## Related questions with answers

A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man's shoes and the merry-go-round is

$μ_S = 0.5,$

how far from the axis of rotation can he stand without sliding?

Solutions

Verified**Given:**

- Angular velocity of a merry-go-round: $\omega=2.5 \,\frac{\text{rad}}{\text{s}}$;
- Coefficient of static friction: $\mu_\text{S}=0.5$;

**Required:**

- Distance from the axis of rotation where man can stand without sliding.

$\begin{align*} \sum_i F_i = F_f &= m a \\ F_f&= \mu_S N \\ &=\mu_S m g \\ a&=\sqrt{{a_c}^2 +{a_t}^2} \\ a_t&=0 \\ \implies a&=a_c \\ &=\omega^2 r \\ \implies \mu_S m g &= m \omega^2 r \\ \implies r&=\dfrac{\mu_S g}{\omega^2} \\ &=\dfrac{\left(0.5\right)\left(9.8\right)}{\left(2.5\right)^2} \\ &=0.78 \text{ m} \end{align*}$

In order to calculate the radius $r$, we will use Newton's second law of linear motion. Note that the only horizontal force acting on the person is friction ($F_f$). It is also important to remark that since there is no tangential acceleration (the angular speed is constant), the total acceleration $a$ is equal to the centripetal acceleration $a_c$, which can be calculated with the angular speed $\omega$. Finally, we replace $F_f$ and $a$ in the equation and then we solve for $r$.

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