A body is weighed with a spring balance in a train at rest, shown a weight `W`. When the train begins to move with a velocity `upsilon` around the equator from west to east and if the angular velocity of the train is `omega` then the weight shown by spring balance is

A train of mass m moves with a velocity upsilon on the equator from east to west. If omega is the angular speed of earth about its axis and R is the radius of the earth then the normal reaction acting on the train is

A train of mass m moves with a velocity upsilon on the equator from east to west. If omega is the angular speed of earth about its axis and R is the radius of the earth then the normal reaction acting on the train is

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Difference between Weight W_(r ) and the gravitational attraction on the body can be given as

Weight of a body depends directly upon acceleration due to gravity g . Value of g depends upon many factors. It depends upon the shape of earth, rotation earth etc. Weight of a body at a pole is more then that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda as per relation given below : g_(rot)=g-Romega^(2)cos^(2)lambda where R is radius of earth and omega is angular velocity of earth. A body of mass m weighs W_(r ) in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It observed that weight W_(m) of the same body in the moving train is different from W_(r ) . Let v_(e ) be the velocity of a point on equator with respect to axis of rotation of earth and R be the radius of the earth. Clearly the relative between earth and trainwill affect the weight of the body. Weight W_(m) of the body can be given as

Two 2 kg weight are attached to spring balance as shown in figure-2.122. What is the reading of the scale ?

A body is suspended on a spring balance in a ship sailing along the equator with a speed v . Show that the scale reading will be very close to W_(0)(1 pm (2 omega v)/(g)) , where omega is the angular speed of the earth and W_(0) is the reading of spring balance when the ship is at rest. Explain the plus or minus sign [Hint : W_(0)=mg-momega^(2)R,W=mg-m((momega+v)^(2))/(R) when the ship is moving from west to east]

A body is suspended on a spring balance in a ship sailing along the equator with a speed v . Show that the scale reading will be very close to W_(0)(1 pm (2 omega v)/(g)) , where omega is the angular speed of the earth and W_(0) is the reading of spring balance when the ship is at rest. Explain the plus or minus sign

A body is weighed by a spring balance to be 1.000 kg at the north pole. How much will it weight at the equator? Account for the earth's rotation only.

From the roof of a train, a metal ball is suspended by a string. When the train moves with uniform velocity, will the string remain vertical?