Question

Figure P6.57 shows a photo of a swing ride at an amusement park. The structure consists of a horizontal, rotating, circular platform of diameter Dfrom which seats of mass mare suspended at the end of mass less chains of length d. When the system rotates at constant speed, the chains swing outward and make an angle $\mathit{\theta }$with the vertical. Consider such a ride with the following parameters: D = 8.00 m, d = 2.50 m, m = 10.0 kg, and $\mathit{\theta }$= 28.0$\mathbf{°}$(a) What is the speed of each seat? (b) Draw a diagram of forces acting on the combination of a seat and a 40.0 - kgchild and (c) find tension in the chain.

Short answer

a) The speed of each seat is 5.19 m/s .

b) The diagram of forces acting on the combination of a seat is shown in step 2.

c) The tension in the chain is 555N.

Step-by-step solution

Centrifugal force

The expression for the centrifugal force is given by,

$F=m\frac{v^2}{r}$

Here$m$ is the mass of the object, v is the speed and r is the radius, F is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

Calculate the speed of each seat

(a) The below figure shows the direction of forces and direction.

Here F is the centrifugal force, D is the diameter, d is the length of the chain, m is the mass of the seats, M is the mass of the child, T is the tension, $\theta$is the angle, $T\cos \theta$is the vertical component of the tension and $T\sin \theta$is the horizontal component of the tension.

From the above figure, the radius is equal to

$r=d\sin \theta +\frac{D}{2}$

Substitute$8.0\mathrm{m}$for $D$,$2.50m$for $d$,${28}^o$for$\theta$in the above equation.

$$\begin{gathered} r=2.50\sin 28°+\frac{8.0}{2} \\ =5.17 \text{m} \end{gathered}$$

From the above figure, the expression for the centrifugal force is

$F=\left(M+m\right)\frac{v^2}{r}$ ....... (1)

Substitute$5.17m$for $r$,$40\mathrm{kg}$for $M$,$10\mathrm{kg}$for m in the above equation.

$$\begin{gathered} F=\left(40+10\right)\frac{v^2}{5.17} \\ =9.67v^2 \end{gathered}$$

Now balance the force in horizontal and vertical direction.

Horizontal direction

$T\sin \theta =F$

Substitute$\left(M+m\right)\frac{v^2}{r}$ for$F$ from equation (1) into the above equation.

$T\sin \theta =\left(M+m\right)\frac{v^2}{r}$ ...... (2)

Vertical direction

$T\cos \theta =\left(M+m\right)g$ ....... (3)

The ratio of equation (2) and equation (3) is

$$\begin{gathered} \frac{T\sin \theta }{T\cos \theta }=\frac{\left(M+m\right)v^2}{r\left(M+m\right)g} \\ \tan \theta =\frac{v^2}{rg} \\ v=\sqrt{rg\tan \theta } \end{gathered}$$

Substitute role="math" localid="1663678381997" $5.17m$for $r$, $9.8m/s^2$for $g$and ${28}^{°}$for $\theta$in the above equation.

$$\begin{gathered} v=\sqrt{\left(5.17\right)\left(9.8\right)\tan {28}^{°}} \\ =5.19m/s \end{gathered}$$

Therefore, the speed of each seat is $5.19m/s$.

Drawing the diagram of forces acting on the combination of a seat

(b)

The total mass is equivalent to the mass of child and seat.

$M_{total}=M+m$

Substitute $40\mathrm{kg}$for $M$and $10kg$for $m$in the above equation.

$$\begin{gathered} M_{total}=M+m \\ =\left(40+10\right) \text{kg} \\ =50 \text{kg} \end{gathered}$$

The total weight of combination, i.e seat and child is,

$W=M_{total}g$

Substitute$50kg$for$M_{total}$and $9.8m/s$.

$$\begin{gathered} W=\left(50\right)\left(9.8\right) \\ =490N \end{gathered}$$

The horizontal component of the tension is $T\sin 28°$and the vertical component of the tension is $T\cos 28°$.

The below diagram shows the direction of the forces acting on the combination of a seat.

Calculating the tension in the chain

(c)

From equation (2).

$T\sin \theta =\left(M+m\right)\frac{v^2}{r}$

As,$T\sin \theta =F$ and $F=9.67v^2$.

$T\sin \theta =9.67v^2$

On solving equation

$T=\frac{9.67v^2}{\sin \theta }$

Substitute $28°$for $\theta$and $5.19m/s$for $v$in the above equation.

$$\begin{gathered} T=\frac{9.67{\left(5.19\right)}^2}{\sin 28°} \\ =555 \text{N} \end{gathered}$$

Therefore, the tension in the chain is $555N$.