GSO001 Problem 19

This problem is solved by formulating the Lagrangian in analytical mechanics and analyzing small oscillations via a Taylor expansion of the effective potential.

The Cartesian coordinates $(x, y, z)$(x,y,z) of the mass point are expressed using the central angle $\theta$θ and time $t$t as:

$$\begin{aligned} x &= R \sin\theta \cos(\omega t) \\ y &= R \sin\theta \sin(\omega t) \\ z &= -R \cos\theta \end{aligned}$$xyz​=Rsinθcos(ωt)=Rsinθsin(ωt)=−Rcosθ​

Differentiating with respect to $t$t to obtain velocity components and computing $v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2$v2=x˙2+y˙​2+z˙2:

$$v^2 = R^2 \dot{\theta}^2 + R^2 \omega^2 \sin^2\theta$$v2=R2θ˙2+R2ω2sin2θ

The kinetic energy $T$T and potential energy $U$U (measured from the origin, not from the lowest point $z = -R$z=−R) are:

$$\begin{aligned} T &= \frac{1}{2} m v^2 = \frac{1}{2} m R^2 \dot{\theta}^2 + \frac{1}{2} m R^2 \omega^2 \sin^2\theta \\ U &= m g z = -m g R \cos\theta \end{aligned}$$TU​=21​mv2=21​mR2θ˙2+21​mR2ω2sin2θ=mgz=−mgRcosθ​

The Lagrangian $L = T - U$L=T−U is:

$$L = \frac{1}{2} m R^2 \dot{\theta}^2 + \frac{1}{2} m R^2 \omega^2 \sin^2\theta + m g R \cos\theta$$L=21​mR2θ˙2+21​mR2ω2sin2θ+mgRcosθ

Applying the Euler-Lagrange equation $\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0$dtd​(∂θ˙∂L​)−∂θ∂L​=0:

$$m R^2 \ddot{\theta} - \left( m R^2 \omega^2 \sin\theta \cos\theta - m g R \sin\theta \right) = 0$$mR2θ¨−(mR2ω2sinθcosθ−mgRsinθ)=0 $$\ddot{\theta} = \sin\theta \left( \omega^2 \cos\theta - \frac{g}{R} \right)$$θ¨=sinθ(ω2cosθ−Rg​)

At the equilibrium position $\theta_0$θ0​, we have $\ddot{\theta} = 0$θ¨=0, so either $\sin\theta_0 = 0$sinθ0​=0 or $\cos\theta_0 = \frac{g}{R\omega^2}$cosθ0​=Rω2g​.

Since $\omega > \sqrt{g/R}$ω>g/R​ implies $\frac{g}{R\omega^2} < 1$Rω2g​<1, the stable equilibrium position in the range $0 < \theta_0 < \pi/2$0<θ0​<π/2 is given by:

$$\cos\theta_0 = \frac{g}{R\omega^2}$$cosθ0​=Rω2g​

Next, setting $\theta = \theta_0 + \phi$θ=θ0​+ϕ (where $\phi$ϕ is small) and linearizing the equation of motion, the angular frequency $\Omega$Ω of small oscillations is obtained from:

$$\Omega^2 = \frac{V_{\text{eff}}''(\theta_0)}{mR^2}$$Ω2=mR2Veff′′​(θ0​)​

Here, the effective potential $V_{\text{eff}}(\theta)$Veff​(θ) corresponds to the negative of the $\theta$θ-dependent part of the Lagrangian:

$$V_{\text{eff}}(\theta) = - \frac{1}{2} m R^2 \omega^2 \sin^2\theta - m g R \cos\theta$$Veff​(θ)=−21​mR2ω2sin2θ−mgRcosθ

Computing the derivatives:

$$V_{\text{eff}}'(\theta) = - m R^2 \omega^2 \sin\theta \cos\theta + m g R \sin\theta$$Veff′​(θ)=−mR2ω2sinθcosθ+mgRsinθ $$V_{\text{eff}}''(\theta) = - m R^2 \omega^2 (\cos^2\theta - \sin^2\theta) + m g R \cos\theta$$Veff′′​(θ)=−mR2ω2(cos2θ−sin2θ)+mgRcosθ

Substituting $\theta = \theta_0$θ=θ0​ and using the relation $m g R = m R^2 \omega^2 \cos\theta_0$mgR=mR2ω2cosθ0​:

$$\begin{aligned} V_{\text{eff}}''(\theta_0) &= - m R^2 \omega^2 (\cos^2\theta_0 - \sin^2\theta_0) + (m R^2 \omega^2 \cos\theta_0) \cos\theta_0 \\ &= m R^2 \omega^2 \sin^2\theta_0 \end{aligned}$$Veff′′​(θ0​)​=−mR2ω2(cos2θ0​−sin2θ0​)+(mR2ω2cosθ0​)cosθ0​=mR2ω2sin2θ0​​

Therefore:

$$\Omega^2 = \omega^2 \sin^2\theta_0 = \omega^2 (1 - \cos^2\theta_0) = \omega^2 - \frac{g^2}{R^2 \omega^2}$$Ω2=ω2sin2θ0​=ω2(1−cos2θ0​)=ω2−R2ω2g2​

Substituting the given numerical values:

• $R \omega^2 = 0.50 \times 7.00^2 = 0.50 \times 49 = 24.5 \text{ m/s}^2$Rω2=0.50×7.002=0.50×49=24.5 m/s2
• $\cos\theta_0 = \frac{g}{R \omega^2} = \frac{9.80}{24.5} = 0.4$cosθ0​=Rω2g​=24.59.80​=0.4
• $\Omega^2 = 49 \times (1 - 0.4^2) = 49 \times (1 - 0.16) = 49 \times 0.84 = 41.16 \text{ rad}^2/\text{s}^2$Ω2=49×(1−0.42)=49×(1−0.16)=49×0.84=41.16 rad2/s2

The required answer is $\Omega^2$Ω2 multiplied by $100$100, giving $4116$4116.

Answer: 4116