if (int_count_times_randomized == 0){
return 60;
} else {
return random_min_max_precision(30, 350, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(2, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
return (true);
A object in uniform circular motion.
[[
namespace_circular_motion.theta = ~theta~ * Math.PI/180;
namespace_circular_motion.omega = namespace_circular_motion.theta / ~t~;
namespace_circular_motion.v = namespace_circular_motion.omega * ~r~;
return "";
]]
An object travels around a circle of radius $~r~ m$. Its angle measured at the center increases by $~theta~^\circ$ in $~t~ s$. Find the angular velocity $\omega$.
First we convert the angle to radians:
$$
\begin{eqnarray}
\theta &=& ~theta~^\circ \times \frac{\pi \ rad}{180^\circ} \\
&=& [[return sf_latex(namespace_circular_motion.theta)]] rad
\end{eqnarray}
$$
Angular velocity
$$
\begin{eqnarray}
\omega &=& \frac{\Delta \theta}{\Delta t} \\
&=& \frac{[[return sf_latex(namespace_circular_motion.theta)]] rad}{~t~ s} \\
&=& [[return sf_latex(namespace_circular_motion.omega)]] rad/s
\end{eqnarray}
$$
It means that every second the angle increases by $[[return sf_latex(namespace_circular_motion.omega)]] rad$.
$\omega = $
return sf_math(namespace_circular_motion.omega)
5%
Select unit for $\omega$:
$m/s$
$Hz$
$rad$
$rad/s$
3
circular_motion || rotation
Exercise - Angular velocity and tangential speed
if (int_count_times_randomized == 0){
return 60;
} else {
return random_min_max_precision(30, 350, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(2, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
return (true);
A object in uniform circular motion.
[[
namespace_circular_motion.theta = ~theta~ * Math.PI/180;
namespace_circular_motion.omega = namespace_circular_motion.theta / ~t~;
namespace_circular_motion.v = namespace_circular_motion.omega * ~r~;
return "";
]]
An object travels around a circle of radius $~r~ m$. Its angle measured at the center increases by $~theta~^\circ$ in $~t~ s$.
Find:
First we convert the angle to radians:
$$
\begin{eqnarray}
\theta &=& ~theta~^\circ \times \frac{\pi \ rad}{180^\circ} \\
&=& [[return sf_latex(namespace_circular_motion.theta)]] rad
\end{eqnarray}
$$
Angular velocity
$$
\begin{eqnarray}
\omega &=& \frac{\Delta \theta}{\Delta t} \\
&=& \frac{[[return sf_latex(namespace_circular_motion.theta)]] rad}{~t~ s} \\
&=& [[return sf_latex(namespace_circular_motion.omega)]] rad/s
\end{eqnarray}
$$
It means that every second the angle increases by $[[return sf_latex(namespace_circular_motion.omega)]] rad$.
Tangential speed
$$
\begin{eqnarray}
v &=& r \omega \\
&=& (~r~ m)([[return sf_latex(namespace_circular_motion.omega)]] rad/s) \\
&=& [[return sf_latex(namespace_circular_motion.v)]] m/s
\end{eqnarray}
$$
$\omega = $
return sf_math(namespace_circular_motion.omega)
5%
$v = $
return sf_math(namespace_circular_motion.v)
5%
Select unit for $\omega$:
$m/s$
$Hz$
$rad$
$rad/s$
3
Select unit for $v$:
$m/s$
$Hz$
$rad$
$rad/s$
0
circular_motion || rotation
Exercise - Angular acceleration basic
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(-10, 10, -1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 8;
} else {
return random_min_max_precision(-10, 10, 0, false, -0.5, 0.5); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(2, 20, 0); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_circular_motion.omega_f = ~omega_i~ + ~delta_omega~;
namespace_circular_motion.alpha = ~delta_omega~ / ~t~;
return "";
]]
An object travels around a circle with an initial angular velocity of $~omega_i~ rad/s$. After $~t~ s$ its angular velocity becomes $[[return namespace_circular_motion.omega_f]] rad/s$.
Find the angular acceleration $\alpha$.
Exercise - Angular velocity, frequency, and period ($\omega$, $f$ and $T$)
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 0;
} else {
return random_min_max_precision(0, 2, 0); //defined in setup_exercise_all.js.
