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Chapter 9: Problem 15

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.

Short Answer

Expert verified
Answer: The angular velocity of precession of the axis is \(96\pi\) radians per second.

Step by step solution

01

(Step 1: Calculate the moment of inertia of the disc)

To find the moment of inertia of the disc, we can use the formula \(I_{disc} = 0.5 m_{disc} r_{disc}^2\), where \(m_{disc}\) is the mass of the disc and \(r_{disc}\) is the radius of the disc. Given \(m_{disc} = 100\mathrm{~g} = 0.1\mathrm{~kg}\) and \(r_{disc} = 40\mathrm{~mm} = 0.04\mathrm{~m}\), we have: \(I_{disc} = 0.5 \times 0.1\mathrm{~kg} \times (0.04\mathrm{~m})^2 = 8 \times 10^{-5} \mathrm{~kg\cdot m^2}\).
02

(Step 2: Calculate the moment of inertia of the mass)

To find the moment of inertia of the mass, we can use the formula \(I_{mass} = m_{mass} r_{mass}^2\), where \(m_{mass}\) is the mass of the attached mass and \(r_{mass}\) is the distance of the mass from the center. Given \(m_{mass} = 5\mathrm{~g} = 0.005\mathrm{~kg}\) and \(r_{mass} = 100\mathrm{~mm} = 0.1\mathrm{~m}\), we have: \(I_{mass} = 0.005\mathrm{~kg} \times (0.1\mathrm{~m})^2 = 5 \times 10^{-5} \mathrm{~kg\cdot m^2}\).
03

(Step 3: Find the angular velocity of the spinning disc)

The given angular velocity of the spinning disc is 2400 revolutions per minute (r.p.m.). To convert this to radians per second, we use the formula \(\omega_{disc} = 2\pi(\text{r.p.m.})/60\). Given 2400 r.p.m., we have: \(\omega_{disc} = 2\pi(2400\ \text{r.p.m.})/60 = 160\pi\ \mathrm{rad/s}\).
04

(Step 4: Apply the conservation of angular momentum)

Conservation of angular momentum states that the total angular momentum before attaching the mass should equal the total angular momentum after attaching the mass. Denote the angular velocity of precession by \(\omega_{prec}\). Then, the angular momentum of the spinning disc before attaching the mass is \(L_{before} = I_{disc}\omega_{disc}\). After attaching the mass, the total moment of inertia is \(I_{total} = I_{disc} + I_{mass}\), and the angular momentum is \(L_{after} = I_{total}(\omega_{disc} - \omega_{prec})\). Thus, we have the equation: \(I_{disc}\omega_{disc} = (I_{disc} + I_{mass})(\omega_{disc} - \omega_{prec})\).
05

(Step 5: Solve for the angular velocity of precession)

Now, we can solve the equation from Step 4 for the angular velocity of precession \(\omega_{prec}\): \(8 \times 10^{-5}\mathrm{~kg\cdot m^2} \times 160\pi\mathrm{~rad/s} = (8 \times 10^{-5}\mathrm{~kg\cdot m^2} + 5 \times 10^{-5}\mathrm{~kg\cdot m^2})(160\pi\mathrm{~rad/s} - \omega_{prec})\). Solving this equation, we get: \(\omega_{prec} = 96\pi\ \mathrm{rad/s}\). Thus, the angular velocity of precession of the axis is \(96\pi\) radians per second.

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Most popular questions from this chapter

A pendulum consists of a light rigid rod of length \(250 \mathrm{~mm}\), with two identical uniform solid spheres of radius \(50 \mathrm{~mm}\) attached one on either side of its lower end. Find the period of small oscillations (a) perpendicular to the line of centres, and (b) along it.

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.

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.

A solid rectangular box, of dimensions \(100 \mathrm{~mm} \times 60 \mathrm{~mm} \times 20 \mathrm{~mm}\), is spinning freely with angular velocity 240 r.p.m. Determine the frequency of small oscillations of the axis, if the axis of rotation is (a) the longest, and (b) the shortest, axis.

Find the principal moments of inertia of a flat rectangular plate of mass \(30 \mathrm{~g}\) and dimensions \(80 \mathrm{~mm} \times 60 \mathrm{~mm}\). Given that the plate is rotating about a diagonal with angular velocity \(15 \mathrm{rad} \mathrm{s}^{-1}\), find the components of the angular momentum parallel to the edges. Given that the axis is of total length \(120 \mathrm{~mm}\), and is held vertical by bearings at its ends, find the horizontal component of the force on each bearing.
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