All Solutions
Page 289: Lesson Check
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
end{align*}
$$
So, the torque is the force distributed over the object that determines how are the object’s particles rotating around a chosen axis which is equal to the product of the force and the distance.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
end{align*}
$$
As we know the moment arm is the length between a joint axis and the line of force acting on that joint. So, the torque is the force distributed over the object that determines how are the object’s particles rotating around a chosen axis which is equal to the product of the force and the distance.
In order to evaluate the angular acceleration, we use the following relation:
$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
tau &= alpha ~ I \
end{align*}
$$
So, the torque is equal to the angular acceleration times the moment of inertia.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
end{align*}
$$
So, the torque is the force distributed over the object that determines how are the object’s particles rotating around a chosen axis which is equal to the product of the force and the distance. So, the object should rotate, if there is a net external torque acting on the object.
Since both systems move with the same angular speed $omega$ and the two masses are positioned far from the axis of rotation $I_{1} > I_{2}$.
Solve for the left system:
As the textbook mentions that the angular momentum is given by the product of an object’s momentum of inertia and its angular speed
$$
begin{align*}
L_{1} &= I_{1} ~ omega_{1} \
&= I_{1} ~ omega
end{align*}
$$
Solve for the right system:
As the textbook mentions that the angular momentum is given by the product of an object’s momentum of inertia and its angular speed
$$
begin{align*}
L_{2} &= I_{2} ~ omega_{2} \
&= I_{2} ~ omega
end{align*}
$$
Since, the momentum of inertia of the left system is greater than the momentum of inertia of the right system $I_{1} > I_{2}$.
So, the angular momentum of the left system is greater than the angular momentum of the right system.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
vec{tau} &= vec{r} times vec{F} \
&= r ~ F ~ sin left( theta right) \
end{align*}
$$
Torque is a vector quantity placed in a plane perpendicular to the planes of the force and the displacement. So, it is not only the force that the torque depends on but although the distance and the angle of the force. Therefore, the classmate is incorrect.
The moment arm of the force is $r = 0.84 mathrm{~m}$. The force that exerted on the object is $F = 5.5 mathrm{~N}$.
$textbf{Required: }$
Finding the producing torque by the force.
Since the only force that acts on the meterstick is the weight. Then the direction of the net force is downward. As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
&=r ~ F ~ sin left( theta right) \
&= 0.84 mathrm{~m} times 5.5 mathrm{~N} times sin left( 90^{circ} right) \
&= 4.62 mathrm{~N cdot m}
end{align*}
$$
So, the producing torque by the force is $4.62 mathrm{~N cdot m}$.
The moment of inertia of the wheel is $I = 0.092 mathrm{~kg cdot cm^{2}}$. The torque that applied to the wheel is $tau = 7.4 mathrm{~N cdot m}$.
$textbf{Required: }$
Finding the angular acceleration of the wheel.
In order to evaluate the angular acceleration of the wheel, we use the following relation:
$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
&= dfrac{ 7.4 mathrm{~N cdot m} }{ 0.092 times 10^{-4} mathrm{~kg cdot m^{2}} } \
&= 8.043 times 10^{5} mathrm{~rad/s^{2}}
end{align*}
$$
So, the angular acceleration of the wheel is $8.043 times 10^{5} mathrm{~rad/s^{2}}$.
The torque that applied to the fan is $tau = 8.3 mathrm{~N cdot m}$.the angular acceleration of the fan is $alpha = 62 mathrm{~rad/s^{2}}$.
$textbf{Required: }$
Finding the moment of inertia of the fan.
In order to evaluate the angular acceleration of the wheel, we use the following relation:
$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
end{align*}
$$
Rearrange and solve for the moment of inertia of the fan:
$$
begin{align*}
I &= dfrac{ tau }{ alpha} \
&=dfrac{ 8.3 mathrm{~N cdot m} }{ 62 mathrm{~rad/s^{2}}} \
&= 0.134 mathrm{~kg cdot m^{2}}
end{align*}
$$
So, the moment of inertia of the fan is $0.134 mathrm{~kg cdot m^{2}}$.
The radius of the disk is $r = 0.58 mathrm{~m}$. The mass of the disk is $m = 6.1 mathrm{~kg}$. The angular acceleration is $alpha = 17 mathrm{~rad/s^{2}}$.
$textbf{Required: }$
Finding the required torque to give the disk the angular acceleration.
In order to evaluate the angular acceleration of the disk, we use the following relation:
$$
begin{align*}
alpha &= dfrac{ tau }{ I } \
tau &= alpha ~ I \
end{align*}
$$
In order to evaluate the moment of inertia of the disk, we use the following relation:
$$
begin{align*}
I &= dfrac{1}{2} ~ m ~ r^{2} \
end{align*}
$$
Rearrange and solve the required torque to give the disk the angular acceleration:
$$
begin{align*}
tau &= alpha ~ I \
&= alpha ~ dfrac{1}{2} ~ m ~ r^{2} \
&= dfrac{1}{2} times 17 mathrm{~rad/s^{2}} times 6.1 mathrm{~kg} times left( 0.58 mathrm{~m} right)^{2} \
&= 17.442 mathrm{~N cdot m}
end{align*}
$$
So, the required torque to give the disk the angular acceleration is $17.442 mathrm{~N cdot m}$.
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