chapter 10 - santiago canyon college 10 lecture notes physics 210 page 1 santiago canyon college...

Chapter 10 Lecture Notes

Physics 210 Santiago Canyon College

Chapter 10

Rotational Kinematics

Above we have a rigid mass that is rotating about a fixed axis. Let’s look at the motion of one of the millions of particles that make up this object.

Our particle is currently at point P as it rotates counterclockwise about the axis. We can see that our particle will trace out a circular path (even if the object we not circular). The displacement of our particle from its starting point (along the reference line) is: = θs r (1) As our object continues to rotate, it will move from some initial angle θi to some final angle θf along our circle of radius ir . Thus the angular displacement of the particle is:

∆θ = θ − θf i As we did in the case of linear motion, we can define the average angular speed:

−∆θ ω = ∆ 1

avgradunits 1 or 1 s

t sec



To find the instantaneous angular speed we take the limit as the time interval goes to zero:

∆ →

∆θ θω = =

∆t 0

dlimt dt

(2)

As the instantaneous angular speed changes, we know that our particle is accelerating. The average angular acceleration is:

−∆ω α = 1 ∆ 2

avg 2

radunits or 1 st s

Following the procedure above, the instantaneous angular acceleration is:

∆ →

∆ω ωα = =

∆t 0

dlimt dt

(3)

When an object is rotating about a fixed axis, every particle on the object rotates through the same angle, has the same angular speed, and the same angular acceleration. The angular displacement θ , angular velocity ω , and angular acceleration α are analogous to the linear displacement x, linear velocity v, and linear acceleration a. Therefore, for constant angular acceleration, we can derive rotational kinematic equations. If we start with the angular acceleration:

ωα = ⇒ ω = α

d d dtdt

Integrating both sides:

ω

ωω = α∫ ∫

f

i

t

0d dt



ω − ω = αf i t (4) If we now look at the angular velocity:

θω = ⇒ θ = ω

d d dtdt

Integrating both sides:

( )θ

θθ = ω = ω + α∫ ∫ ∫

f

i

t t

i0 0d dt t

θ − θ = ω + α 2f i i

1t t2

(5)

We can also show that: ( )ω = ω + α θ − θ2 2

f i f i2 (6) Example 1: A CD has a playing time of 74 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolutions per minute (rpm). At the end of the music, the CD is rotating at 210 rpm. Find the magnitude of the average acceleration of the CD. Express your answer in rad/s2. Solution:



Example 2: A floor polisher has a rotating disk of radius 15 cm. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 45 s, in order to buff an especially scuffed area of the floor. How far (in meters) does a spot on the outer edge of the disk moving during this time? Solution:



Rotational Energy Recall that the kinetic energy of a particle is:

= 21K mv2

If we assume that a rotating object is made up of many particles, then the kinetic energy of the object would be:

= ∑ 2i i

i

1K m v2

Substituting the fact that = ωv r into the above gives:

= ω∑ 2 2i i i

i

1K m r2

We know that all particles that make up a rigid object rotate at the same rate. Therefore:

= ω = ω∑2 2 2i i

i

1 1K m r I2 2

(7)

I is the moment of inertia. Calculating the Moment of Inertia The moment of inertia about an axis tells us how much a rotating object resists changes in its angular motion. If we have a system of discrete masses:

= ∑ 2i i

iI m r

For a continuous system, we imagine that the object is divided into many elements of mass ∆ im . Now the moment of inertia is:



∆∑ 2

i ii

I r m

Now we take the limit as the mass of each element goes to zero:

∆ →= ∆ =∑ ∫

i

2 2i im 0 i

I lim r m r dm (8)

Imagine that we have a uniform rod of mass M and length L. The mass per unit length of the rod is:

µ =ML

Since the mass is linearly distributed, the differential mass element is:

= µ =Mdm dx dxL

The rod is going to rotate about an axis through the center of mass. The moment of inertia is:



−

−

=

=

=

=

∫

∫

2

L22

L2

L3 2

L2

2

I x dm

M x dxL

M x3L

ML12

What if the axis of rotation does not pass through the center of mass of the object?



We could redo the integral calculation but there is an easier way. We can use the parallel-axis theorem, which states:

= + 2CMI I MD

Let’s say that we wanted to rotate the previous rod about one end instead of the center. In this case:

=LD2

Now the moment of inertia is:

= + = + = + =

22 2 2 22

CMML L ML ML MLI I MD M12 2 12 4 3

Is this correct? We can check this by redoing the inertia integral:

=

=

=

=

∫

∫

2

L 2

0

L3

02

I r dm

M x dxL

M xL 3

ML3



Example 3: Use the parallel axis theorem to find the moment of inertia of a sold sphere of mass M and radius R about an axis which is tangent to the sphere. Solution:



Example 4: A 1.0 m diameter wagon wheel consists of a thin rim having a mass of 8.0 kg and six spokes, each having a mass of 1.2 kg. Determine the moment of inertia of the wagon wheel for rotation about its axis. Solution:



Torque The tendency of a force to cause an object to rotate is measured by a vector quantity called torque.

