2\], where \(I\) represents the moment of inertia and \(\omega\) represents the angular velocity. This tells us that the kinetic energy of an object depends not just on how fast it's spinning \(\omega\), but also on how its mass is distributed \(I\).

Initially, the disc-insect system has a certain kinetic energy based on their joint moment of inertia and the speed of rotation. When the insect moves inward, towards the axis, it affects both the system's total moment of inertia and kinetic energy. Contrary to what one might think, even though no external forces are doing work on the system, the kinetic energy changes! This is due to the energy that the insect itself expends crawling inward, which brings us to the law of conservation of energy.
Conservation of Energy
Conservation of energy is a fundamental principle that states that energy cannot be created or destroyed, only transformed from one form to another. In the case of our rotating system, this concept assures that as the insect moves, the energy within the system changes form but the total energy remains constant.

The insect crawling toward the center of the disc does work against the centripetal force, and this work gets transformed into increased kinetic energy as the system speeds up. This is where the gain in kinetic energy originates. When the insect moves back out, the process reverses: kinetic energy is converted back to potential energy, specifically the potential energy associated with the insect being farther from the axis of rotation. There’s a clear exchange between kinetic and potential energy while the overall energy within the system remains unchanged, showcasing the conservation of energy in action.

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Most popular questions from this chapter

The axis of a gyroscope is free to rotate within a smooth horizontal circle in colatitude \(\lambda\). Due to the Coriolis force, there is a couple on the gyroscope. To find the effect of this couple, use the equation for the rate of change of angular momentum in a frame rotating with the Earth (e.g., that of Fig. 5.7), \(\dot{\boldsymbol{J}}+\boldsymbol{\Omega} \wedge \boldsymbol{J}=\boldsymbol{G}\), where \(\boldsymbol{G}\) is the couple restraining the axis from leaving the horizontal plane, and \(\boldsymbol{\Omega}\) is the Earth's angular velocity. (Neglect terms of order \(\Omega^{2}\), in particular the contribution of \(\boldsymbol{\Omega}\) to \(\boldsymbol{J} .\) ) From the component along the axis, show that the angular velocity \(\omega\) about the axis is constant; from the vertical component show that the angle \(\varphi\) between the axis and east obeys the equation \(I_{1} \ddot{\varphi}-I_{3} \omega \Omega \sin \lambda \cos \varphi=0\) Show that the stable position is with the axis pointing north. Determine the period of small oscillations about this direction if the gyroscope is a flat circular disc spinning at 6000 r.p.m. in latitude \(30^{\circ} \mathrm{N}\). Explain why this system is sensitive to the horizontal component of \(\boldsymbol{\Omega}\), and describe the effect qualitatively from the point of view of an inertial observer.

A uniformly charged sphere is spinning freely with angular velocity \(\omega\) in a uniform magnetic field \(\boldsymbol{B}\). Taking the \(z\) axis in the direction of \(\boldsymbol{\omega}\), and \(\boldsymbol{B}\) in the \(x z\)-plane, write down the moment about the centre of the magnetic force on a particle at \(r\). Evaluate the total moment of the magnetic force on the sphere, and show that it is equal to \((q / 2 M) \boldsymbol{J} \wedge \boldsymbol{B}\), where \(q\) and \(M\) are the total charge and mass, respectively. Hence show that the axis will precess around the direction of the magnetic field with precessional angular velocity equal to the Larmor frequency of \(\S 5.5\). What difference would it make if the charge distribution were spherically symmetric, but non- uniform?

Calculate the principal moments of inertia of a uniform, solid cone of vertical height \(h\), and base radius \(a\) about its vertex. For what value of the ratio \(h / a\) is every axis through the vertex a principal axis? For this case, find the position of the centre of mass and the principal moments of inertia about it.

A gyroscope consisting of a uniform circular disc of mass \(100 \mathrm{~g}\) and radius \(40 \mathrm{~mm}\) is pivoted so that its centre of mass is fixed, and is spinning about its axis at 2400 r.p.m. A \(5 \mathrm{~g}\) mass is attached to the axis at a distance of \(100 \mathrm{~mm}\) from the centre. Find the angular velocity of precession of the axis.

Find the principal moments of inertia of a uniform solid cube of mass \(m\) and edge length \(2 a\) (a) with respect to the mid-point of an edge, and (b) with respect to a vertex.

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