Question

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

Answer 1

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 = (m*v_i^2)/r`

So, if `v_i` doubles, `F_c` would quadruple. The original `F_c` was `(m*v_i^2)/r` and it increased to `(4*m*v_i^2)/r`. That is an increase of `(3*m*v_i^2)/r` or 300%.

I hope this helps,
Steve

Answer 2

`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=\frac{mv^2}{r}`

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

`F'_c=\frac{m(2v)^2}{r}=\frac{4mv^2}{r}`

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

`\frac{F'_{c}-F_c}{F_c}\times 100`

`=\frac{\frac{4mv^2}{r}-\frac{mv^2}{r}}{\frac{mv^2}{r}}\times 100`

`=\frac{4-1}{1}\times 100`

`=300`

hence the centripetal force `F_c` will increase by `300 %`