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Q57P

Expert-verifiedFound in: Page 175

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**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 D **** from which seats of mass m **** are 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${\mathbf{\xb0}}$**** (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 - kg **** child and (c) find tension in the chain.**

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.

**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.**

(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\mathrm{cos}\theta $is the vertical component of the tension and $T\mathrm{sin}\theta $ is the horizontal component of the tension.

From the above figure, the radius is equal to

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

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

$r=2.50\mathrm{sin}28\xb0+\frac{8.0}{2}\phantom{\rule{0ex}{0ex}}=5.17\u200a\text{m}$

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.

$F=\left(40+10\right)\frac{{v}^{2}}{5.17}\phantom{\rule{0ex}{0ex}}=9.67{v}^{2}$

Now balance the force in horizontal and vertical direction.

Horizontal direction

$T\mathrm{sin}\theta =F$

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

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

Vertical direction

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

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

$\frac{T\mathrm{sin}\theta}{T\mathrm{cos}\theta}=\frac{\left(M+m\right){v}^{2}}{r\left(M+m\right)g}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{{v}^{2}}{rg}\phantom{\rule{0ex}{0ex}}v=\sqrt{rg\mathrm{tan}\theta}$

Substitute role="math" localid="1663678381997" $5.17m$ for $r$, $9.8m/{s}^{2}$ for $g$ and ${28}^{\xb0}$ for $\theta $in the above equation.

$v=\sqrt{\left(5.17\right)\left(9.8\right)\mathrm{tan}{28}^{\xb0}}\phantom{\rule{0ex}{0ex}}=5.19m/s$

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

(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.

${M}_{total}=M+m\phantom{\rule{0ex}{0ex}}=\left(40+10\right)\u200a\text{kg}\phantom{\rule{0ex}{0ex}}=50\u200a\text{kg}$

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

$W={M}_{total}g$

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

$W=\left(50\right)\left(9.8\right)\phantom{\rule{0ex}{0ex}}=490N$

The horizontal component of the tension is $T\mathrm{sin}28\xb0$ and the vertical component of the tension is $T\mathrm{cos}28\xb0$.

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

(c)

From equation (2).

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

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

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

On solving equation

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

Substitute $28\xb0$ for $\theta $ and $5.19m/s$ for $v$ in the above equation.

$T=\frac{9.67{\left(5.19\right)}^{2}}{\mathrm{sin}28\xb0}\phantom{\rule{0ex}{0ex}}=555\u200a\text{N}$

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

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