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Class
Date
Concept-Development
Practice Page
10-2
The Flying Pig
The toy pig flies in a circle at constant speed. This
arrangement is called a conical pendulum because the
supporting string sweeps out a cone. Neglecting the action
of its flapping wings, only two forces act on the pig—
gravitational mg, and string tension T.
Vector Component Analysis:
Note that vector T can be resolved into two components—
horizontal Tx , and vertical Ty. These vector components
are dashed to distinguish them from the tension vector T.
Circle the correct answers.
1. If T were somehow replaced with Tx and Ty the pig
(would) (would not) behave identically to being
supported by T.
2. Since the pig doesn’t accelerate vertically, compared with the magnitude of mg, component Ty,
must be (greater) (less) (equal and opposite).
3. The velocity of the pig at any instant is (along the radius of) (tangent to) its circular path.
4. Since the pig continues in circular motion, component Tx must be a
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(centripetal) (centrifugal) (nonexistent) force, which equals (zero) (mv2/r).
Furthermore, Tx is (along the radius of) (tangent to) the circle swept out.
Vector Resultant Analysis:
5. Rather than resolving T into horizontal and
vertical components, use your pencil to
sketch the resultant of mg and T using the
parallelogram rule.
6. The resultant lies in a (horizontal) (vertical)
direction, and (toward) (away from) the center
of the circular path. The resultant of mg and T is
a (centripetal) (centrifugal) force.
CONCEPTUAL PHYSICS
Chapter 10
Circular Motion
55
Banked Airplanes
An airplane banks as it turns along a horizontal circular path in the air. Except for the thrust of its
engines and air resistance, the two significant forces on the plane are gravitational mg (vertical), and
lift L (perpendicular to the wings).
Vector Component Analysis:
With a ruler and a pencil, resolve vector L
into two perpendicular components,
horizontal Lx, and vertical Ly. Make these
vectors dashed to distinguish them from L.
Circle the correct answers.
1. The velocity of the airplane at any instant is
(along the radius of) (tangent to) its circular path.
2. If L were somehow replaced with Lx and Ly, the
airplane (would) (would not) behave the same as
being supported by L.
3. Since the airplane doesn’t accelerate vertically, component Ly must be
(greater than) (less than) (equal and opposite to) mg.
4. Since the plane continues in circular motion, component Lx must equal (zero) (mv2/r) , and be a
(centripetal) (centrifugal) (nonexistent) force. Furthermore, Lx is
Vector Resultant Analysis:
5. Rather than resolving L into horizontal and vertical
components, use your pencil to sketch the resultant
of mg and L using the parallelogram rule.
6. The resultant lies in a (horizontal) (vertical)
direction, and (toward) (away from) the center of
the circular path. The resultant of mg and L is a
(centripetal) (centrifugal) force.
7. The resultant of mg and L is the same as (Lx) (Ly).
8. Challenge: Explain in your own words why the resultant
of two vectors can be the same as a single component of one of them.
For any pair of vectors to be added, if ⌺Vy = 0, and ⌺Vx ≠ 0, the resultant
will be ⌺Vx.
CONCEPTUAL PHYSICS
56
Chapter 10
Circular Motion
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(along the radius of) (tangent to) the circular path.
Name
Class
Date
Banked Track
A car rounds a banked curve with just the right speed so
that it has no tendency to slide down or up the banked
road surface. Shown below are two main forces that act
on the car perpendicular to its motion—gravitational
mg and the normal force n (the support force of the
surface).
Vector Component Analysis:
Note that vector n is resolved into two
perpendicular components, horizontal nx and
vertical ny. As usual, these vectors are dashed
to distinguish them from n.
Circle the correct answers.
1. If n were somehow replaced with
nx and ny, the car (would) (would not)
behave identically to being supported by n.
2. Since the car doesn’t accelerate vertically, component ny must be
(greater than) (equal and opposite to) (less than) mg.
© Pearson Education, Inc., or its affiliate(s). All rights reserved.
3. The velocity of the car at any instant is (along the radius of) (tangent to) its circular path.
4. Since the car continues in uniform circular motion, component nx must equal (zero) (mv2/r)
and be a (centripetal) (centrifugal) (nonexistent) force. Furthermore, nx is
(along the radius of) (tangent to) the circular path.
Vector Resultant Analysis:
5. Rather than resolving n into horizontal
and vertical components, use your pencil
to sketch the resultant of mg and n using
the parallelogram rule.
6. The resultant lies in a (horizontal) (vertical)
direction, and (toward) (away from) the
center of the circular path. The resultant of
mg and n is a (centripetal) (centrifugal) force.
7. The resultant of mg and n is the same
as (nx) (ny), and provides the
(centripetal) (centrifugal) force.
CONCEPTUAL PHYSICS
Chapter 10
Circular Motion
57
Leaning
When turning a corner on a bicycle, everyone knows that you’ve got to lean “into the curve.” What is
the physics of this leaning? It involves torque, friction, and centripetal force (mv2/r).
First, consider the simple case of riding a bicycle along a straight-line path.
Except for the force that propels the bike forward (friction of the road in the
direction of motion) and air resistance (friction of air against the direction of
motion), only two significant forces act: weight mg and the normal force n. (The
vectors are drawn side-by-side, but actually lie along a single vertical line.)
Circle the correct answers.
1. Since there is no vertical acceleration, we can say that the magnitude of
(n > mg) (n < mg) (n = mg) , which means that in the vertical direction,
(⌺Fy > 0) (⌺Fy < 0) (⌺Fy = 0).
2. Since the bike doesn’t rotate or change in its rotational state, then the total
torque is (zero) (not zero).
Now consider the same bike rounding a corner. In order to safely make the turn, the bicyclist leans in
the direction of the turn. A force of friction pushes sideways on the tire toward the center of the curve.
R
3. The friction force, f, provides the centripetal force that produces a
curved path. Then (f = mv2/r) (f ≠ mv2/r).
4. Consider the net torque about the center of mass (CM) of the
bike-rider system. Gravity produces no torque about this point, but n
(clockwise) (counterclockwise) rotation, and the one involving f
tends to produce (clockwise) (counterclockwise) rotation.
These torques cancel each other when the resultant of vectors n
and f pass through the CM.
5. With your pencil, use the parallelogram rule and sketch in the resultant of vectors n and f. Label your resultant R. Note the R passes
through the center of mass of the bike-rider system. That means that
R produces (a clockwise) (a counterclockwise) (no) torque about
the CM. Therefore the bike-rider system
(topples clockwise) (topples counterclockwise) (doesn’t topple).
CONCEPTUAL PHYSICS
58
Chapter 10
Circular Motion
© Pearson Education, Inc., or its affiliate(s). All rights reserved.
and f do. The torque involving n tends to produce