We want to look at the tangential forces on the object above:

= = αt tF ma m r

= α2trF mr

τ = α 2mr

The net torque on an object is:

τ = α = α = α∑ ∑ ∑2 2i i i i

i im r m r I

τ = α∑ I (9) Equation (9) is called Newton’s 2nd law for rotation. To compute the torque we want to look at the tangential forces applied at a point.



In the diagram above, the tangential force is:

= φtF Fsin Since this force is applied a distance r from the point of rotation, the torque applied is:

τ = = φtF r Fsin r Instead of looking at the tangential force, we can simply look at the line along which the force is applied (called the line of action).

Now to compute the torque we need to find the perpendicular distance from the point of rotation to the line of action (this is called the lever arm). The length of the lever arm for the previous diagram is:

φrsinl =



Now the applied torque is:

τ = φF Frsinl =

Two blocks having masses 1m and 2m are connected to each other by a light cord that passes over a frictionless pulley that has a moment of inertia I and radius R. What are the tensions in the string and the acceleration of the system? New steps because of the rotation of the pulley: We must draw a free body diagram for the masses and the pulley. Tensions may not be uniform throughout the rope due to the net torque required for the pulley to rotate.

From the free body diagram for 1m : = =∑ x 1 1F T m a (10)



From the free body diagram for 2m : = − =∑ y 2 2 2F m g T m a (11) Now we apply Newton’s 2nd law for rotation to the pulley:

( ) τ = − = α = ⇒ − =

∑ 2 1 2 1 2

a IaT R T R I I T TR R

(12)

Adding equation (10) to equation (11): ( )− + = + α1 2 2 1 2T T m g m m (13) Now we add equation (13) to equation (12):

= + + ⇒ = + +

22 1 2 2

1 2 2

m gIm g m m a aIR m m

R

(14)

Substituting equation (14) into equation (10) gives:

= = + +

1 21 1

1 2 2

m m gT m aIm m

R

Finally substitute equation (14) into equation (11) :

( ) + = − =+ +

2 1 2

2 2

1 2 2

Im g mRT m g a

Im mR



Example 5:

A 4 kg block resting on a frictionless horizontal ledge is attached to a string that passes over a pulley and is attached to a hanging block of mass 2 kg. The pulley is a uniform disk of radius 8 cm and 0.6 kg mass.

a. Find the speed of the 2 kg block after it falls from rest a distance 2.5 m.

b. What is the angular velocity of the pulley at this time? c. What is the acceleration of each block? d. What are the tensions in the string?

Solution:



Example 6: The system shown below is released from rest. The 30 kg block is 2 m above the ledge. The pulley is a uniform disk with a radius of 10 cm and a mass of 5 kg. Find

a. the speed of the 30 kg block just before it hits the ledge, b. the angular speed of the pulley at that time, c. the tensions in the strings, and d. the time it takes for the 30 kg block to reach the ledge.

Solution:



Rolling Without Slipping

As the cylinder above rolls along a straight path, the center of mass moves in a straight line (a point on the surface moves in a path called a cycloid). As our cylinder rotates through an angle φ , the center of mass moves a distance = φs R . Therefore, the linear speed of the center of mass is:

( ) φ= = φ = = ωcm

ds d dv R R Rdt dt dt

The acceleration of the center of mass is:

ω= = = αcm

cmdv da R R

dt dt

Our cylinder is moving linearly and rotating at the same time, this means that the kinetic energy is:

= + ω2 2cm cm

1 1K Mv I2 2



Example 7: A 0.1 kg yo-yo consists of two solid disks of radius 10 cm joined together by a massless rod of radius 1 cm and a string wrapped around the rod. One end of the string is held fixed and is under constant tension T as the yo-yo is released. Find the acceleration of the yo-yo and the tension T. Solution:



Example 8: A uniform sphere rolls without slipping down an incline. What must be the angle of the incline if the linear acceleration of the center of mass of the sphere is 0.2g? Solution:



Vector Cross Products The vector (cross) product of ( ) ×

A and B A B produces a third vector

C of magnitude: = φC ABsin (15) Cross products are a way of multiplying two vectors to produce another vector (unlike the dot product which multiplies two vectors together to produce a scalar). The resultant vector will be orthogonal to the plane which contains the two original vectors. The direction of the resultant vector can be determined in two ways: The right hand rule: When using the right hand rule, the fingers of the right hand point in the direction of the first vector with the palm of the right hand facing in the direction of the second vector. Then the thumb of the right hand will point in the direction of the resultant vector.

