Chapter 10, Problem 113P

(a)

To determine

The magnitude of tension in the ribbon.

Answer to Problem 113P

The magnitude of tension in the ribbon is $0.0513\text{N}$.

Explanation of Solution

The density of the air is $1.29\text{kg}/{\text{m}}^{\text{3}}$, the density of helium is $0.179\text{kg}/{\text{m}}^{\text{3}}$, radius of the balloon is $12.0\text{cm}$, mass of the balloon is $2.80\times {10}^{-3}\text{kg}$ and acceleration due to gravity is $9.80\text{m}/{\text{s}}^{\text{2}}$.

Write the expression to calculate the total mass of the balloon and helium.

$M=\rho \left(\frac{4}{3}\pi r^3\right)+m$

Here, $\rho$ is the density of the helium, M is the total mass of the balloon and helium, m is the mass of the balloon alone and r is the radius of the balloon.

Write the expression for the tension in the ribbon.

$T=F_b-Mg$

Here, T is the tension in the ribbon, $F_b$ is the buoyant force and g is the acceleration due to gravity.

Write the expression for buoyant force.

$F_b={\rho }_0\left(\frac{4}{3}\pi r^3\right)g$

Here, ${\rho }_0$ is the density of the air.

Rewrite the equation for T using the above equation.

$T={\rho }_0\left(\frac{4}{3}\pi r^3\right)g-Mg$

Substitute the expression for M in the above equation to rewrite.

$\begin{array}{l}T={\rho }_0\left(\frac{4}{3}\pi r^3\right)g-\left(\rho \left(\frac{4}{3}\pi r^3\right)+m\right)g\\ T=\left(\left(\frac{4}{3}\pi r^3\right)\left({\rho }_0-\rho \right)-m\right)g\end{array}$

Substitute $1.29\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_0$, $0.179\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $12.0\text{cm}$ for r, $2.80\times {10}^{-3}\text{kg}$ for m and $9.80\text{m}/{\text{s}}^{\text{2}}$ for g in the above equation to calculate T.

$\begin{array}{c}T=\left(\left(\frac{4}{3}\pi {\left(12.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^3\right)\left(1.29\text{kg}/{\text{m}}^{\text{3}}-0.179\text{kg}/{\text{m}}^{\text{3}}\right)-2.80\times {10}^{-3}\text{kg}\right)9.80\text{m}/{\text{s}}^{\text{2}}\\ =0.0513\text{N}\end{array}$

Conclusion:

Therefore, the magnitude of tension in the ribbon is $0.0513\text{N}$.

(b)

To determine

The magnitude of time period of oscillation.

Answer to Problem 113P

The magnitude of time period of oscillation is $2.69\text{s}$.

Explanation of Solution

The length of the ribbon is $2.30\text{m}$.

Write the expression to calculate the spring constant.

$k\approx \frac{T}{l}$

Here, k is the spring constant and l is the length of the ribbon.

Write the expression to calculate the time period.

$T^{\prime }=2\pi \sqrt{\frac{M}{k}}$

Here, $T^{\prime }$ is the time period.

Use $k\approx \frac{T}{l}$ and the expression for M in the above equation to rewrite.

$T^{\prime }=2\pi \sqrt{\frac{\left(\rho \left(\frac{4}{3}\pi r^3\right)+m\right)l}{T}}$

Substitute $2.80\times {10}^{-3}\text{kg}$ for m, $0.179\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $12.0\text{cm}$ for r, $0.0513\text{N}$ for T and $2.30\text{m}$ for l in the above equation to calculate $T^{\prime }$.

$\begin{array}{c}{T}^{\prime }=\sqrt{\frac{\left(0.179\text{kg}/{\text{m}}^{\text{3}}\left(\frac{4}{3}\pi {\left(12.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^3\right)+2.80\times {10}^{-3}\text{kg}\right)2.30\text{m}}{0.0513\text{N}}}\\ =2\pi \sqrt{\frac{\left(0.0041\text{kg}\right)2.30\text{m}}{0.0513\text{N}}}\\ =2.69\text{s}\end{array}$

Conclusion:

Therefore, the magnitude of time period of oscillation is $2.69\text{s}$.