# Concept Question: Normal Force ```Concept Question: Normal Force
Consider a person standing in an elevator that
is accelerating upward. The upward normal
force N exerted by the elevator floor on the
person is
1.  larger than
2.  identical to
3.  smaller than
the downward force of gravity on the person.
Concept Question: Normal Force
Answer 1. The normal force on the person is greater than
the gravitational force on the person because the normal
force must also accelerate the person, as well as oppose
the gravitational force. Thus N – mg = ma implies that
N &gt; mg. So the magnitude of upward normal force is larger
than the magnitude of the downward gravitational force.
Concept Q.: Constraints and Pulleys
A massless rope is attached to the ceiling at one end and held
stationary with tension T at the other end. Through a system of
frictionless pulleys it supports a body of mass m. What is the
value of m?
1.
4T / g
2.
4T sin θ / g
3.
(2 + 2 )T / g
4.
(2 + 2 )T sin θ / g
5.
5T / g
Con. Q. Ans.: Constraints and Pulleys
Answer 3. The object is static, so the
vertical sum of the forces acting on the
object is zero
2T + 2T sin 45 − mg = 0
Thus the value of the mass is
m = (2 + 2 )T / g
Concept Question: Angular Speed
Object A sits at the outer edge (rim) of a merry-go-round, and
object B sits halfway between the rim and the axis of rotation.
The merry-go-round makes a complete revolution once every
thirty seconds. The magnitude of the angular velocity of Object
B is
1.
2.
3.
4.
half the magnitude of the angular velocity of Object A .
the same as the magnitude of the angular velocity of Object
A.
twice the the magnitude of the angular velocity of Object A .
impossible to determine.
Concept Question Ans.: Angular Speed
Object A sits at the outer edge (rim) of a merry-go-round, and
object B sits halfway between the rim and the axis of rotation.
The merry-go-round makes a complete revolution once every
thirty seconds. The magnitude of the angular velocity of Object
B is
Answer: 2. All points in a rigid body rotate with the same
angular velocity.
Concept Question: Circular
Motion
An object moves counter-clockwise along the circular path shown
below. As it moves along the path its acceleration vector
continuously points toward point S. The object
1.
speeds up at P, Q, and R.
2.
3.
4.
5.
6.
7.
slows down at P, Q, and R.
speeds up at P and slows down at R.
slows down at P and speeds up at R.
speeds up at Q.
slows down at Q.
No object can execute such a motion.
Concept Question Circular
Answer 3. At the point P the
acceleration has a positive tangential
component so it is speeding up. At the
point S the acceleration has a zero
tangential component so it is moving at
a constant speed. At the Point R the
acceleration has a negative tangential
component so it is slowing down.
Concept Question: Circular
Motion and Force
A pendulum bob swings down and is moving fast at the lowest point
in its swing. T is the tension in the string, W is the gravitational force
exerted on the pendulum bob. Which free-body diagram below best
represents the forces exerted on the pendulum bob at the lowest
point? The lengths of the arrows represent the relative magnitude of
the forces.
Concept Question Circular
Answer d. The bob is undergoing circular motion. It is accelerating
towards the center. Newton’s Second Law gives
T − W = mrω 2
hence T = W + mrω 2
so T &gt; W .
Concept Q.: Pushing a Baseball Bat
A baseball bat is pushed with a force F. You may
assume that the push is instantaneous. Which of the
following locations will the force produce an acceleration
of the center of mass with the largest magnitude?
1.
2.
3.
4.
Pushing at Position 1.
Pushing at Position 2 (center of mass).
Pushing at Position 3.
Pushing at 1,2,and 3 all produce the same magnitude of
acceleration of the center of mass/
Con. Q. Ans. : Pushing a Baseball Bat
Answer 4. The external force is equal to the total
mass times the instantaneous acceleration of the
center-of-mass. It doesn’t matter where the external
force acts with regards to the center-of-mass
acceleration.
Concept Question: Losing Mass But
Not Momentum
Consider an ice skater gliding on ice holding a bag of
sand that is leaking straight down with respect to the
moving skater. As a result of the leaking sand, the speed
of the skater
1.  increases.
2.  does not change.
3.  decreases.
4.  not sure.
Con. Q. Ans.: Losing Mass But Not
Momentum
Answer 2. The sand leaves the bag with the same
horizontal speed as the skater. The momentum of the
system of the skater and sand does not change and so
the speed of the skater does not change.
