# If the velocity of body is doubled then % increase in centripetal force will be?

Jul 21, 2018

That is an increase of 300%.

#### Explanation:

Since you are asking about centripetal force, I assume you are talking about a body moving in a circular path. In that case, speed is a better term than velocity. Or even better would be "instantaneous velocity", ${v}_{i}$. I say that because doubling velocity would imply it continues moving in a straight line.

The formula for centripetal force, ${F}_{c}$ is

${F}_{c} = \frac{m \cdot {v}_{i}^{2}}{r}$

So, if ${v}_{i}$ doubles, ${F}_{c}$ would quadruple. The original ${F}_{c}$ was $\frac{m \cdot {v}_{i}^{2}}{r}$ and it increased to $\frac{4 \cdot m \cdot {v}_{i}^{2}}{r}$. That is an increase of $\frac{3 \cdot m \cdot {v}_{i}^{2}}{r}$ or 300%.

I hope this helps,
Steve

300 %

#### Explanation:

The centripetal force ${F}_{c}$ on a body with mass $m$ & linear velocity $v$ moving on a circular path of radius $r$ is given as

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

Now, the centripetal force $F {'}_{c}$ on the same body with mass $m$ & with double linear velocity $2 v$ moving on the same circular path of radius $r$ is given as

$F {'}_{c} = \setminus \frac{m {\left(2 v\right)}^{2}}{r} = \setminus \frac{4 m {v}^{2}}{r}$

Now, the percent (%) increase in the centripetal force

$\setminus \frac{F {'}_{c} - {F}_{c}}{{F}_{c}} \setminus \times 100$

$= \setminus \frac{\setminus \frac{4 m {v}^{2}}{r} - \setminus \frac{m {v}^{2}}{r}}{\setminus \frac{m {v}^{2}}{r}} \setminus \times 100$

$= \setminus \frac{4 - 1}{1} \setminus \times 100$

$= 300$

hence the centripetal force ${F}_{c}$ will increase by 300 %