Question

A 42.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 26.0 m.

(a) What is the centripetal acceleration of the child?

magnitude___ m/s²

(b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? magnitude ___ N

(c) What force does the seat exert on the child at the highest point of the ride?magnitude___  N

(d) What force does the seat exert on the child when the child is halfway between the top and bottom? (Assume the Ferris wheel is rotating clockwise and the child is moving upward.)

magnitude ___ N

direction___  °counter-clockwise from the horizontal

Answer 1

Solution;

Using the analysis;

(a)

From the relationship of centripetal acceleration and angular velocity;

$a_c=rw^2$

Here;

$D=26m$ ;$r=13m$

$w=4rev/min$

Convert as;

$w=4×\frac{2π}{60}rad/s$

Substitute back into the equation;

$a_c=13×(4×\frac{2π}{60})^2$

$a_c=2.28m/s^2$

(b)

From Newton's Law;

$\sum F_x=0$

(Taking +x-axis as positive)

$F_c-F_{sx}=0$

$F_c=F_{sx}$

But ; $F_c=ma_c$

Substitute the value of m=42kg and $a_c=2.28m/s^2$

$F_{sx}=42×2.28$

$F_{sx}=95.8N$

Again from Newton's law;

$\sum F_y=0$

$F_{sy}-F_g=0$

$F_{sy}=F_g$

But; $F_g=mg$

$F{sy}=mg$

Substitute the values;

$F_{sy}=42×9.81$

$F_{sy}=412.02N$

The resultant force acting on the child due to the seat at lowest point is;

$F_s=\sqrt{F_{sx}^2+F_{sy}^2+2F_{sx}F_{sy}cos\theta}$

Substitute the various values,take $\theta=0°$

$F_s=\sqrt{95.8^2+412.02^2+2(95.8)(412.02)cos(0)}$

Magnitude:$F_s=507.82N$

Direction:Up towards the centre of the path.

(c)

At highest point of the ride;

Take $\theta=180^°$

$F_s=\sqrt{95.8^2+412.02^2+2(95.8×412.02cos(180)}$

$F_s=316.22N$

(d)

Force when the child is hall way bottom and top;

Take $\theta=90°$

$F_s=\sqrt{95.8^2+412.02^2+2(95.8×412.02×cos(90)}$

$F_s=423.01N$

Direction of the net force acting on the child is given as;

$\theta=tan^{-1}(\frac{F_{sy}}{F_{sx}})$

Substituting the values;

$\theta=tan^{-1}(\frac{412.02}{95.8})$

$\theta=76.91°$