}
return (true);
[[
namespace_circular_motion.f = ~omega~/(2*Math.PI);
namespace_circular_motion.period = 2*Math.PI/~omega~;
return "";
]]
The value of one of the variables (angular velocity $\omega$, frequency $f$, and period $T$) below has been given to you. Use that information to calculate the others.
Hint: Use $\omega = \frac{2\pi}{T} = 2\pi f$.
Solution
Rearrange $\omega = \frac{2\pi}{T} = 2\pi f$ to solve for what you need.
When $\omega$ is given, we have:
$$
\begin{eqnarray}
T &=& \frac{2\pi}{\omega} \\
&=& \frac{2\pi}{~omega~ rad/s} \\
&=& [[return sf_latex(namespace_circular_motion.period)]] s
\end{eqnarray}
$$
$$
\begin{eqnarray}
f &=& \frac{\omega}{2\pi} \\
&=& \frac{~omega~ rad/s}{2\pi} \\
&=& [[return sf_latex(namespace_circular_motion.f)]] Hz
\end{eqnarray}
$$
When $f$ is given, we have:
$$
\begin{eqnarray}
\omega &=& 2\pi f \\
&=& 2\pi ([[return sf_latex(namespace_circular_motion.f)]] Hz) \\
&=& ~omega~ rad/s
\end{eqnarray}
$$
$$
\begin{eqnarray}
T &=& \frac{1}{f} \\
&=& \frac{1}{[[return sf_latex(namespace_circular_motion.f)]] Hz} \\
&=& [[return sf_latex(namespace_circular_motion.period)]] s
\end{eqnarray}
$$
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(5, 100, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(2, 8, 0); //defined in setup_exercise_all.js
}
return (true);
A object moves around a circle of radius $r = ~r~ m$ a total of $~count_revolution~$ time[[return (~count_revolution~>1)? "s":""]] every $~t_total~ s$.
Find:
The time it takes for the object to go around 1 circle is the period $T$.
$\omega = \frac{2\pi}{T} = 2\pi f$.
$v = \omega r$.
Solution
Period
The period is the time it takes to go around the circle once:
$$
\begin{eqnarray}
T &=& \frac{~t_total~ s}{~count_revolution~} \\
&=& [[return sf_latex(namespace_circular_motion.period)]] s
\end{eqnarray}
$$
Once we have $T = [[return sf_latex(namespace_circular_motion.period)]] s$, we can find everything else using $\omega = \frac{2\pi}{T} = 2\pi f$ and $v = \omega r$.
$$
\begin{eqnarray}
v &=& \omega r \\
&=& ([[return sf_latex(namespace_circular_motion.omega)]] rad/s) (~r~ m) \\
&=& [[return sf_latex(namespace_circular_motion.v)]] m/s
\end{eqnarray}
$$
The number of revolutions per second is the frequency $f$.
$\omega = \frac{2\pi}{T} = 2\pi f$.
$v = \omega r$.
Solution
Frequency $f$
By definition, the number of revolutions per second is the frequency $f$, so we have $f = [[return namespace_circular_motion.f]] Hz$ (or $rps$).
Once we have $f$, we can find everything else using $\omega = \frac{2\pi}{T} = 2\pi f$ and $v = \omega r$.
Period
The period is the time it takes to go around the circle once:
$$
\begin{eqnarray}
T &=& \frac{1}{f} \\
&=& \frac{1}{~count_revolution~ Hz} \\
&=& [[return sf_latex(namespace_circular_motion.period)]] s
\end{eqnarray}
$$
$$
\begin{eqnarray}
v &=& \omega r \\
&=& ([[return sf_latex(namespace_circular_motion.omega)]] rad/s) (~r~ m) \\
&=& [[return sf_latex(namespace_circular_motion.v)]] m/s
\end{eqnarray}
$$
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 0;
} else {
return random_min_max_precision(0, 2, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 4;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js.
}
return (true);
An object of mass $~m~ kg$ moves around a circle of radius $~r~ m$ with
[[
namespace_circular_motion.omega = ~omega~;
namespace_circular_motion.f = ~omega~/(2*Math.PI);
namespace_circular_motion.period = 2*Math.PI/~omega~;
namespace_circular_motion.a_cent = ~r~ * ~omega~*~omega~;
namespace_circular_motion.f_cent = ~m~ * namespace_circular_motion.a_cent;
namespace_circular_motion.v = ~r~ * ~omega~;
if (~int_variable_choice~ == 0){
return `angular velocity $~omega~ rad/s$.`;
} else if (~int_variable_choice~ == 1){
return `frequency $${sf_latex(namespace_circular_motion.f)} Hz$.`;
} else if (~int_variable_choice~ == 2){
return `period $${sf_latex(namespace_circular_motion.period)} s$.`;
} else {
return "";
}
]]
Find:
The centripetal acceleration $a_{cent}$.
