Osm Physics by Ashish

Foucault’s pendulum is a long, heavy simple pendulum designed to demonstrate that the Earth rotates. Léon Foucault first publicly demonstrated it in 1851 in the Panthéon in Paris (a 28 kg bob on a ~67 m wire). In an inertial frame the plane of oscillation stays fixed (by inertia). Because the Earth rotates beneath it, an observer on Earth sees the plane of swing slowly rotate (precess). It consists of a heavy bob (mass) suspended from a long, flexible wire. The key feature is that it is free to swing in any vertical plane (isotropic suspension). As the pendulum swings, the plane of oscillation appears to rotate relative to the ground. Hence, Foucault’s pendulum provides direct experimental evidence of the rotation of the Earth. Angular Velocity of Foucault’s Pendulum If the pendulum is situated at latitude \(\lambda\), the angular velocity of rotation of its plane of oscillation relative to the Earth is \[ \boxed{\omega...

If a body of mass \(m\) is thrown vertically upward with a velocity \(u\) from a certain place of the earth. After \(t\) seconds the velocity of a body will be— \[ \boxed{ \vec{v}' = \hat{k}'(u-gt) } \qquad ...(1) \] If angle of latitude at this place is \(\lambda\) and the angular velocity of rotational reference frame is \(\vec{\omega}\), then \[ \boxed{ \vec{\omega} = \hat{j}'\omega\cos\lambda + \hat{k}'\omega\sin\lambda } \qquad ...(2) \] Coriolis force acting on the body is \[ \vec{F}_c = -2m \left( \vec{\omega}\times\vec{v}' \right) \] \[ = -2m \begin{vmatrix} \hat{i}' & \hat{j}' & \hat{k}'\\ 0 & \omega\cos\lambda & \omega\sin\lambda\\ 0 & 0 & u-gt \end{vmatrix} \] \[ \boxed{ \vec{F}_c = -\hat{i}'\,2m(u-gt)\omega\cos\lambda } \qquad ...(3) \] Thus the displacement of the body due to Coriolis force acting toward west in the northern hemisphere is toward west. From ...

Let a body of mass \(m\) is falling downward from a height \(h\) due to gravity. After \(t\) seconds its velocity will be \[ \boxed{ \vec{v}' = -\hat{k}'gt } \qquad ...(1) \] where \(\hat{k}'\) is the unit vector vertically upward. If angle of latitude at this place is \(\lambda\) and the angular velocity of rotational reference frame is \(\vec{\omega}\), then \[ \boxed{ \vec{\omega} = \hat{j}'\omega\cos\lambda + \hat{k}'\omega\sin\lambda } \qquad ...(2) \] Thus, Coriolis force acting on the body is \[ \vec{F}_c = -2m \left( \vec{\omega}\times\vec{v}' \right) \] \[ = -2m \begin{vmatrix} \hat{i}' & \hat{j}' & \hat{k}'\\ 0 & \omega\cos\lambda & \omega\sin\lambda\\ 0 & 0 & -gt \end{vmatrix} \] \[ \boxed{ \vec{F}_c = \hat{i}'\,2mgt\omega\cos\lambda } \qquad ...(3) \] Since the direction of \(\hat{i}'\) is towards east, therefore the body falling downward in the nort...

Effects of Centrifugal and Coriolis Forces on Earth Let us assume that a particle of mass \(m\) is moving horizontally on earth with a velocity \(\vec{v}'\) at the place P whose angle of latitude is \(\lambda\). Draw a reference frame S' at point P whose X'-axis is towards east, Y'-axis is towards north and Z'-axis is vertically upwards. This reference frame is rotating with an angular velocity \(\vec{\omega}\) as earth is rotating about its north-south axis. Therefore, \[ \boxed{ \vec{\omega} = \hat{j}'\omega\cos\lambda + \hat{k}'\omega\sin\lambda } \qquad ...(1) \] and the velocity of particle is \[ \boxed{ \vec{v}' = \hat{i}'v_x' + \hat{j}'v_y' } \qquad ...(2) \] Thus the Coriolis force \[ \vec{F}_c = -2m \left( \vec{\omega}\times\vec{v}' \right) \] \[ = -2m \begin{vmatrix} \hat{i}' & \hat{j}' & \hat{k}'\\ 0 & \omega\cos\lambda & \omega\sin\lambda\\ v_x' ...