Question

A $30\text{kg}$ boy is riding a merry-go-round. If the linear speed and angular speed of the boy are $6\text{m/s}$ and $2\text{rad/s}$ respectively. The magnitude of centripetal force on the boy is

• A. $60\text{N}$
• B. $40\text{N}$
• C. $360\text{N}$
• D. $1080\text{N}$

Answer 1

The correct option is C $360\text{N}$

Given, mass of boy $\left(m\right)=30\text{kg}$
linear speed $\left(v\right)=6\text{m/s}$
Angular speed $\left(\omega \right)=2\text{rad/s}$

Radius of curvature $\left(r\right)=\frac{v}{\omega }=3\text{m}$
To find centripetal force: we know that,

$F_c=\frac{m{v}^2}{r}=\frac{30\times (6{)}^2}{3}=\frac{30\times 36}{3}=360\text{N}$

Hence option C is the correct answer