All Solutions
Page 305: Standardized Test Prep
The radius of the wheel is $R = 0.1 mathrm{~m}$. The axle of rotation is $r = 0.02 mathrm{~m}$.
Solve for the force at $A$:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{A} &= R ~ F_{A} \
&= 0.1 mathrm{~m} cdot 20 mathrm{~N} \
&= boxed{2 mathrm{~N cdot m}}
end{align*}
$$
Solve for the force at $B$:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{B} &= r ~ F_{B} \
&= 0.02 mathrm{~m} cdot 50 mathrm{~N} \
&= boxed{1 mathrm{~N cdot m}}
end{align*}
$$
Solve for the force at $C$:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{C} &=R ~ F_{C} \
&= 0.1 mathrm{~m} cdot 30 mathrm{~N} \
&= boxed{3 mathrm{~N cdot m}}
end{align*}
$$
Solve for the force at $D$:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau_{D} &=R ~ F_{D} \
&= 0.1 mathrm{~m} cdot 10 mathrm{~N} \
&= boxed{1 mathrm{~N cdot m}}
end{align*}
$$
So, the correct answer is (C) force C
The radius of the wheel is $R = 0.1 mathrm{~m}$. The axle of rotation is $r = 0.02 mathrm{~m}$.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
end{align*}
$$
In order to evaluate the total torque on the system, we use the superposition principle:
$$
begin{align*}
vec{tau_{t}} &= vec{tau_{A}} + vec{tau_{B}} + vec{tau_{C}} + vec{tau_{D}} \
&= 0.1 mathrm{~m} cdot 20 mathrm{~N} ~ hat{k} + 0.02 mathrm{~m} cdot 50 mathrm{~N} ~ hat{k} – 0.1 mathrm{~m} cdot 30 mathrm{~N} ~ hat{k} – 0.1 mathrm{~m} cdot 10 mathrm{~N} ~ hat{k} \
&= boxed{ – 1 mathrm{~N cdot m} ~ hat{k}}
end{align*}
$$
So, the net torque on the system is $1 mathrm{~N cdot m}$, clockwise.
The radius of the wheel is $R = 0.1 mathrm{~m}$. The axle of rotation is $r = 0.02 mathrm{~m}$.
Solve for the net torque equals zero:
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
tau &= r ~ F \
end{align*}
$$
In order to evaluate the total torque on the system, we use the superposition principle:
$$
begin{align*}
vec{tau_{t}} &= vec{tau_{A}} + vec{tau_{B}} + vec{tau_{C}} + vec{tau_{D}} \
&= 0 mathrm{~N cdot m} ~ hat{k} \
end{align*}
$$
Rearrange and solve for the torque at C:
$$
begin{align*}
tau_{C} &= – vec{tau_{A}} – vec{tau_{B}} – vec{tau_{D}} \
&= – 0.1 mathrm{~m} cdot 20 mathrm{~N} ~ hat{k} – 0.02 mathrm{~m} cdot 50 mathrm{~N} ~ hat{k} + 0.1 mathrm{~m} cdot 10 mathrm{~N} ~ hat{k} \
&= – 2 mathrm{~N cdot m} ~ hat{k} \
end{align*}
$$
Rearrange and solve for the force at point C:
$$
begin{align*}
tau_{C} &= R ~ F_{C} \
F_{C} &= dfrac{ tau_{C} }{ R } \
&= dfrac{ – 2 mathrm{~N cdot m} ~ hat{k} }{ 0.1 mathrm{~m} } \
&= boxed{- 20 mathrm{~N} ~ hat{k}}
end{align*}
$$
So, the magnitude of the force at point C is $20 mathrm{~N}$. Therefore, the correct answer is (B) $20mathrm{~N}$
The radius of the ball is $dfrac{ 0.2 mathrm{~m} }{2} = 0.1 mathrm{~m}$. The angular velocity of the ball is $omega = 4 mathrm{~rad/s}$.
$textbf{Required: }$
Finding the linear speed of the center of mass.
As the textbook mentions, the tangential speed of a rotating object is given by
$$
begin{align*}
v &= r ~ omega \
&= 0.1 mathrm{~m} cdot 4 mathrm{~rad/s} \
&= boxed{0.4 mathrm{~m/s}}
end{align*}
$$
So, the correct answer is (C) $0.4 mathrm{~m/s}$
Length of the rod is one meter.
Let the distance of the mass $m_2$ from the point of support is $r_2:m$. Then the distance of $m_1$ from the point of support is $(1-r_2):m$.
