Question

(a) a 19.0 kg child is riding a playground merry - go - round that is rotating at 50.0 rpm. what centripetal force, in n, must she exert to stay on if she is 2.00 m from its center? n (b) what centripetal force, in n, does she need to stay on an amusement park merry - go - round that rotates at 3.00 rpm if she is 7.80 m from its center? n (c) compare each force with her weight. \\( \frac { \text { force from part } ( a ) } { \text { weight } } = \\) \\( \frac { \text { force from part } ( b ) } { \text { weight } } = \\)

Explanation:

Step1: Convert rotational speed to angular velocity

The formula for angular velocity \(\omega\) (in \(rad/s\)) from rotational speed \(n\) (in \(rpm\)) is \(\omega=\frac{2\pi n}{60}\).
For part (a): \(n = 50.0\ rpm\), so \(\omega_a=\frac{2\pi\times50.0}{60}\approx5.24\ rad/s\)
For part (b): \(n = 3.00\ rpm\), so \(\omega_b=\frac{2\pi\times3.00}{60}= 0.314\ rad/s\)

Step2: Calculate centripetal force

The formula for centripetal force \(F_c = m\omega^{2}r\), where \(m = 19.0\ kg\)
For part (a): \(r = 2.00\ m\), \(F_{ca}=19.0\times(5.24)^{2}\times2.00\)

$$F_{ca}=19.0\times27.46\times2.00 = 1043.5\ N$$

For part (b): \(r = 7.80\ m\), \(F_{cb}=19.0\times(0.314)^{2}\times7.80\)

$$F_{cb}=19.0\times0.0986\times7.80=14.6\ N$$

Step3: Calculate weight

The formula for weight \(W=mg\), where \(g = 9.8\ m/s^{2}\), \(W=19.0\times9.8 = 186.2\ N\)

Step4: Calculate the ratios

For part (a): \(\frac{F_{ca}}{W}=\frac{1043.5}{186.2}\approx5.60\)
For part (b): \(\frac{F_{cb}}{W}=\frac{14.6}{186.2}\approx0.0784\)

Answer:

(a) \(1043.5\ N\)
(b) \(14.6\ N\)
(c) \(\frac{\text{force from part (a)}}{\text{weight}}\approx5.60\), \(\frac{\text{force from part (b)}}{\text{weight}}\approx0.0784\)