Advertisements
Advertisements
Question
A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west. When the ship is stationary with respect to water the tension in the string is T0. (a) Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T0 and the earth's attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string? Angular speed of earth's rotation is ω and radius of the earth is R.
Advertisements
Solution
(a) Speed of the ship due to rotation of the Earth is v = ωR, where R is the radius of the Earth and ω is its angular speed.
(b) The tension in the string is given by
T0 = mg − mω2R
∴ T0 − mg = mω2R
(c) Let the ship move with a speed v.
Then the tension in the string is given by
\[T = mg - m \omega_1^2 R\]
\[ = T_0 - \frac{\left( v - \omega R \right)^2}{R^2}R\]
\[ = T_0 - \frac{\left( v^2 + \omega^2 R^2 - 2\omega Rv \right)}{R}R\]
\[ \therefore T = T_0 + 2 \ \omega { vm }\]
APPEARS IN
RELATED QUESTIONS
Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?
A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth? If yes, where will it land? If no, how can an astronaut make it land on the earth?
Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth's surface. Will any satellite moving at this height be a geostationary satellite? Will any satellite moving at this height have a time period of 24 hours?
A spacecraft consumes more fuel in going from the earth to the moon than it takes for a return trip. Comment on this statement.
The time period of an earth-satellite in circular orbit is independent of
State the conditions for various possible orbits of satellite depending upon the horizontal/tangential speed of projection.
Answer the following question in detail.
Why an astronaut in an orbiting satellite has a feeling of weightlessness?
Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.
Given: G = 6.67 × 10-11 Nm2/kg2
R = 6400 km, M = 6 × 1024 kg
Solve the following problem.
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface.
(ME = 5.98 × 1024 kg, R = 6.4 × 106 m)
The kinetic energy of a revolving satellite (mass m) at a height equal to thrice the radius of the earth (R) is ______.
There is no atmosphere on moon because ____________.
Reason of weightlessness in a satellite is ____________.
Assuming that the earth is revolving around the sun in circular orbit of radius 'R', the angular momentum is directly proportional to rn. The value of 'n' is ______.
Two satellites of masses m and 4m orbit the earth in circular orbits of radii 8r and r respectively. The ratio of their orbital speeds is ____________.
The ratio of energy required to raise a satellite to a height `(2R)/3` above earth's surface to that required to put it into the orbit at the same height is ______.
R = radius of the earth
A satellite is revolving in a circular orbit around the earth has total energy 'E'. Its potential energy in that orbit is ______.
A geostationary satellite is orbiting the earth at a height 6R above the surface of the earth, where R is the radius of the earth. This time period of another satellite at a height (2.5 R) from the surface of the earth is ______.
Show the nature of the following graph for a satellite orbiting the earth.
- KE vs orbital radius R
- PE vs orbital radius R
- TE vs orbital radius R.
A satellite revolves around a planet very close to its surface. By what maximum factor can its kinetic energy be increased suddenly, such that it revolves in orbit in the same way?