The tangential speed $v$.
Hint:
Use $\omega = \frac{2\pi}{T} = 2\pi f$ to find $\omega$ first.
$a_{cent} = r \omega^2$.
$v = r \omega$.
Solution
Find $\omega$
When $f$ is given, we can first find $\omega$:
$$
\begin{eqnarray}
\omega &=& 2\pi f \\
&=& 2\pi ([[return sf_latex(namespace_circular_motion.f)]] Hz) \\
&=& ~omega~ rad/s
\end{eqnarray}
$$
Find $\omega$
When $T$ is given, we can first find $\omega$:
$$
\begin{eqnarray}
\omega &=& \frac{2\pi}{T} \\
&=& \frac{2\pi}{[[return sf_latex(namespace_circular_motion.period)]] s} \\
&=& ~omega~ rad/s
\end{eqnarray}
$$
$$
\begin{eqnarray}
v &=& r \omega \\
&=& (~r~ m)(~omega~ rad/s) \\
&=& [[return sf_latex(namespace_circular_motion.v)]] m/s
\end{eqnarray}
$$
$a_{cent} = $
return sf_math(namespace_circular_motion.a_cent);
5%
Unit:
return "m/s^2";
not_number
$v=$
return sf_math(namespace_circular_motion.v);
5%
Unit:
return "m/s";
not_number
circular_motion || rotation
Exercise - Centripetal force basic
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 0;
} else {
return random_min_max_precision(0, 2, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 4;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js.
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js.
}
return (true);
An object of mass $~m~ kg$ moves around a circle of radius $~r~ m$ with
[[
namespace_circular_motion.omega = ~omega~;
namespace_circular_motion.f = ~omega~/(2*Math.PI);
namespace_circular_motion.period = 2*Math.PI/~omega~;
namespace_circular_motion.a_cent = ~r~ * ~omega~*~omega~;
namespace_circular_motion.f_cent = ~m~ * namespace_circular_motion.a_cent;
namespace_circular_motion.v = ~r~ * ~omega~;
if (~int_variable_choice~ == 0){
return `angular velocity $~omega~ rad/s$.`;
} else if (~int_variable_choice~ == 1){
return `frequency $${sf_latex(namespace_circular_motion.f)} Hz$.`;
} else if (~int_variable_choice~ == 2){
return `period $${sf_latex(namespace_circular_motion.period)} s$.`;
} else {
return "";
}
]]
Find:
The centripetal acceleration $a_{cent}$.
The centripetal force $F_{cent}$.
The tangential speed $v$.
Hint:
Use $\omega = \frac{2\pi}{T} = 2\pi f$ to find $\omega$ first.
$a_{cent} = r \omega^2$.
$F_{cent} = m a_{cent} = m r \omega^2$.
$v = r \omega$.
Solution
Find $\omega$
When $f$ is given, we can first find $\omega$:
$$
\begin{eqnarray}
\omega &=& 2\pi f \\
&=& 2\pi ([[return sf_latex(namespace_circular_motion.f)]] Hz) \\
&=& ~omega~ rad/s
\end{eqnarray}
$$
Find $\omega$
When $T$ is given, we can first find $\omega$:
$$
\begin{eqnarray}
\omega &=& \frac{2\pi}{T} \\
&=& \frac{2\pi}{[[return sf_latex(namespace_circular_motion.period)]] s} \\
&=& ~omega~ rad/s
\end{eqnarray}
$$
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(1, 7, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 12;
} else {
return random_min_max_precision(10, 30, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 50;
} else {
return random_min_max_precision(30, 100, 0); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_circular_motion.v = Math.sqrt(~r~ * 0.01 * ~tension~ / ~m~);
return "";
]]
A ball of mass $~m~ kg$ is tied to a string $~r~ cm$ (not in meters!) long and swung around in circles. The string can withstand a maximum tension of $~tension~N$.
