Question

A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acceleration of the stone and it's centripetal force?​

Answer 1

force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

Answer 2

Answer:

Acceleration: $100\; {\rm m\cdot s^{-2}}$ assuming that the radius of the rotation is $1\; {\rm m}$.

Centripetal force: $15\; {\rm N}$.

Explanation:

In a circular motion, if the tangential velocity is $v$ and the radius of the motion is $r$, the centripetal acceleration of the motion would be $a = (v^{2} / r)$.

In this question, it is implied that for this circular motion, $v = 10\; {\rm m\cdot s^{-1}}$ while $r = 1\; {\rm m}$. Thus, the (centripetal) acceleration would be:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}$.

Note that the unit of mass in this question is gram, whereas the standard unit for mass should be ${\rm kg}$ (so as to leverage the fact that $1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}$.) Apply unit conversion: $m = 150\; {\rm g} = 0.150\; {\rm kg}$.

Using that fact that $(\text{net force}) = (\text{mass}) \, (\text{acceleration})$:

$\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}$.