A car is moving with constant linear acceleration a along horizontal X-axis. A solid sphere of mass M and radius R is found rolling without slipping on the horizontal floor of the car in the same direction as seen from an inertial frame outside the car. The acceleration of the sphere in the inertia frame is
The Hamiltonian of a particle is $$H = \frac{{{p^2}}}{{2m}} + pq,$$ where q is generalised coordinate and p is the corresponding canonical momentum. The Lagrangian is
A rod of length L0 makes an angle θ0 with the Y-axis in its rest frame while the rest frame moves to the right along the X-axis with relativistic speed v with respect to lab frame. If $$\gamma = {\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)^{ - \frac{1}{2}}},$$ the angle in the lab frame is
A particle of mass m moves in a potential V(x) = $$\frac{1}{2}$$ mω2x2 + $$\frac{1}{2}$$ mμv2, where x is the position coordinate, v is the speed and ω, μ are constants. The canonical momentum of the particle is
A particle of mass 2 kg is moving such that at time t second. Its position in metre is given by $$\overrightarrow {\bf{r}} \left( t \right) = 5{\bf{\hat i}} - 2{t^2}{\bf{\hat j}}.$$ The angular momentum of the particle at t = 2 s about the origin in kg-m2/s, is
The Lagrangian of a particle moving in a plane under the influence of central potential is given by $$L = \frac{1}{2}m\left( {{{\dot r}^2} = {r^2}{{\dot \theta }^2}} \right) - V\left( r \right).$$ The generalised momenta corresponding to r and θ are given by
The Lagrangian of a particle of mass m moving in a plane is given by L = $$\frac{1}{2}$$ [m(vx2 + vy2)] + a(xvy - yvx) where, vx and vy are velocity components and a is constant, The canonical momenta of the particle are given by
Three particles of mass m each situated at x1(t), x2(t) and x3(t) respectively are connected by two spring constants k and unstretched lengths $$l$$. The system is free to oscillate only in one-dimension along the straight line joining all the three particles. The Lagrangian of the system is
