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The angular speed of earth around its ow...

The angular speed of earth around its own axis is

A

`(pi)/(43200) `rad/s

B

`(pi)/(3600)`rad/s

C

`(pi)/(86400)` rad/s

D

`(pi)/(1800)`rad/s

Text Solution

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The correct Answer is:
To find the angular speed of the Earth around its own axis, we can follow these steps: ### Step 1: Understand the Concept of Angular Speed Angular speed (ω) is defined as the rate of change of angular displacement with respect to time. It is usually measured in radians per second. ### Step 2: Identify the Time Period The Earth completes one full rotation around its axis in 24 hours. We need to convert this time period into seconds for our calculations. \[ \text{Time period (T)} = 24 \text{ hours} = 24 \times 60 \times 60 \text{ seconds} \] ### Step 3: Calculate the Time Period in Seconds Now, we calculate the time period in seconds: \[ T = 24 \times 60 \times 60 = 86400 \text{ seconds} \] ### Step 4: Use the Formula for Angular Speed The formula for angular speed is given by: \[ \omega = \frac{2\pi}{T} \] ### Step 5: Substitute the Time Period into the Formula Now, we substitute the value of T into the formula: \[ \omega = \frac{2\pi}{86400} \] ### Step 6: Calculate the Angular Speed Now we can perform the calculation: \[ \omega = \frac{2\pi}{86400} \approx \frac{6.2832}{86400} \approx 7.272 \times 10^{-5} \text{ radians per second} \] ### Final Answer Thus, the angular speed of the Earth around its own axis is approximately: \[ \omega \approx 7.272 \times 10^{-5} \text{ radians per second} \] ---
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Knowledge Check

  • A satellite moves estwards very near the surface of the Earth in equitorial plane with speed (v_(0)) . Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R =radius of the Earth and omega be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the earth.

    A
    `(4piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))`
    B
    `(2piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`
    C
    `(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`
    D
    `(2piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))`
  • 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
    `mg[1-((omega R - 2 upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
    B
    `mg[1- 2((omega R - upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
    C
    `mg[1- ((omega R - 2 upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
    D
    `mg[1-2 ((omega R - upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
  • What should be the angular velocity of earth about its own axis so that the weight of the body at the equator would become 3/4 th of its present value ?

    A
    `1/400 rads^(-1)`
    B
    `1/800 rad s^(-1)`
    C
    `1/1600 rad s^(-1)`
    D
    `1/3200 rad s^(-1)`
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