All Solutions
Page 326: Lesson Check
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
end{align*}
$$
According to Newton’s second law:
$$
begin{align*}
sum F_{y} &= m ~ a_{cp} \
&= f_{cp} \
&= m ~ dfrac{v^{2} }{ r } \
end{align*}
$$
So, the direction of the force that produces the circular motion (centripetal force) is always directed toward the center of the circle. Also, the force that produces the circular motion (centripetal force) is directly proportional to the mass of the object and the square of the speed of the object. Also, the force that produces the circular motion (centripetal force) is inversely proportional to the radius of the circular path.
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
end{align*}
$$
According to Newton’s second law:
$$
begin{align*}
sum F_{y} &= m ~ a_{cp} \
&= f_{cp} \
&= m ~ dfrac{v^{2} }{ r } \
end{align*}
$$
So, the force that produces the circular motion (centripetal force) is directly proportional to the mass of the object and the square of the speed of the object. Also, the force that produces the circular motion (centripetal force) is inversely proportional to the radius of the circular path.
So, the direction of the force that produces the circular motion (centripetal force) is always directed toward the center of the circle.
The direction of the force that produces the circular motion (centripetal force) is always directed toward the center of the circle.
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
end{align*}
$$
In order to evaluate the linear accerlation, we use the following relation:
$$
begin{align*}
a &= dfrac{ d v}{ d t } \
&= dfrac{ d^{2} x }{ d t^{2} } \
end{align*}
$$
As we see from the previous relations, The main difference in the centripetal acceleration is a time-independent quantity. But the unit of both accelerations are the same. The regular acceleration unit is $mathrm{m/s^{2}}$ and the centripetal acceleration unit is $dfrac{ left( mathrm{~m/s} right)^{2} }{ mathrm{~m} } = dfrac{ mathrm{~m^{2}/s^{2}} }{ m } = mathrm{~m/s^{2}}$. Also, we know that the centripetal acceleration direction is perpendicular to the linear acceleration direction.
In order to evaluate the linear acceleration, we use the following relation:
$$
begin{align*}
a &= dfrac{ d v}{ d t } \
&= dfrac{ d^{2} x }{ d t^{2} } \
end{align*}
$$
As the previous relation said any change in velocity that occurs will make a change in the linear acceleration. Also, we know that the velocity and the acceleration are vectors that determine by both quantity and direction. So, any change in the direction of the car by the steering wheel will make a change in direction of acceleration of the car. So, the answer is “Yes, the steering wheel can produce an acceleration.”
Solve for the first case:
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,1} &= dfrac{ v_{1}^{2} }{ r_{1}} \
&= dfrac{ v^{2} }{ r } \
end{align*}
$$
Solve for the second case (the speed and the radius are doubled):
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp} &= dfrac{ v_{2}^{2} }{ r_{2}} \
&= dfrac{ left( 2~ v right)^{2} }{ 2 ~ r } \
&= dfrac{4 v^{2} }{ 2 ~ r } \
&= dfrac{ 2 v^{2} }{ r } \
&= 2 ~ a_{cp,1}
end{align*}
$$
So, the centripetal acceleration is doubled due to both the speed and the radius are doubled.
Solve for the object $A$:
As Newton’s second law for circular motion, the centripetal force is given by the mass time the centripetal acceleration:
$$
begin{align*}
f_{cp,A} &= m ~ a_{cp} \
&= m ~ dfrac{ v^{2} }{ r} \
&= 60 mathrm{~kg} times dfrac{ left( 1 mathrm{~m/s} right)^{2} }{ 5 mathrm{~m} } \
&= 12 mathrm{~N}
end{align*}
$$
So, the centripetal force of the object $A$ is $12 mathrm{~N}$.
Solve for the object $B$:
As Newton’s second law for circular motion, the centripetal force is given by the mass time the centripetal acceleration:
$$
begin{align*}
f_{cp,B} &= m ~ a_{cp} \
&= m ~ dfrac{ v^{2} }{ r} \
&= 5 mathrm{~kg} times dfrac{ left( 2 mathrm{~m/s} right)^{2} }{ 2 mathrm{~m} } \
&= 10 mathrm{~N}
end{align*}
$$
So, the centripetal force of the object $B$ is $10 mathrm{~N}$.
