How Do Changes in Location Affect Circular Motion and Weight?

[Clari]

1. An object of mass 2.00kg is i)at the Poles, ii) at the Equator. Assuming the Earth is a perfect sphere of raius 6.4 x 10^6 m, caulculate the change in weight of the mass when taken from the Poles to the equator.

I solve it in this way: At the Poles, mg - R = mv^2 / r...R = mg - mv^2 / r...At the equator, R' = mv^2 / r...since reaction gives the sensatin of weight, change in weight of the mass = R- R'...= mg - 2mv^2 /r.....= 19.86N( I get v by concerning the mean angular speed of the Earth assuming it takes 24.0 hours to rotate about its axis)...But the answer is 0.068N...so there must be something wrong in my steps

2. An object of mass 0.5 kg is rotated in a horizontal circle by a string 1m long. The maximum tension in the string before it breaks is 50N. What is the greatest number of revolutions per second of the object?

I know the number of revolutions per second, T = 2pi root(l/g) = 1.99 s...sorry, am I missing something?

3.)A stone of mass 500 g is attached to a string of length 50m which will break if the tension in it exceeds 20 N. The stone is whirled in a vertical circle, the axis of rotation eing at a height of 100cm above the ground. The angular speed is very slowly increased until the string breaks. Where will the stone hit the ground?

first, I found the speed of the stone at which the string breaks. that is T - mg = mv^2 /r...20 - 0.5 *10 = 0.5 v^2 / 0.5...v = 3.87 m/s
then by s = ut +1/2 gt^2... 100 - 50 = 3.87 t +5t^2...t = 4.1s...distance at which the stone hit the ground = v*t = 3.87 * 4.1 = 15.9 m...but the answer is 122cm away...>_<

Please help...Thank you~

 

[Parth Dave]

1. The North and South poles lie on the axis of rotation for the Earth. If you are on the axis of rotation, are you rotating?

 

[Doc Al]

1. An object of mass 2.00kg is i)at the Poles, ii) at the Equator. Assuming the Earth is a perfect sphere of raius 6.4 x 10^6 m, caulculate the change in weight of the mass when taken from the Poles to the equator.

I solve it in this way: At the Poles, mg - R = mv^2 / r...R = mg - mv^2 / r...At the equator, R' = mv^2 / r...since reaction gives the sensatin of weight, change in weight of the mass = R- R'...= mg - 2mv^2 /r.....= 19.86N( I get v by concerning the mean angular speed of the Earth assuming it takes 24.0 hours to rotate about its axis)...But the answer is 0.068N...so there must be something wrong in my steps

It may be clearer if you write the centripetal acceleration in terms of angular velocity: a_c = \omega^2 r, where r is the distance to the axis of rotation. At the equator, you have maximum centripetal acceleration (r = radius of earth); at the poles, r = 0.

2. An object of mass 0.5 kg is rotated in a horizontal circle by a string 1m long. The maximum tension in the string before it breaks is 50N. What is the greatest number of revolutions per second of the object?

I know the number of revolutions per second, T = 2pi root(l/g) = 1.99 s...sorry, am I missing something?

Not sure where you got that equation; it looks like the period of a simple pendulum. For this problem, get an approximate answer by assuming that the tension equals the centripetal force.

3.)A stone of mass 500 g is attached to a string of length 50m which will break if the tension in it exceeds 20 N. The stone is whirled in a vertical circle, the axis of rotation eing at a height of 100cm above the ground. The angular speed is very slowly increased until the string breaks. Where will the stone hit the ground?

first, I found the speed of the stone at which the string breaks. that is T - mg = mv^2 /r...20 - 0.5 *10 = 0.5 v^2 / 0.5...v = 3.87 m/s

It looks like you used the wrong radius. It should be 50m, not 0.5.