How fast (in terms of the tangential speed $v$) do you have to spin the object in order to break the string?
Hint:
$F_{cent} = \frac{m v^2}{r}$.
Solution
The centripetal force on the ball is provided by the tension $Tension$, so we can identify $F_{cent}=Tension$ (I don't want to use $T$ for tension here because $T$ is also the notation for period):
$$
\begin{eqnarray}
Tension &=& F_{cent} = \frac{mv^2}{r} \\
v &=& \sqrt{\frac{r (Tension)}{m}} \\
&=& \sqrt{\frac{(~r~ \times 0.01 m) (~tension~ N)}{~m~ kg}} \\
&=& [[return sf_latex(namespace_circular_motion.v)]] m/s
\end{eqnarray}
$$
$v= $
return sf_math(namespace_circular_motion.v)
5%
$m/s$
rotation
Exercise - An accelerating wheel 1
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(5, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 4;
} else {
return random_min_max_precision(1, 7, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 5;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 0.2;
} else {
return random_min_max_precision(0.1, 0.5, 1); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_circular_motion.theta = ~omega_0~ * ~t~ + 0.5 * ~alpha~ * ~t~*~t~;
namespace_circular_motion.s = namespace_circular_motion.theta * ~r~;
return "";
]]
A wheel of radius $~r~m$ was originally rolling on the ground at $~omega_0~ rad/s$. It then accelerated at $~alpha~ rad/s^2$ for $~t~ s$.
Find for this time period:
The angle of rotation (i.e. the angular displacement) $\theta$.
The distance $s$ it traveled on the ground.
Hint: Use the angular version of the equations of motion.
Solution
The angle
The information given:
$\theta_0=0 rad$
$\omega_0=~omega_0~ rad/s$
$\theta=?$
$\omega=$ missing
$\alpha= ~alpha~ rad/s^2$
$t=~t~ s$
Note that "$\theta=?$" above means we are looking for $\theta$. Out of the final variables $\theta, \omega, \alpha, t$, only $\omega$
does not appear in the list above. This means we should use the $\omega$-equation:
$$
\begin{eqnarray}
\theta-\theta_0 &=& \omega_0 t + \frac{1}{2} \alpha t^2 \\
\Rightarrow \theta &=& (~omega_0~ rad/s) (~t~ s) + \frac{1}{2} (~alpha~ rad/s^2) (~t~ s)^2 \\
&=& [[return sf_latex(namespace_circular_motion.theta)]] rad
\end{eqnarray}
$$
Distance traveled
$$
\begin{eqnarray}
s &=& r \theta \\
&=& (~r~ m)([[return sf_latex(namespace_circular_motion.theta)]] rad) \\
&=& [[return sf_latex(namespace_circular_motion.s)]]m
\end{eqnarray}
$$
$\theta= $
return sf_math(namespace_circular_motion.theta)
5%
$rad$ $s= $
return sf_math(namespace_circular_motion.s)
5%
$m$
rotation
Exercise - An accelerating wheel 2
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(2, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 4;
} else {
return random_min_max_precision(2, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 80;
} else {
return random_min_max_precision(10, 100, -1); //defined in setup_exercise_all.js
}
return (true);
[[
namespace_circular_motion.alpha = (~omega_f~*~omega_f~ - ~omega_0~*~omega_0~) / (2 * ~theta~);
return "";
]]
A wheel was originally rolling on the ground at $~omega_0~ rad/s$. Its final angular velocity became $~omega_f~ rad/s$ after it rotated $~theta~ rad$.
Find the angular acceleration of the wheel.
Hint: Use the angular version of the equations of motion.
Solution
The information given:
$\theta_0=0 rad$
$\omega_0=~omega_0~ rad/s$
$\theta= ~theta~ rad$
$\omega= ~omega_f~ rad/s$
$\alpha=$ ?
