Day: October 26, 2022

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202210261208 Dynamics Figures (Elementary) Q8

215. A cylindrical tube with a radius r is connected by means of spokes to two hoops with a radius R. The mass of both the hoops is M. The mass of the tube and the spokes in comparison with the mass M can be neglected. A string passed over a weightless pulley is wound around the tube. A weight with a mass m is attached to the end of the string.

Find the acceleration a=|\mathbf{a}| of the weight, the tension T=|\mathbf{T}| of the string and the force of friction f=|\mathbf{f}| acting between the hoops and the surface. (Assume that the hoops do not slip.) Also determine k the coefficient of friction at which the hoops will slip.

Extracted from B. Bukhovtsev et al. (1978). Problems In Elementary Physics.

Setup.

The kinetic energy T of the dumbbell M is in two parts, translational (/linear) and rotational (/angular), i.e.,

\begin{aligned} \textrm{KE}_{\textrm{translational}} & = \frac{1}{2}Mv_M^2 \\ \textrm{KE}_{\textrm{rotational}} & = \frac{1}{2}I\omega^2 \\ \end{aligned}

where the velocity of its centre of gravity (CG) is v_M, and its moment of inertia for discs I=MR^2. It has no potential energy V of its position h=0.

The kinetic energy T of the weight m is given by \textrm{KE}=\frac{1}{2}mv_m^2, and its potential energy V by \textrm{PE}=mgh.

The Lagrangian \mathcal{L} of a system is obtained from

\mathcal{L}=T-V,

whereby the Euler–Lagrange equation is given as

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{x}}\bigg)-\frac{\partial\mathcal{L}}{\partial x}=0}

Note the relationship between r, R, v_M, v_m, and \omega, i.e.,

\begin{aligned} v_m & = r\omega \\ v_M & = R\omega \\ \end{aligned}

Also beware that forces F of any kinds, e.g., weight \mathbf{W}=m\mathbf{g}, tension \mathbf{T}, normal reaction \mathbf{N}, and friction \mathbf{f}, have not been taken into consideration.

(to be continued)