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 80.0 cm, calculate

A.the angular speed of the rotating hoop and

B.the speed of its center.

Answer 1

Concept and reason

The concepts required to solve this problem are the expression for angular speed and conservation of energy principle.

Initially, calculate the angular speed of the rotating hoop by using the conservation of energy principle. Finally, calculate the speed at the center of the hoop using the angular speed obtained in the step 1.

Fundamentals

Speed at the center is given by,

$v = r\omega$

Here, $\omega$ is the angular speed of the rotating hoop and $r$ is the radius of the hoop.

Conservation of energy is given by,

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

Here, $I$ is the moment of inertia of the hoop, ${\omega _1}$ is the initial angular speed, m is the mass of the hoop, ${v_1}$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)

Calculate the angular speed of the rotating hoop as follows:

Substitute 0 for ${\omega _1}$ and 0 for ${v_1}$ in the equation $\frac{1}{2}{I_1}{\omega _1}^2 + \frac{1}{2}m{v_1}^2 + mg{h_1} = \frac{1}{2}I{\omega _2}^2 + \frac{1}{2}m{v_2}^2 + mg{h_2}$.

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

Substitute $h$ for $\left( {{h_1} - {h_2}} \right)$ in the above equation.

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

Angular speed is given by,

${\omega _2} = \frac{{\sqrt {gh} }}{r}$.

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

$\begin{array}{c}\\{\omega _2} = \frac{{\sqrt {\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {80{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)} \right)} }}{{8\;{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)}}\\\\ = 35.02{\rm{ rad/s}}\\\end{array}$

Therefore, the angular speed of the rotating hoop is $35.02{\rm{ rad/s}}$.

(b)

Calculate the speed at the center as follows:

Substitute ${\rm{35 rad/s}}$ for $\omega$ and $8{\rm{ cm}}$ for $r$ in the equation $v = r\omega$.

$\begin{array}{c}\\v = \left( {35.02{\rm{ rad/s}}} \right)\left( {\left( {8{\rm{ cm}}} \right)\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)\\\\ = 2.8{\rm{ m/s}}\\\end{array}$

Therefore, the speed at the center of the hoop is $2.8{\rm{ m/s}}$.

Ans: Part a

Therefore, the angular speed of the rotating hoop is $35.02{\rm{ rad/s}}$.