Question

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (the figure ). After the hoop has descended 55.0 cm, calculate
a) the angular speed of the rotating hoop in rad/s
b) the speed of its center. in m/s
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Please answer the questions specific. thank you!

Answer 1

Concept and reason

The concepts required to solve this question are conservation of energy and kinematic equations.

Initially, write the expression of the conservation of energy and substitute the values of initial rotational speed and initial speed in the expression. Then, rearrange the expression for angular speed. Next, calculate the angular speed of the rotating hoop. Finally, calculate the speed using the angular speed and the radius of the hoop.

Fundamentals

The expression of the conservation of energy for the system is,

$\frac{1}{2}I{\omega _1}^2 + \frac{1}{2}m{v_1}^2 + mg{h_1} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 + mg{h_2}$

Here, $I$ is the moment of inertia of the hoop, ${\omega _1}$ is the initial angular speed, m is the mass, ${v_1}$is the initial speed, g is the acceleration due to gravity, ${h_1}$ is the initial height, ${\omega _2}$ is the final angular speed, ${v_2}$is the final speed, and ${h_2}$ is the final height.

(a)

The expression of the conservation of energy for the system is,

$\frac{1}{2}{I_1}{\omega _1}^2 + \frac{1}{2}m{v_1}^2 + mg{h_1} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 + mg{h_2}$

Substitute 0 for ${\omega _1}$ and 0 for ${v_1}$.

$\begin{array}{c}\\\frac{1}{2}{I_1}{\left( 0 \right)^2} + \frac{1}{2}m{\left( 0 \right)^2} + mg{h_1} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 + mg{h_2}\\\\mg{h_1} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 + mg{h_2}\\\end{array}$

Rearrange the above equation.

$\begin{array}{c}\\mg{h_1} - mg{h_2} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2\\\\mg\left( {{h_1} - {h_2}} \right) = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2\\\end{array}$

Substitute $h$ for $\left( {{h_1} - {h_2}} \right)$.

$\frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 = mgh$

Substitute $m{r^2}$ for I, $r{\omega _2}$ for ${v_2}$ and rearrange the equation for ${\omega _2}$.

$\begin{array}{c}\\\frac{1}{2}\left( {m{r^2}} \right){\omega _2}^2 + \frac{1}{2}m{\left( {r{\omega _2}} \right)^2} = mgh\\\\\frac{1}{2}{r^2}{\omega _2}^2 + \frac{1}{2}{\left( {r{\omega _2}} \right)^2} = gh\\\\{r^2}{\omega _2}^2 = gh\\\\{\omega _2} = \frac{{\sqrt {gh} }}{r}\\\end{array}$

Substitute $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, 55 cm for h, and 8.0 cm for r.

$\begin{array}{c}\\{\omega _2} = \frac{{\sqrt {\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {55{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)} }}{{8.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}\\\\ = 29.02{\rm{ rad/s}}\\\end{array}$

(b)

The speed of the rotating hoop at its center is,

$v = r\omega$

Substitute $29.02{\rm{ rad/s}}$ for $\omega$ and 8.0 cm for r.

$\begin{array}{c}\\v = \left( {29.02{\rm{ rad/s}}} \right) \times \left( {\left( {8.0{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)\\\\ = 2.32{\rm{ m/s}}\\\end{array}$

Ans: Part a

The angular speed of the angular hoop is $29.02{\rm{ rad/s}}$.