# The tension in a 2 m length of string that whirls a 1 kg mass at 4 m/s in a horizontal circle is calculated to be 8 N. How do you alculate the tension for the following case: twice the mass, speed, and distance?

Apr 11, 2018

Calculate the answer using the centripetal force equation.

#### Explanation:

To start off, notice that the problem deals with horizontal rotation.

From that, we can see that the force that points to the center of the circle is the Tension on the string.

Thus, Tension is the Centripetal Force that keeps the mass moving in a circle

Then, using the equation for centripetal force,

${F}_{\text{centripetal}} = m {v}^{2} / r$

$T = {F}_{\text{centripetal}}$

$T = m {v}^{2} / r$

To check, we can substitute the values from the given.

$T = \left(1 k g\right) {\left(4 \frac{m}{s}\right)}^{2} / \left(2 m\right)$

$T = \left(1 k g\right) \frac{16 {m}^{2} / {s}^{2}}{2 m}$

$T = 8 N$

and

$8 N = 8 N$.

Thus if we double every given that we have,

$T = \left(2 k g\right) {\left(8 \frac{m}{s}\right)}^{2} / \left(4 m\right)$

$T = \left(2 k g\right) \frac{64 {m}^{2} / {s}^{2}}{4 m}$

$T = 64 N$