Question

A body of mass 6 kg moves in a (counterclockwise) circular path of radius 2 meters, making one revolution every 5 seconds. You may assume the circle is in the xy-plane, and so you may ignore the third component.

Compute the centripetal force acting on the body.

Answer 1

1. Draw a picture
2. What are the forces acting on a body
3. What is the centripetal force

Hint: Newton's second law of motion

Answer 2

@iambatman i have no clue what to do

Answer 3

F = ma

Answer 4

You can construct a position vector for the object at the time t in seconds. To get one revolution per 8 seconds, we want the argument of the sines and cosines to be 2pi t/8 (so each 8 seconds gives 2pi units. For a radius of 6 m, the path is

p(t) = 6 cos(2pi t/8) i + 6 sin(2pi t/8) j

The velocity will be tangent to the circular path and the acceleration toward the center. (The set up above takes the path to be the circle x^2 + y^2 = 6^2 in the xy-plane.)

a(t) = p '' (t) = (4 pi^2/64)(6) ( -cos(2pi t/8) i - sin(2pi t/8) j )

The fact that the path is circular simplifies things greatly because you don't have to decompose the acceleration into tangential and normal components. The acceleration points to the center.

Multiply by mass m to get force.

The magnitue is mass times ||a|| = m (4 pi^2/64)(6) = 3m pi^2/8.

Answer 5

wow, thank you for the answer, will take me a while to digest and understand