$t=$ missing
Note that "$\alpha=?$" above means we are looking for $\alpha$. Out of the final variables $\theta, \omega, \alpha, t$, only $t$
does not appear in the list above. This means we should use the $t$-equation:
$$
\begin{eqnarray}
\omega^2 &=& \omega_0^2 +2 \alpha \theta \\
\Rightarrow \alpha &=& \frac{\omega^2 - \omega_0^2}{2 \theta} \\
&=& \frac{(~omega_f~ rad/s)^2 - (~omega_0~ rad/s)^2}{2 (~theta~ rad) } \\
&=& [[return sf_latex(namespace_circular_motion.alpha)]] rad/s^2
\end{eqnarray}
$$
$\alpha= $
return sf_math(namespace_circular_motion.alpha)
5%
Select unit for $\alpha$:
$m/s$
$Hz$
$rad$
$rad/s$
$rad/s^2$
4
rotation || moment_of_inertia
Exercise - Moment of inertia basic
if (int_count_times_randomized == 0){
return 1;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return -3;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 1;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 1;
} else {
return random_min_max_precision(-3, 3, 0, false, -0.5, 0.5); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
return (true);
Rotating a disc about the center.Rotating a disc about the edge.
[[
namespace_circular_motion.i_cm = 0.5 * ~m~ * ~r~*~r~;
namespace_circular_motion.i_edge = namespace_circular_motion.i_cm + ~m~ * ~r~*~r~;
return "";
]]
A disc of radius $r = ~r~ m$ and mass $m = ~m~ kg$ has moment of inertia about its center given by $I_{CM} = \frac{1}{2} m r^2$.
Calculate the moment of inertia $I_{CM}$.
Find the moment of inertia $I_{edge}$ if the axis of rotation is shifted to the edge.
We need to shift the axis by $d = r = ~r~ m$ to move it to the edge:
$$
\begin{eqnarray}
I_{edge} &=& \frac{1}{2} m r^2 + m d^2 \\
&=& \frac{1}{2} m r^2 + m r^2 \\
&=& \frac{3}{2} m r^2 \\
&=& \frac{3}{2} (~m~ kg) (~r~ m)^2 \\
&=& [[return sf_latex(namespace_circular_motion.i_edge)]] kgm^2
\end{eqnarray}
$$
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
return (true);
Rotating a rod about the center.Rotating a rod about the edge.
[[
namespace_circular_motion.i_cm = (1/12) * ~m~ * ~l~*~l~;
namespace_circular_motion.i_edge = namespace_circular_motion.i_cm + ~m~ * (~l~/2)*(~l~/2);
return "";
]]
A rod of length $l = ~l~ m$ and mass $m = ~m~ kg$ has moment of inertia about its center given by $I_{CM} = \frac{1}{12} m l^2$.
Calculate the moment of inertia $I_{CM}$.
Find the moment of inertia $I_{edge}$ if the axis of rotation is shifted to the edge.
We need to shift the axis by $d = \frac{l}{2} = [[return sf_latex(~l~ / 2)]] m$ to move it to the edge:
$$
\begin{eqnarray}
I_{edge} &=& \frac{1}{12} m l^2 + m d^2 \\
&=& \frac{1}{12} m l^2 + m (\frac{l}{2})^2 \\
&=& \frac{1}{12} m l^2 + \frac{1}{4} m l^2 \\
&=& \frac{1}{3} m l^2 \\
&=& \frac{1}{3} (~m~ kg) (~l~ m)^2 \\
&=& [[return sf_latex(namespace_circular_motion.i_edge)]] kgm^2
\end{eqnarray}
$$
$I_{CM}= $
return sf_math(namespace_circular_motion.i_cm)
5%
$I_{edge}= $
return sf_math(namespace_circular_motion.i_edge)
5%
Select unit for moment of inertia:
$Nm$
$N$
$kgm$
$kgm^2$
$kgm/s$
3
rotation || moment_of_inertia || rotational_ke
Exercise - Rotational kinetic energy basic
if (int_count_times_randomized == 0){
return 1;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(1, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return -3;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
return 0;
if (int_count_times_randomized == 0){
return 1;
} else {
return random_min_max_precision(-3, 3, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(2, 9, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 6;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(2, 5, 0); //defined in setup_exercise_all.js
}
return (true);
Rotating a disc about the center.
[[
namespace_circular_motion.i_cm = 0.5 * ~m~ * ~r~*~r~;
namespace_circular_motion.ke_rotational = 0.5*namespace_circular_motion.i_cm * ~omega~*~omega~;
return "";
]]
A disc of radius $r = ~r~ m$ and mass $m = ~m~ kg$ has moment of inertia about its center given by $I= \frac{1}{2} m r^2$.
Calculate the moment of inertia $I$.
Find the rotational kinetic energy $KE_{rot}$ if it rotates at $\omega = ~omega~ rad/s$.
