All Solutions
Page 275: Lesson Check
As the problem mentions, the angular position of the rotating wheel is $Delta theta = 32^{circ}$ which is counterclockwise direction. As we know the counterclockwise direction is defined as positive and clockwise is negative for rotational variables. So, Counterclockwise is the positive rotation direction and clockwise is the negative direction. So, the given angular position is positive.
As the problem mentions, the wheels of the car moves from right to left which is counterclockwise direction. As we know the counterclockwise direction is defined as positive and clockwise is negative for rotational variables. So, Counterclockwise is the positive rotation direction and clockwise is the negative direction. So, the given angular position is positive.
As we know the average angular velocity $omega_{avg}$ is given by the angular displacement divided by the time interval which the displacement occurs.
$$
begin{align*}
omega_{avg} &= dfrac{ Delta theta }{ Delta t }
end{align*}
$$
So, the direction of the angular speed is the same as the direction of the angular position. So, the angular velocity is positive as the angular position is positive.
The initial angular velocity of the object is $omega_{i} = 1 mathrm{~rad/s}$. The angular acceleration of the object is $alpha_{avg} = – 0.5 mathrm{~rad/s^{2}}$.
$textbf{Required: }$
Finding if the final angular speed increasing or decreasing.
Solve for the time interval of $1 mathrm{~s}$:
As the textbook mentions, the average angular acceleration is given by the change in angular velocity within a given time interval
$$
begin{align*}
alpha_{avg} &= dfrac{ Delta omega }{ Delta t } \
&= dfrac{ omega_{f} – omega_{i} }{Delta t } \
&= dfrac{ 7.9 mathrm{~rad/s} – 5.2 mathrm{~rad/s} }{ 1.3 mathrm{~s} } \
&= 2.077 mathrm{~rad/s^{2}}
end{align*}
$$
Rearrange and solve for the final angular speed:
$$
begin{align*}
omega_{f} &= alpha_{avg} ~ Delta t + omega_{i} \
&= – 0.5 mathrm{~rad/s^{2}} times 1 mathrm{~s} + 1 mathrm{~rad/s} \
&= 0.5 mathrm{~rad/s}
end{align*}
$$
So, the final speed decreasing for all time intervals since the time interval is always positive.
The angular position is $theta = 2 mathrm{~rad}$.
$textbf{Required: }$
Fidning the arc length.
As we know the relation between the angles and the arc length of any circle is given by the following relation:
$$
begin{align*}
s &= r ~ theta \
&= 2 r
end{align*}
$$
So, the arc length, in this case, is equal to the diameter of the circle.
In order to evaluate the angular speed, we use the following relation:
$$begin{align*}
omega &= dfrac{ 2 pi }{ T } \
end{align*}$$
So, the angular speed is only depending on the periodic time.
$$omega_h = frac{2 pi}{12 text{ hrs}}$$
$$omega_m = frac{2 pi}{1 text{ hrs}}$$
$$
begin{aligned}
frac{omega_h}{omega_m} &= frac{frac{2pi}{12}}{frac{2 pi}{1}}\
&= frac{1}{12}\
end{aligned}$$
Hence, it follows:
$$boxed{omega_h = frac{1}{12} omega_m}$$
The time interval that the minute hand of the clock takes to rotate is $Delta t = 15 mathrm{~min} = dfrac{1}{4} mathrm{~hr}$.
$textbf{Required: }$
Finding the angular position of the minute hand.
In order to evaluate the angular speed of the one revolution of the minute hand, we use the following relation:
$$
begin{align*}
omega &= dfrac{ 2 pi }{ T } \
&= dfrac{ 2 pi }{ 1 mathrm{~hr} } \
&= 2 pi mathrm{~rad/hr}
end{align*}
$$
So, the angular speed is only depending on the periodic time. As we know that the periodic time of the minute hand is $1 mathrm{~hr}$. As we know the average angular velocity $omega_{avg}$ is given by the angular displacement divided by the time interval which the displacement occurs.
$$
begin{align*}
omega &= dfrac{ Delta theta }{ Delta t } \
end{align*}
$$
Rearrrange and solve for the angular position of the minute hand:
$$
begin{align*}
Delta theta &= omega ~ Delta t \
&= 2 pi mathrm{~rad/hr} times dfrac{1}{4} mathrm{~hr} \
&= dfrac{ pi }{2} mathrm{~rad}
end{align*}
$$
So, the angular position of the minute hand is $dfrac{ pi }{2} mathrm{~rad}$.
We know that the Earth completes one cycle of rotation in one year, so the period is:
$$T = 1 text{ year}$$
In order to evaluate the angular speed of the one revolution of the Earth, we use the following relation:
$$begin{align*}
omega &= dfrac{ 2 pi }{ T } \
&= dfrac{ 2 pi }{ 365.25 cdot 24 cdot 60 cdot 60 mathrm{~s} } \
&= 1.991 cdot 10^{-7} mathrm{~rad/s}
end{align*}$$
So, the angular speed of the one revolution of the Earth is $1.991 cdot 10^{-7} mathrm{~rad/s}$.
The radius of the wheels is $r = 0.62 mathrm{~m}$. The angular speed of the bicycle wheel is $omega = 21 mathrm{~rad/s}$.
$textbf{Required: }$
Finding the linear speed of the given point.
As the textbook mentions, the tangential speed of a rotating object is given by
$$
begin{align*}
v &= r ~ omega \
&= 0.62 mathrm{~m} times 21 mathrm{~rad/s} \
&= 13.02 mathrm{~m/s}
end{align*}
$$
So, the linear speed of the given point is $13.02 mathrm{~m/s}$.
The initial angular speed of the propeller is $omega_{i} = 5.1 mathrm{~rad/s}$. The angular acceleration of the propeller is $alpha = 1.6 mathrm{~rad/s^{2}}$. The time interval that the propeller takes to reach to the final angular velocity is $Delta t = 3 mathrm{~s}$.
As the textbook mentions, the average angular acceleration is given by the change in angular velocity within a given time interval
$$
begin{align*}
alpha_{avg} &= dfrac{ Delta omega }{ Delta t } \
&= dfrac{ omega_{f} – omega_{i} }{Delta t } \
end{align*}
$$
Rearrange and solve for the final angular speed:
$$
begin{align*}
omega_{f} &= alpha_{avg} ~ Delta t + omega_{i} \
&= 1.6 mathrm{~rad/s^{2}} times 3 mathrm{~s} + 5.1 mathrm{~rad/s} \
&= 9.9 mathrm{~rad/s}
end{align*}
$$
So, the final speed of the propeller is $9.9 mathrm{~rad/s}$.