Concept Question: Pushing Carts
Consider two carts, of masses m and 2m, at rest on an air
track. If you push one cart for 3 seconds and then the other
for the same length of time, exerting equal force on each,
the kinetic energy of the light cart is
1) larger than
2) equal to
3) smaller than
the kinetic energy of the heavy car.
Concept Question Ans. : Pushing
Carts
Answer 1. The kinetic energy of an object can be written as
1 2 p2
K = mv =
2
2m
Because the impulse is the same for the two carts, the
change in momentum is the same. Both start from rest so
they both have the same final momentum. Since the mass
of the lighter cart is smaller than the mass of the heavier
cart, the kinetic energy of the light cart is larger than the
kinetic energy of the heavy cart.
Concept Question: Potential Energy and
Inverse Square Gravity
A comet is speeding along a hyperbolic orbit toward the Sun.
While the comet is moving away from the Sun, the change in
potential energy of the Sun-comet system is:
(1) positive
(2) zero
(3) negative
Energy and Inverse Square Gravity
Answer: 1. The displacement of the
comet has a component in the
opposite direction as the force on the
comet so the work done is negative.
(The comet's acceleration is always
toward the Sun; when the comet
moves away from the Sun, the work is
negative.) Therefore the change in
potential energy of the Sun-comet
system is positive.
Concept Question: Recoil
Suppose you are on a cart, initially at rest on a track
with very little friction. You throw balls at a partition
that is rigidly mounted on the cart. After the balls
bounce straight back as shown in the figure, is the
cart
1. moving to the right?
2.  moving to the left?
3.  at rest.
Concept Question: Recoil
are no horizontal external
forces acting on the system,
the momentum of the cart,
person and balls must be
constant. All the balls
bounce back to the right,
then in order to keep the
momentum constant, the
cart must move forward.
Concept Q.: Moment of Inertia Same
Masses
All of the objects below have the same mass. Which of the objects&quot;
has the largest moment of inertia about the axis shown?&quot;
(1) Hollow Cylinder
(2) Solid Cylinder
(3)Thin-walled Hollow Cylinder
Concept Q. Ans.: Moment of Inertia
Same Masses
All of the objects below have the same mass. Which of the objects&quot;
has the largest moment of inertia about the axis shown?&quot;
Answer 3. The mass distribution for the thin-hollow walled cylinder
is furthest from the axis, hence it’s moment of inertia is largest.
Concept Question: Kinetic Energy
A disk with mass M and radius R is spinning with
angular speed ω about an axis that passes through
the rim of the disk perpendicular to its plane.
Moment of inertia about cm is (1/2)M R2. Its total
kinetic energy is:
1. (1/4)M R2 ω2
4. (1/4)M Rω2
2. (1/2)M R2 ω2
5. (1/2)M Rω2
3. (3/4)M R2 ω2
6. (1/4)M Rω
Concept Q. Ans.: Kinetic Energy
Answer 3. The parallel axis theorem states the moment
of inertia about an axis passing perpendicular to the
plane of the disc and passing through a point on the
edge of the disc is equal to
I edge = I cm + mR 2
The moment of inertia about an axis passing
perpendicular to the plane of the disc and passing
through the center of mass of the disc is equal to
I cm = (1 / 2)mR 2
Therefore
I edge = (3 / 2)mR 2
The kinetic energy is then
K = (1 / 2)I edgeω 2 = (3 / 4)mR 2ω 2
Concept Q. Mag. of Angular Momentum
In the above situation where a
particle is moving in the x-y plane
with a constant velocity, the
magnitude of the angular
momentum
about the point S (the origin)
1)  decreases then increases
2)  increases then decreases
3)  is constant
4)  is zero because this is not
circular motion
Concept Q. Ans.: Mag. of Ang. Mom.
Solution 3. As the particle moves in the positive xdirection, the perpendicular distance from the origin to
the line of motion does not change and so the
magnitude of the angular momentum about the origin is
constant.
Concept Q.: Angular Momentum of
Disk
A disk with mass M and radius R is spinning with angular
speed ω about an axis that passes through the rim of the disk
perpendicular to its plane. The magnitude of its angular
momentum is:
1
1.
M R 2ω 2
4
2.
1
M R 2ω 2
2
3
3.
M R 2ω 2
2
1
4.
M R 2ω
4
5.
1
M R 2ω
2
3
6.
M R 2ω
2
Concept Q. Ans.: Ang. Mom. of
Disk
Answer 6. The moment of inertia of the disk about an axis
that passes through the rim of the disk perpendicular to its
plane is I = Icm + MR2 = (3/2)MR2 . So the magnitude of its
angular momentum is L = (3/2)MR2 ω.
Concept Q.: Change in Angular Mom.
•
1.
2.
3.
A person spins
a tennis ball on a string in a

