problem with centripetal force?

Question

You swing an 0.8366 kg mass held by a string in a horizontal circle of 0.6137 m radius making a turn every 2.1285 seconds. What is the centripetal force?

The answer is 4.473901428206534, but what formula do u use to get it, how do u do it?

Answer 1

F=mr(2 x pi x f)^2
Standard equation in any text book (omega = 2 x pi x f).

Answer 2

With all due respect, I'm afraid that the preceding answer is incomplete because (i) ' f ' is not defined in terms of the available data, and (ii) The question of UNITS is not addressed or clarified.

The centripetral force is 4.4739... "kilograms" of force =
(9.81 x 4.4739...) Newtons = 43.9 N.

Here's how you do it:

The centripetal acceleration is

a = w^2 r,

where the angular velocity w (usually a Greek 'omega') is given by:

w = (2 pi) / P, where P is the period in seconds.

(Angular velocity ' w ' is the ubiquitous rotational speed measure used in dynamical physics, rather that ' f ,' the "circular" rotational speed measured in "cycles per second.")

The force required = m a = m w^2 r.

The numerical value of this is 4.4739... somethings, but the question is what "somethings" ?

So far, you have been working only in units of "kilograms" as a measure of weight or force. As someone educated in this kind of material over 50 years ago, that's fine by me. However, physicists now draw a clear distinction between the kilogram as a (passive) MASS, and its dynamical equivalent 9.81 NEWTONS as a weight or force.

(In one second, a Newton accelerates 1 kilogram from rest to a velocity of 1 metre per second.)

So 4.4739... "kilograms" of force = (9.81 x 4.4739...) Newtons
= 43.9 N.

NOTE : The answer CANNOT be given to more than 3 significant figures, as the input of the figure '9.81' was itself limited to 3 significant figures.

Answer 3

the first one isn't easily a Centripetal stress venture. it really is a Newton's huge-spread regulation of Gravitation venture. basically use the formulation : acceleration by using gravity = (G x mass of earth or the planet) / (a million.5 x radius of earth)^2 the second one one is yet another NULG venture, yet with a given top remote from the earth's floor. Use: acceleration by using gravity = (G x mass of earth) / (radius of earth + 580000 m)^2 the price of G (gravitational consistent) in our image voltaic gadget is 6.sixty seven x 10^-11 Nm^2/kg take into interest the classic values: Newtons, meters, and kilograms. The third one is about centripetal stress. acceleration by using gravity in the international = velocity of satellite tv for pc^2 / radius of earth Deriving: velocity = sq. root of (acceleration x radius) acceleration by using gravity in the international = 9.8 m/s^2

Answer 4

The magnitude of the normal acceleration is
a = v²/r where v is the velocity (in m/s) of the mass, r is the radius (in meters) of the circle, and a is the centripetal acceleration im m/s². Now remember that
F = ma and you're home free.

And the answer you gave is the correct one.

HTH ☺


Doug