Solve for the object $C$:
As Newton’s second law for circular motion, the centripetal force is given by the mass time the centripetal acceleration:
$$
begin{align*}
f_{cp,C} &= m ~ a_{cp} \
&= m ~ dfrac{ v^{2} }{ r} \
&= 24 mathrm{~kg} times dfrac{ left( 2 mathrm{~m/s} right)^{2} }{ 8 mathrm{~m} } \
&= 12 mathrm{~N}
end{align*}
$$
So, the centripetal force of the object $c$ is $12 mathrm{~N}$.
Solve for the object $D$:
As Newton’s second law for circular motion, the centripetal force is given by the mass time the centripetal acceleration:
$$
begin{align*}
f_{cp,D} &= m ~ a_{cp} \
&= m ~ dfrac{ v^{2} }{ r} \
&= 2 mathrm{~kg} times dfrac{ left( 4 mathrm{~m/s} right)^{2} }{ 1 mathrm{~m} } \
&= 32 mathrm{~N}
end{align*}
$$
So, the centripetal force of the object $d$ is $32 mathrm{~N}$.
Therefore, the ranking is $D > A = C >B$.
Solve for the object $A$:
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 1 mathrm{~m/s} right)^{2} }{ 5 mathrm{~m} } \
&= 0.2 mathrm{~m/s^{2}}
end{align*}
$$
So, the centripetal acceleration of the object $A$ is $0.2 mathrm{~N}$.
Solve for the object $B$:
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,B} &= m ~ dfrac{ v^{2} }{ r} \
&= dfrac{ left( 2 mathrm{~m/s} right)^{2} }{ 2 mathrm{~m} } \
&= 2 mathrm{~m/s^{2}}
end{align*}
$$
So, the centripetal acceleration of the object $B$ is $2 mathrm{~m/s^{2}}$.
Solve for the object $C$:
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,C} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 2 mathrm{~m/s} right)^{2} }{ 8 mathrm{~m} } \
&= 0.5 mathrm{~m/s^{2}}
end{align*}
$$
So, the centripetal acceleration of the object $c$ is $0.5 mathrm{~m/s^{2}}$.
Solve for the object $D$:
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,D} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 4 mathrm{~m/s} right)^{2} }{ 1 mathrm{~m} } \
&= 16 mathrm{~m/s^{2}}
end{align*}
$$
So, the centripetal acceleration of the object $d$ is $0.5 mathrm{~m/s^{2}}$.
Therefore, the ranking is $D > B > C >A$.
The radius of the vertical circle is $r = 1.3 mathrm{~m}$.
$textbf{Required:}$
Finding the bucket’s speed.
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 1 mathrm{~m/s} right)^{2} }{ 5 mathrm{~m} } \
&= 0.2 mathrm{~m/s^{2}}
end{align*}
$$
According to Newton’s second law in the vertical direction:
$$
begin{align*}
sum F_{y} &= m ~ a_{cp} \
&= m ~ g \
&= m ~ dfrac{v^{2} }{ r } \
end{align*}
$$
Rearrange and solve for the bucket’s speed:
$$
begin{align*}
v &= sqrt{ g ~ r } \
&= sqrt{ 9.8 mathrm{~m/s^{2}} times 1.3 mathrm{~m} } \
&= 3.569 mathrm{~m/s}
end{align*}
$$
So, the bucket’s speed is $3.569 mathrm{~m/s}$.
The centripetal acceleration of the car is $a_{cp} = 4.4 mathrm~{m/s^{2}}$. The car’s speed is $v = 15 mathrm{~m/s}$.
$textbf{Required: }$
Finding the radius of the corner.
As textbook mentions the centripetal acceleration is given by the following relation
$$
begin{align*}
a_{cp,A} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 1 mathrm{~m/s} right)^{2} }{ 5 mathrm{~m} } \
&= 0.2 mathrm{~m/s^{2}}
end{align*}
$$
Rearrange and solve for the radius of the corner:
$$
begin{align*}
r &= dfrac{ v^{2} }{a_{cp,A}} \
&= dfrac{ left( 15 mathrm{~m/s} right)^{2} }{ 4.4 mathrm{~m/s^{2}} } \
&= 51.136 mathrm{~m}
end{align*}
$$
So, the radius of the corner is $51.136 mathrm{~m}$.