Determinant Method: To use the determinant method, we first write our two vectors in component form:

= + +

x y zˆˆ ˆA A i A j A k

= + +

x y zˆˆ ˆB B i B j B k



Now the cross-product is:

= × =

x y z

x y z

ˆˆ î j kC A B A A A

B B B

( ) ( ) ( )

= − +

= − + − + −

y z x yx z

x zy z x y

y z z y z x x z x y y x

A A A AA A ˆˆ î j kB BB B B B

ˆˆ î A B A B j A B A B k A B A B

To see how this works, consider the following two vectors:

= + − = − ˆ ˆˆ ˆ Â i 2j 3k and B 4j k

The cross product is:

− −= − = − +

− −−

ˆˆ î j k2 3 1 3 1 2ˆˆ ˆC 1 2 3 i j k4 1 0 1 0 4

0 4 1

= + + ˆˆ ˆ10i j 4k

Properties of Cross Products Cross products obey the following properties:

1. × =

A A 0 2. × = − ×

A B B A 3. ( ) ( ) ( )× + = × + ×

A B C A B A C

4. ( ) × = × + ×

d dB dAA B A Bdt dt dt



Example 9: Find = ×

C A B for each of the following: a. = + = − ˆˆ ˆ Â 2i 4k B i j

b. = = − ˆˆ Â 3j B i 2k

c. = + − = + ˆ ˆˆ ˆ Â i 2j k B 2i 3k

Solution:



Angular Momentum Angular Momentum of a Particle We have previously defined the linear momentum as:

= p mv

But what if the particle is moving in a circle? If a particle is moving along a circular path of radius r, we can define an angular version of momentum.

= ×

L r p L is the angular momentum and has a magnitude: = φL mvrsin (16) Since the angular momentum is defined using the cross product, the direction of the angular momentum vector is determined using the right hand rule. How does the angular momentum relate to torque? We have previously seen that the torque is defined as:

τ = × ⇒ τ = ×∑ ∑

r F r F We have previously seen that Newton’s 2nd law can be written as:

=∑

dpFdt

Combining these two results:

τ = ×∑

dprdt

(17)



Now we have related the net torque to the linear momentum. To finish relating it to the angular momentum, we need to start with:

= ×

L r p Take the derivative of both sides of the equation:

( )

( )

= ×

= × + ×

= × + ×

dL d r pdt dt

dp drr pdt dtdpr v mvdt

= ×

dprdt

(18)

Looking at equations (17) and (18) we can see that:

= τ∑

dLdt

Angular Momentum of a Rigid Object Imagine that we have a rigid object in the xy plane rotating about the z axis. We know that any point on the object moves in a circular path; therefore, the angular momentum of some piece im is:

( )= = ω = ω2i i i i i i i i iL m r v m r r m r

To find the total angular momentum, we sum over all the masses:

= = ω = ω∑ ∑ 2i i i

i iL L m r I



If we take the derivative of both sides of the equation above:

ω= ω + = α

dL dI dI Idt dt dt

Combining with previous results:

τ = α =∑ dLIdt

Conservation of Angular Momentum We have previously shown that torque and angular momentum are related by:

τ =∑

sysdLdt

If the net torque acting on the system is zero, then we have:

= ⇒ =

syssys

dL0 L constant

dt

This is the law of conservation of angular momentum and it means that the initial and final angular momenta of a system are equal. From our previous work we know that = ωL I for rigid objects. Therefore, we can rewrite the conservation of angular momentum as:

ω = ωi i f fI I How can we use this? Let’s look at a merry-go-round. A merry-go-round consists of a horizontal platform in the shape of a circular disk that rotates about a frictionless, vertical axis. The platform has a mass M = 100 kg with a radius R = 2.0 m. A person who is initially standing at the outer rim of the disk, walks slowly towards its center. That mass of the person is



m = 60 kg and the system’s initial angular speed is 2.0 rad/s. What is the angular speed when the person is 0.5 m from the center? Find the initial moment of inertia of the merry-go-round plus the person:

= +

= +

i Disk person, initial

2 2

I I I

1 MR mR2

Find the final moment of inertia for the system:

= +

= +

f Disk person, final

2 2

I I I

1 MR mr2

Apply the conservation of angular momentum:

ω = ωi i f fI I

+ ω = ω = ω =

+

2 2

if i i

2 2f

1 MR mRI rad2 4.11I sMR mr2

Is kinetic energy conserved?