The condition for balance is
$m_1 (1-r_2) = m_2 r_2$
$m_1 – m_1 r_2 = m_2 r_2$
$m_1 r_2 + m_2 r_2 = m_1$
$r_2(m_1 + m_2) = m_1$
or
$r_2 = dfrac{m_1}{m_1+m_2}$
Substituting the values of $m_1$ and $m_2$,
$r_2 = dfrac{0.2}{0.2 +0.3} = dfrac{0.2}{0.5} = 0.4:m = 40:cm$
So the point of support is $40:cm$ from $300:g$ object.
As the textbook mentions that the rotational kinetic energy is given by
$$
begin{align*}
K.E_{rot} &= dfrac{1}{2} ~ I ~ omega^{2} \
end{align*}
$$
In order to evaluate the total kinetic energy of the ball, we use the following relation:
$$
begin{align*}
K.E_{t} &= dfrac{1}{2} m ~ v^{2} + K.E_{rot} \
&= dfrac{1}{2} m ~ v^{2} + dfrac{1}{2} ~ I ~ omega^{2} \
end{align*}
$$
As the textbook mentions, the tangential speed of a rotating object is given by
$$
begin{align*}
v &= r ~ omega \
omega &= dfrac{ v}{r}
end{align*}
$$
Since the problem assigns that the momentum of inertia equals to $dfrac{2}{5} ~ m ~ r^{2}$. Substituting from the previous calculations in the total kinetic energy equation:
$$
begin{align*}
K.E_{t} &= dfrac{1}{2} m ~ v^{2} + dfrac{1}{2} ~ I ~ omega^{2} \
&= dfrac{1}{2} m ~ v^{2} + dfrac{1}{2} ~ dfrac{2}{5} ~ m ~ r^{2} ~ dfrac{ v^{2}}{r^{2}} \
&= dfrac{1}{2} m ~ v^{2} + dfrac{1}{5} ~ m ~ v^{2} \
&= boxed{0.7 ~ m ~ v^{2}} \
end{align*}
$$
So, the correct answer is (D) $0.7 ~ m ~ v^{2}$
The radius of the turntable is $r = 0.08 mathrm{~m}$. The angular speed of the turntable is $omega = 45 mathrm{~rpm}$.
$textbf{Required: }$
Finding the linear speed.
As the textbook mentions, the tangential speed of a rotating object is given by
$$
begin{align*}
v &= r ~ omega \
&= 0.08 mathrm{~m} cdot 45 mathrm{~rpm} \
&= 0.08 mathrm{~m} cdot 45 cdot dfrac{ 2 pi }{ 60} mathrm{~rad/s} \
&= boxed{0.39 mathrm{~m/s}}
end{align*}
$$
So, the correct answer is (D) $0.38 mathrm{~m/s}$.
As the textbook mentions, the angular velocity of a rotating object is given by
$$
begin{align*}
vec{omega} &= dfrac{ vec{v} }{r} \
end{align*}
$$
Since the linear velocity is a vector, then the angular velocity is a vector.
As the textbook mentions, the angular acceleration is given by the change in angular velocity within a given time interval
$$
begin{align*}
vec{alpha} &= dfrac{ Delta vec{omega} }{ Delta t } \
&= dfrac{ vec{omega_{f}} – vec{omega_{i}} }{Delta t } \
end{align*}
$$
Since the angular velocity is a vector, then the angular acceleration is a vector, too.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
vec{tau} &= r ~ vec{F} \
end{align*}
$$
So, the torque is a vector while the force is a vector.
As the textbook mentions that the moment of inertia of the hollow sphere is given by
$$
begin{align*}
I &= dfrac{2}{3} ~ m ~ r^{2} \
end{align*}
$$
Since the radius and the mass of the object are scalars, then the moment of inertia is a scalar quantity. Therefore, the correct answer is (B) moment of inertia.
In order to evaulate the angular position, we use the following relation:
$$
begin{align*}
Delta theta &= theta_{f} – theta_{i} \
end{align*}
$$
So, the angular position is a scalar.
As the textbook mentions, the angular speed of a rotating object is given by
$$
begin{align*}
omega &= dfrac{ v }{r} \
end{align*}
$$
Since the linear speed is a scalar and equals the magnitude of the linear velocity, then the angular speed is a scalar.
As the textbook mentions that the torque is given by the product of force and distance
$$
begin{align*}
vec{tau} &= r ~ vec{F} \
end{align*}
$$
So, the torque is a vector while the force is a vector.
As the textbook mentions that the moment of inertia of the hollow sphere is given by
$$
begin{align*}
I &= dfrac{2}{3} ~ m ~ r^{2} \
end{align*}
$$
Since the radius and the mass of the object are scalars, then the moment of inertia is a scalar quantity. Therefore, the correct answer is (C) torque.