horizontal circle withv velocity
(so that the axis of
rotation is vertical). At the point indicated below, the
ball is given a sharp F
blow (force ) in the forward
direction. This causes a change in angular
ΔLmomentum
in the
r̂
θ̂
k̂
direction
direction
direction
Concept Q. Ans.: Change in Ang. Mom.
circle points in the positive
k̂ -direction. The change in
the angular momentum about the center of the circle is
proportional to the angular impulse about the center of
the circle which is in the direction of the torque
Concept Question: Rolling
Without Slipping
If a wheel of radius R rolls without slipping through an
angle θ, what is the relationship between the distance
the wheel rolls, x, and the product Rθ?
1. x &gt; Rθ.
2. x = Rθ.
3. x &lt; Rθ.
Concept Q. Ans. : Rolling
Without Slipping
Answer 2. Rolling without slipping condition, x = Rθ.
Concept Question: Pulling a Yo-Yo
1
Concept Q. Ans.: Pulling a Yo-Yo 1
Answer 1. For forces below a
fixed maximum value, the torque
the ground and yo-yo is only due
to the force F and produces an
angular acceleration directed
into the plane of the figure.
Hence the cylinder rolls to the
right, in the direction of F,
winding up the string.The torque
about the center of mass due to
the force of friction is larger in
magnitude than the torque due
to the pulling force.
Concept Question: Cylinder Rolling
Down Inclined Plane
A cylinder is rolling without slipping down an inclined plane.
The friction at the contact point P is
1.  Static and points up the inclined plane.
2.  Static and points down the inclined plane.
3.  Kinetic and points up the inclined plane.
4.  Kinetic and points down the inclined plane.
5.  Zero because it is rolling without slipping.
35
Concept Q. Ans.: Cylinder Rolling
Down Inclined Plane
Answer 1. The friction at the contact point P is static and
points up the inclined plane. This friction produces a torque
about the center of mass that points into the plane of the
figure. This torque produces an angular acceleration into
the plane, increasing the angular speed as the cylinder rolls
down.
36
Concept Question: Angular Collisions
A long narrow uniform stick lies motionless on ice
(assume the ice provides a frictionless surface).
The center of mass of the stick is the same as the
geometric center (at the midpoint of the stick). A
puck (with putty on one side) slides without spinning
on the ice toward the stick, hits one end of the stick,
and attaches to it.
Which quantities are constant?
1.  Angular momentum of puck about center of mass of
stick.
2.  Momentum of stick and ball.
3.  Angular momentum of stick and ball about any
point.
4.  Mechanical energy of stick and ball.
5.  None of the above 1-4.
6.  Three of the above 1.4
7.  Two of the above 1-4.
Concept Q. Ans.: Angular Collisions
(2) and (3) are correct. There are no external
forces acting on this system so the momentum of
the center of mass is constant (1). There are no
external torques acting on the system so the
angular momentum of the system about any point
is constant (3) . However there is a collision force
acting on the puck, so the torque about the center
of the mass of the stick on the puck is non-zero,
hence the angular momentum of puck about
center of mass of stick is not constant. The
mechanical energy is not constant because the
collision between the puck and stick is inelastic.
Concept Question: SHM Velocity
A block of mass m is attached to a spring with spring constant
k is free to slide along a horizontal frictionless surface.
At t = 0 the block-spring system is stretched an amount
x0 &gt; 0 from the equilibrium position and is released from rest.
What is the x -component of the velocity of the block when it
first comes back to the equilibrium?
4
T
1.
vx = −x0
3.
k
vx = −
x0
m
2.
4.
vx = x0
4
T
vx =
k
x0
m
Answer 3. The particle starts with potential energy. When it
first returns to equilibrium it now has only kinetic energy. Since
the energy of the block-spring system is constant:
(1 / 2)mvx2 = (1 / 2)kx02
Take positive root because object is moving in negative xdirection when it first comes back to equilibrium, and xo &gt; 0, we
require that vx,o &lt; 0, therefore
k
vx = −
x0
m
Concept Question: Energy Diagram 3
A particle with total mechanical energy
E has position x &gt; 0 at t = 0
1) escapes to infinity
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
Concept Q. Ans.: Energy Diagram 3
Solution 3. The range of motion
for the particle is limited to the
regions in which the kinetic energy
is either zero or positive, so the
particle is confined to move around
the region surrounding a . The
motion will be periodic but not
simple harmonic motion because
the potential energy function is not
a quadratic function and only for
functions will the motion be simple
harmonic. Hence the particle
oscillates around a.
42
Concept
Question: Rotating Vector