Hint:
$KE_{rot} = \frac{1}{2} I \omega^2$.
Solution
$I$ about the center of mass
$$
\begin{eqnarray}
I &=& \frac{1}{2} m r^2 \\
&=& \frac{1}{2} (~m~ kg) (~r~ m)^2 \\
&=& [[return sf_latex(namespace_circular_motion.i_cm)]] kgm^2
\end{eqnarray}
$$
A sphere of mass $m$ and radius $r$ was initially at rest at the top of a hill as shown. Find its final speed when it rolled to the bottom of the hill. You will need $I_{\text{sphere}} = \frac{2}{5} m r^2$.
Solution
We will use conservation of energy. We must take into account the fact that it was rolling, meaning both linear \(KE\) and rotational \(KE\) are present:
Position
\(PE\)
\(KE_{\text{linear}}\)
\(KE_{\text{rotational}}\)
\(E_{\text{total}}\)
Top
\(mgh_1\)
\( 0 \)
\( 0 \)
\(mgh_1\)
Bottom
\(mgh_2\)
\( \frac{1}{2} m v_2^2 \)
\( \frac{1}{2} I \omega_2^2\)
\(mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{2} I \omega_2^2\)
Before equating the energy, however, we need to simplify the final \(KE_{\text{rotational}}\) in order to express everything in terms of \(v_2\):
$$
\begin{eqnarray}
KE_{\text{rotational}} &=& \frac{1}{2} I \omega_2^2 \\
&=& \frac{1}{2} (\frac{2}{5} m r^2) (\frac{v_2}{r})^2\\
&=& \frac{1}{5} m v_2^2
\end{eqnarray}
$$
This gives:
$$
\begin{eqnarray}
mgh_1 &=& mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{5} m v_2^2 \\
\Rightarrow mgh_1 - mgh_2 &=& \frac{7}{10} m v_2^2\\
\Rightarrow g(h_1 - h_2) &=& \frac{7}{10} v_2^2\\
v_2 &=& \sqrt{\frac{10}{7}g(h_1 - h_2) }
\end{eqnarray}
$$
Note that \(m\) and \(r\) both canceled out.
rotation || rotational_ke || energy
Exercise - Cylinder rolling down a hill
if (int_count_times_randomized == 0){
return 10;
} else {
return random_min_max_precision(7, 20, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(1, 5, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 0.5;
} else {
return random_min_max_precision(0.1, 0.9, 1); //defined in setup_exercise_all.js
}
return (true);
Cylinder rolling down a hill
[[
namespace_circular_motion.v_2 = Math.sqrt((4/3)*(9.8)*(~h_1~ - ~h_2~) );
return "";
]]
A cylinder of mass $m$ and radius $r$ was initially at rest at the top of a hill as shown.
Given:
$h_1 = ~h_1~ m$.
$h_2 = ~h_2~ m$.
$r = ~r~ m$
Find its final speed $v$ when it rolled to the bottom of the hill. You will need $I_{\text{cylinder}} = \frac{1}{2} m r^2$.
