# An object with a mass of 12  kg is revolving around a point at a distance of 16  m. If the object is making revolutions at a frequency of 15 Hz, what is the centripetal force acting on the object?

May 26, 2016

There are (at least) two ways to answer this: using linear velocity and angular velocity. I have completed both below to increase your knowledge and options. The answer, using both methods, is $F = 1.7 \times {10}^{6}$ $N$

#### Explanation:

Linear velocity method:

We have the expression $a = {v}^{2} / r$ for the centripetal acceleration. Combining this with Newton's Second Law, $F = m a$, we end up with:

$F = \frac{m {v}^{2}}{r}$

We know the mass and radius, but need to calculate the linear velocity (a vector that is at a tangent to the circular motion. The speed is constant but the velocity is continually changing as the direction of the motion changes).

The object is rotating 15 times each second, and the circumference of a circle of radius $r$ is $2 \pi r$. One circumference is $2 \pi \times 16 = 100.5$ $m$. The object is traveling 15 rotations each second, or $1508$ $m {s}^{-} 1$.

Then $F = \frac{m {v}^{2}}{r} = \frac{12 \times {1508}^{2}}{16} = 1 , 705 , 548$ $N$

Angular velocity method:

Similarly, for the centripetal acceleration using angular velocity, we have $a = {\omega}^{2} r$, and therefore $F = m {\omega}^{2} r$.

Again, we know the mass and the radius. The object is rotating at $15$ $H z$, and each rotation is $2 \pi$ radians, so the angular velocity is $30 \pi = 94.25$ $r a {\mathrm{ds}}^{-} 1$.

Then $F = m {\omega}^{2} r = 12 \times {94.25}^{2} \times 16 = 1 , 705 , 468$ $N$

This is slightly different from the other answer, but only at the 5th significant digit. Remember that we had the data only to 2 significant digits, so this is within the rounding error.

For all practical purposes, both methods yield the same answer.