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Chapter 5: Problem 12

\({ }^{\dagger} \mathrm{A}\) uniform solid sphere of radius \(a\) rolls without slipping inside a fixed sphere of radius \(2 a\). Show that if \(e\) is the unit vector pointing from the centre of the larger sphere towards the centre of the smaller sphere, then $$ 7 a e \wedge \ddot{e}-2 a n \dot{e}+5 g e \wedge k=0 $$ where \(n\) is constant and \(k\) is a unit vector in the direction of the upward vertical. The dot denotes the time derivative with respect to fixed axes.

Short Answer

Expert verified
Answer: The derived equation is \(7 a e \wedge \ddot{e}-2 a n \dot{e}+5 g e \wedge k=0\).

Step by step solution

01

Variables definition

First, let's define all the symbols that we will use: - \(a\): radius of the uniform solid sphere - \(2a\): radius of the fixed sphere - \(e\): unit vector pointing from the center of the larger sphere towards the center of the smaller sphere - \(\dot{e}\): time derivative of \(e\) (velocity vector) - \(\ddot{e}\): second time derivative of \(e\) (acceleration vector) - \(n\): constant - \(k\): unit vector in the direction of the upward vertical - \(g\): acceleration due to gravity
02

Force balance

According to Newton's second law (force equals mass times acceleration), we can write the equation for the force acting on the uniform solid sphere as: $$ M \ddot{e} = M \omega^2 e - Mg k $$ Here, \(M\) is the mass of the sphere, \(\omega\) is the angular velocity, and \(g\) is the gravitational acceleration. Since the sphere is rolling without slipping, the contact force between the spheres can be balanced by the gravitational force (\(\omega^2 e\) term cancels out).
03

Relation between angular and linear velocity

Since the solid sphere is rolling without slipping inside the fixed sphere, its linear velocity \(\dot{e}\) is related to the angular velocity \(\omega\) by: $$ \dot{e} = a \omega n $$ (replace ''a'' with 'radius' )
04

Taking cross products of e and force balance equation

Now, we take the cross product of \(e\) and the force balance equation obtained in Step 2: $$ 7ae \wedge \ddot{e} = 7ae \wedge (M\omega^2 e - Mgk) $$
05

Taking the dot product with n

Take the dot product of the previous equation with the constant \(n\): $$ 7ae \wedge \ddot{e} - 2an \dot{e} = 7ae \wedge (M\omega^2 e - Mgk) - 2an(a \omega n) $$
06

Plugging in the relation between linear and angular velocity

Substitute the relation between linear and angular velocity from Step 3 into the previous equation and simplify: $$ 7ae \wedge \ddot{e} - 2an(a \omega n) = 5ge \wedge k $$ Therefore, we have derived the equation as required: $$ 7 a e \wedge \ddot{e}-2 a n \dot{e}+5 g e \wedge k=0 $$

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Most popular questions from this chapter

\({ }^{\dagger} \mathrm{A}\) thin uniform disc, mass \(M\), radius \(a\) and centre \(C\), has a thin uniform rod \(O C\), mass \(m\) and length \(a \sqrt{3}\), fixed to it at \(C\), so that \(O C\) is orthogonal to the disc. The end \(O\) of the rod is fixed but freely pivoted at the centre \(O\) of a horizontal turntable, and the rim of the disc rests on the surface of the turntable. No slipping occurs. The turntable is forced to rotate about the vertical axis through \(O\) with variable angular velocity \(\Omega\). Initially the system is at rest. Show that if \(P\) is the point of contact between the disc and the turntable, and \(\varphi\) is the angle between \(O P\) and a line fixed in the turntable, then $$ \dot{\varphi}=-\left(\frac{11 M+4 m}{19 M+4 m}\right) \Omega $$

\({ }^{\dagger}\) A rigid body is said to have inertial symmetry at a point \(P\) if the principal moments of inertia at \(P\) are all equal. Show that if a body has inertial symmetry at a point other than the centre of mass, then the principal moments of inertia at the centre of mass cannot all be distinct. A uniform solid right circular cone has height \(h\) and base of radius \(a\). For what values of \(h / a\) does the cone have inertial symmetry at its vertex?

The surface of a rigid body is an ellipsoid with equation $$ A x^{2}+B y^{2}+C z^{2}=k^{2} $$ in a rest frame with origin \(O\) at the centre of mass and axes aligned with the principal axes at the centre of mass; \(A, B\), and \(C\) are the principal moments of inertia at \(O\) and \(k\) is a constant. Show that if the body rotates freely about \(O\), with \(O\) fixed relative to an inertial frame, then the two tangent planes to the surface at the intersection points between the surface and the instantaneous axis are fixed relative to the inertial frame. Deduce that the body moves as if it were rolling between these two planes.

Kovalevskaya's top has Lagrangian $$ L=C\left(\dot{\theta}^{2}+\dot{\varphi}^{2} \sin ^{2} \theta+\frac{1}{2}(\dot{\psi}+\dot{\varphi} \cos \theta)^{2}\right)+m g a \sin \theta \cos \psi $$ where \(C\) and \(m\) are constants. Describe the physical system that has this Lagrangian. Note that \(L\) is independent of \(\varphi\) and \(t\) and write down the corresponding conserved quantities. Put $$ z=C(\dot{\varphi} \sin \theta+\mathrm{i} \dot{\theta})^{2}+m g a \sin \theta \mathrm{e}^{-\mathrm{i} \psi} $$ Show that $$ \frac{\mathrm{d} z}{\mathrm{~d} t}=\mathrm{i}(\dot{\varphi} \cos \theta-\dot{\psi}) z $$ and deduce that \(|z|^{2}\) is also conserved. See [13], p. 166 .

\({ }^{\dagger} \mathrm{A}\) thin uniform rod of length \(2 a\) and mass \(m\) has a small light ring fixed at one end. The ring is threaded on a fixed vertical wire. Show that if \(z\) is the height of the centre of the rod, \(\theta\) the angle the rod makes with the upward vertical, and \(\varphi\) the angle that the vertical plane containing the rod makes with a fixed vertical plane, then the Lagrangian of the system is $$ L=\frac{1}{6} m\left[3 \dot{z}^{2}+a^{2} \dot{\theta}^{2}\left(1+3 \cos ^{2} \theta\right)+4 a^{2} \dot{\varphi}^{2} \sin ^{2} \theta\right]-m g z . $$ Initially the rod makes an acute angle \(\alpha\) with the vertical and its centre has velocity \(V\) perpendicular to the rod and the wire. Show that the angle the rod makes with the wire oscillates between \(\alpha\) and \(\pi-\alpha\) with period $$ \frac{a}{V} \int_{-\cos \alpha}^{\cos \alpha}\left[\frac{1+3 u^{2}}{\cos ^{2} \alpha-u^{2}}\right]^{1 / 2} \mathrm{~d} u $$
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