= ω = + ω = 2 2 2 2

i i i i1 1 1K I MR mR 880 J2 2 2

= ω = + ω =

2 2 2 2f f f f

1 1 1K I MR mr 1800 J2 2 2



Example 10: You stand on a frictionless platform that is rotating with an angular speed of 1.5 rev/s. Your arms are outstretched, and you hold heavy weights in each hand. The moment of inertia of you, the extended weights, and the platform is ∗ 26 kg m . When you pull the weights in towards your body, the moment of inertia decreases to ∗ 21.8 kg m .

a. What is the resulting angular speed of the platform? b. What is the change in kinetic energy of the system? c. Where did the increase in energy come from?

Solution:



Example 11: The rotational inertia of a collapsing spinning star drops to 1/3 its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy? Solution:



Equilibrium Consider the following objects:

1. A book resting on a table 2. A book sliding along a frictionless table at a constant speed 3. A bicycle wheel moving along a straight path at a constant speed

For each of these objects, the following are true:

1. The linear momentum ( )p of the center of mass is constant.

2. The angular momentum ( )L is constant. These three objects are in equilibrium. The two requirements for equilibrium are:

p = constant and

L = constant Recall that Newton’s 2nd law for translational motion can be written as:

= =∑

dpF 0dt

because the linear momentum is constant

Similarly, from Newton’s 2nd law for rotational motion we have:

τ = =∑

dL 0dt

because the angular momentum is constant

Therefore, the conditions for equilibrium are:

=∑

F 0 and τ =∑ 0 If the linear and angular momenta are both zero, then our object is in static equilibrium.



Examples A uniform beam of length L whose mass m is 1.8 kg rests with its ends on two digital scales. A uniform block whose mass M is 2.7 kg rests on

the beam, its center a distance L4

from the beam’s left end. What do the

scales read?

Our conditions for equilibrium are:

= τ =∑ ∑

F 0 & 0 If we look at the vertical forces: = + − − =∑ y L RF F F Mg mg 0 (19) This equation has two unknowns ( ) L RF & F so we will need to look at the torque equation to solve the system. Since our object is not rotating, we can choose any perpendicular axis as our axis of rotation. In this case we will choose the position of the left scale. Therefore, our net torque equation is (assuming rotations that would cause clockwise rotations are negative):

LF RF

Mg mg Left

Right



τ = − − + =

∑ RL LMg mg F L 04 2

(20)

Solving for RF :

( ) ( )( ) = + = + =

2

R

m9.8g sF 2m M 2 1.8 kg 2.7 kg 15 N4 4

Substituting this into equation (19) gives:

( ) ( ) = + − = + − =

L R 2

mF M m g F 2.7 kg 1.8 kg 9.8 15 N 29 Ns

Example 12:

A 90 N board 12 m long rests on two supports, each 1 m from the end of the board. A 360 N block is placed on the board 3 m from one end. Find the force exerted by each support on the board. Solution:



Example 13:

A 10 m beam of mass 300 kg extends over a ledge as shown. The beam is not attached, but simply rests on the surface. A 60 kg student intends to position the beam so that he can walk to the end of it. How far from the edge of the ledge can the beam extend? Solution:



Example 14:

The diving board shown in the figure has a mass of 30 kg. Find the force on the supports when a 70 kg diver stands at the end of the diving board. Solution:



Ladder Problems A uniform 5 m ladder weighing 60 N leans against a frictionless vertical wall. The foot of the ladder is 3 m from the wall. What is the minimum coefficient of static friction necessary between the ladder and the floor if the ladder is not to slip?

Write the sum of forces equations:

= − =∑ x 1F F 0sf

= − =∑ yF n w 0 Write the sum of torques equation about the foot of the ladder:

( ) ( )τ = − =∑ 1F 4 m w 1.5 m 0 Solve the torque equation for 1F :

( )= = = 11.5 1.5F w 60 N 22.5 N4.0 4.0



Substitute the value for 1F into our equation for the net horizontal force:

= µ = ⇒ µ = = =

1

s 1 sF 22.5 Nw F 0.375w 60 Nsf

Example 15: Romeo takes a uniform 10 m ladder and leans it against the smooth (frictionless) wall of the Capulet residence. The ladder’s mass is 22.0 kg and the bottom rests on the ground 2.8 m from the wall. When Romeo, whose mass is 70 kg, gets 90 percent of the way to the top, the ladder begins to slip. What is the coefficient of static friction between the ground and the ladder? Solution:



Example 16: A uniform ladder rests against a frictionless vertical wall. The coefficient of static friction between the ladder and the floor is 0.3. What is the smallest angle at which the ladder will remain stationary? Solution:

chapter 10 - santiago canyon college 10 lecture notes physics 210 page 1 santiago canyon college...

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