Hint:
$mgh_1 = mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{2} I \omega_2^2$
Solution
First we work out the rotational $KE$:
$$
\begin{eqnarray}
KE_{\text{rotational}} &=& \frac{1}{2} I \omega_2^2 \\
&=& \frac{1}{2} (\frac{1}{2} m r^2) (\frac{v_2}{r})^2\\
&=& \frac{1}{4} m v_2^2
\end{eqnarray}
$$
Using $\frac{1}{4} m v_2^2$ as the final rotational kinetic energy:
$$
\begin{eqnarray}
mgh_1 &=& mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{2} I \omega_2^2 \\
\Rightarrow mgh_1 &=& mgh_2 + \frac{1}{2} m v_2^2 + \frac{1}{4} m v_2^2 \\
\Rightarrow mgh_1 - mgh_2 &=& \frac{3}{4} m v_2^2\\
\Rightarrow g(h_1 - h_2) &=& \frac{3}{4} v_2^2\\
v_2 &=& \sqrt{\frac{4}{3}g(h_1 - h_2) } \\
&=& \sqrt{\frac{4}{3}(9.8)(~h_1~ - ~h_2~) } \\
&=& [[return sf_latex(namespace_circular_motion.v_2)]] m/s
\end{eqnarray}
$$
Angular velocity
We can then use \(\omega = \frac{v}{r}\):
$$
\begin{eqnarray}
\omega_2 &=& \frac{[[return sf_latex(namespace_circular_motion.v_2)]] m/s}{~r~ m} \\
&=& [[return sf_latex(namespace_circular_motion.v_2/~r~)]] rad/s
\end{eqnarray}
$$
$v_2= $
return sf_math(namespace_circular_motion.v_2)
5%
$m/s$ $\omega_2= $
return sf_math(namespace_circular_motion.v_2/~r~)
5%
$rad/s$
circular_motion || rotation || vector
Exercise - Circular motion in $\hat i$, $\hat j$
if (int_count_times_randomized == 0){
return 3;
} else {
return random_min_max_precision(2, 10, 0); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 2;
} else {
return random_min_max_precision(1.1, 5, 1); //defined in setup_exercise_all.js
}
if (int_count_times_randomized == 0){
return 1.2;
} else {
return random_min_max_precision(2, 20, 1); //defined in setup_exercise_all.js
}
return (true);
An object moves in the counter-clockwise direction around a circle of radius $R = ~R~ m$ at $~omega~ rad/s$, starting from the positive $x$-axis at $0s$. Express your answers below for the object at $t = ~t~ s$ in terms of $\hat i$ and $\hat j$.
What is the total angle traveled during $~t~ s$ in radians?
What is the total angle in degrees?
Find the position vector.
Find the velocity vector.
Find the acceleration vector.
Sketch a picture showing the directions of the three vectors above. The lengths of the vectors can be arbitray (no need to submit).
theta = ~omega~ * ~t~;
theta_in_degree = theta * 180 / Math.PI;
theta_in_degree_remainder = theta_in_degree % 360;
s = new Class_Vector(~R~ * Math.cos(theta), ~R~ * Math.sin(theta));
v = new Class_Vector(-1 * ~omega~ * ~R~ * Math.sin(theta), ~omega~ * ~R~ * Math.cos(theta));
a = s.times_scalar(-1 * ~omega~ * ~omega~);
Hint:
$$
\begin{eqnarray}
\theta &=& \omega t \\
\vec s &=& R \cos \theta \hat i + R \sin \theta \hat j
\end{eqnarray}
$$
Solution
The directions of $\vec s$. Not to scale.
The directions of $\vec v$ and $\vec a$. Not to scale.
Angle
In radians:
$$
\begin{eqnarray}
\theta &=& \omega t = (~omega~ rad/s) (~t~ s) \\
&=& [[return sf_latex(theta)]] rad
\end{eqnarray}
$$
In degrees:
$$
\begin{eqnarray}
\theta &=& [[return sf_latex(theta)]] (\frac{180^\circ}{\pi}) \\
&=& [[return sf_latex(theta_in_degree)]] ^\circ
\end{eqnarray}
$$
Taking out multiples of $360^\circ$, we have the angle $[[return sf_latex(theta_in_degree_remainder)]] ^\circ$.
Position
$$
\begin{eqnarray}
\vec s &=& R \cos \omega t \hat i + R \sin \omega t \hat j \\
&=& R \cos \theta \hat i + R \sin \theta \hat j \\
&=& [[return s.string_latex(2, false)]]
\end{eqnarray}
$$
Velocity
$$
\begin{eqnarray}
\vec v &=& \frac{d \vec s}{dt} \\
&=& \frac{d}{dt} (R \cos \omega t \hat i + R \sin \omega t \hat j) \\
&=& -\omega R \sin \omega t \hat i + \omega R \cos \omega t \hat j \\
&=& -\omega R \sin \theta \hat i + \omega R \cos \theta \hat j \\
&=& [[return v.string_latex(2, false)]]
\end{eqnarray}
$$
Acceleration
$$
\begin{eqnarray}
\vec a &=& \frac{d \vec v}{dt} \\
&=& \frac{d}{dt} (-\omega R \sin \omega t \hat i + \omega R \cos \omega t \hat j) \\
&=& -\omega^2 R \cos \omega t \hat i - \omega^2 R \sin \omega t \hat j \\
&=& -\omega^2 R \cos \theta \hat i - \omega^2 R \sin \theta \hat j \\
&=& -\omega^2 \vec s \\
&=& [[return a.string_latex(2, false)]]
\end{eqnarray}
$$