A vector A(t) of fixed length A is rotating about the z-axis at
an angular speed
ω. At t = 0 it is pointing in the positive y
direction. dA(t) / dt is given by
1.
2.
3.
4.
5.
6.
7.
8.

dA(t) / dt = ω Asin(ω t) î − ω Acos(ω t) ĵ

dA(t) / dt = −ω Asin(ω t) î + ω Acos(ω t) ĵ

dA(t) / dt = ω Asin(ω t) î − ω Acos(ω t) ĵ

dA(t) / dt = −ω Asin(ω t) î − ω Acos(ω t) ĵ

dA(t) / dt = ω Acos(ω t) î + ω Asin(ω t) ĵ

dA(t) / dt = −ω Acos(ω t) î + ω Asin(ω t) ĵ

dA(t) / dt = ω Acos(ω t) î − ω Asin(ω t) ĵ

dA(t) / dt = −ω Acos(ω t) î − ω Asin(ω t) ĵ
Concept Q. Ans. :
Time Derivative of Rotating Vector

A(t) = − Asin(ω t) î + Acos(ω t) ĵ
Therefore

dA / dt = −ω Ascos(ω t) î − ω Asin(ω t) ĵ

and is perpendicular to A(t)
Concept Question: Rotating Rod
Consider a massless rod of length I with two point-like objects of mass m
at each end, rotating about the vertical z-axis with angular speed ω. There
is a sleeve on the axis of rotation. At the moment shown in the figure, two
forces are acting on the sleeve. The direction of the change of the angular
momentum about the center of mass points
1.  along the z-axis.
ω
2.  along the line formed by the rod.
3.  into the plane of the figure.
4.  out of the plane of the figure.
the center of mass points into
the plane of the figure.
Therefore the direction of the
change of the angular
mass points into the plane of
the figure.
Concept Question: Stabilizing a Turning Car
When making a turn every car
has a tendency to roll over
because its center of mass is
above the plane where the
Imagine a race car going counter
clockwise on a circular track. It
could mitigate this effect by
mounting a gyroscope on the car.
To be effective the angular
velocity vector of the gyro should
point
2) behind
3) to the left
4) to the right
5) up
6) down
Concept Q. Ans.: Stabilizing a Turning Car