1. Rotation of a Rigid Body
Chapter 12
Rotation of a Rigid Body

2. Vector (or “cross”) Product
Cross Product is a vector perpendicular to the plane of vectors A and B

3. Cross Product of Vectors
Right hand rule: Curl your right hand around the center of rotation with the fingers going from the first vector to the second vector and the thumb will be pointing in the torque direction
We have learned the dot product of two vectors which is in effect a scalar multiplication of a vector. There is another vector product called the cross product and the product is a vector. A common example is the turning of a wrench in which we have a position vector and a force vector and the product of the two is torque, which we will discuss next.
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Using the parallelogram rule and properties of a parallelogram we can get:
Drawing perpendiculars from ehe ends of A and B we see using trig that one is Asinθ and the other Bsinθ.
When talking about torque we normally write the cross product as |Position x Force|
The cross product vector is perpendicular to the plane of the two vectors and its direction is determined by the Right Hand Rule.
Notice from the diagram that AxB and BxA have the same magnitude but opposite directions.
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4. Cross Product Problem Csin(110) C 110° D The magnitude of E is Cdsinθ.
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Magnitude of of E is 1.88 Nm
What is the sign?
We would normally by convention consider the z-axis coming out of the page. From this view the rotation is CCW or positive.
For CxD to make sense the direction of C would have to reverse 180 degrees.
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5. Test your Understanding
Which way will it rotate once the support is removed?
1. Clockwise.
2. Counter-clockwise.
3. Not at all.
4. Not sure what will happen.

6. Torque
If the forces are equal, which will open the heavy door more easily?
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7. Interpretation of torque
Measures tendency of any force to cause rotation
Torque is defined with respect to some origin – must talk about “torque of force about point X”, etc.
Torques can cause clockwise (+) or anticlockwise rotation (-) about pivot point

8. Torque con’t
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9. Torque con’t
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10. Definition of Torque:
where is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force If r and F are not perpendicular then:
where q is the angle between the vectors and .
Torque is defined as the cross product between the position vector, r, and the force vector F.
Torque is the rotational equivalent of force.

11. Definition of Torque: Torque can be interpreted in two ways: Click
F sin θ is the component of F perpendicular to the position vector r.
r sin θ is the component of the position vector perpendicular to the force. This component is called the moment arm.
Most think in terms of the former and resolve the force vector into perpendicular and parallel components.

12. Torque Problem
Adrienne (50 kg) and Bo (90kg) are playing on a 100 kg rigid plank resting on the supports seen below. If Adrienne stands on the left end, can Bo walk all the way to the right end with out the plank tipping over? If not, How far can he get past the support on the right?
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13. Torque Problem con’t 2m 3m 4m 50kg 100kg 90kg N1 N2 x
If the plank tips it will be around the right support. If we sum the torques around that support using x as the distance from the right support at which the plank would tip, we get:
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When we solve for x we get 3.3 m, therefore he cannot get to the end of the plank without tipping.
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14. Moments M1 M2 d1 d2
Suppose we have masses m1 and m2 on the seesaw at distances d1 and d2, respectively, from the fulcrum, when does the seesaw balance?
By Archimedes’ Law of the lever, this occurs when
m1d1 + m2d2 = 0
Moments or Moments of Force are synonymous with Torque
m1d1 = m2d2 is another way of saying that the sum of the torques is zero. Or Net Torque is zero.
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15. Moments con’t M1 M2 x1 x2
If we place a coordinate system so that 0 is at the fulcrum and if we let xi be the coordinate at which is placed then:
m1x1 + m2x2 = m1d1 + m2d2 = 0
Use of number-line causes confusion. X1 is a negative number and the m1x1product is negative; this implies CCW torque is negative, which by convention it isn’t. The key is understanding that CW and CCW net torques must be equal or there will be rotation.
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16. Moments con’t M1 M4 x1 x4 M2 M3 x2 x3 M5 x5
More generally, if we place masses m1, m2, …, mr at points x1, x2, …. , xr, respectively, then the see saw balances with the fulcrum at the origin, if and only if
m1x1 + m2x2 + …+ mrxr = 0
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17. Moments con’t M1 M4 x1 x4 M2 M3 x2 x3 M5 x5
Now, suppose that we place masses m1, m2, … , mr at points x1, x2, … xr, respectively, then where should we place the fulcrum so that the seesaw balances?
The answer is that we place the fulcrum at x-bar where:
m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )= 0
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18. Moments con’t M1 M4 x1 x4 M2 M3 x2 x3 M5 x5
x-bar
m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )
is called the moment about x-bar.
Moment is from the Greek word for movement, not time.
If positive, movement is counter-clockwise, negative it is clockwise.
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19. Moments con’t x1 x4 x2 x3 x5 Suppose that We want to solve for M1 M4
x-bar
Suppose that
We want to solve for
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20. Center of Mass M1 M4 x1 x4 M2 M3 x2 x3 M5 x5
x-bar
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21. Center of Mass con’t
Suppose m1, m2, … , mr are masses located at points (x1, y1), (x2, y2), … , (xr, yr). M1 M4 M2 M3
The moment about the y-axis is:
The moment about the x-axis is:
Center of Mass is
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22. Center of Mass con’t
Now lets find the center of mass of a thin plate with uniform density, ρ.
First we need the mass of the plate:
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23. Center of Mass con’t Next we need the moments of the region:
f(x)
g(x) ∆x
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Given the dx slice shown on the diagram and uniform density. The moment if the portion above the x-axis is ½ f(x) times the mass of the slice, ∆m, which gives a positive moment around the x-axis. The portion below the x-axis gives a negative moment. The mass of each section is ρ(Area) or ρf(x)∆x for the upper portion and ρg(x)∆x for the lower section.
Integrating from a to b gives the moment around the x-axis.
The moment around the y-axis is a little simpler. It is the x times the mass of the slice, or x∆m.
Integrating gives
To find the centers of mass we divide the moments by the total mass and we get the following:
Note that the rho’s cancel and the denominator is the area, giving the final expression.
For the x-bar we get a simpler expression
To find the center of mass we divide by mass:
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24. Center of Mass Problem
Determine the center of mass of the region bounded by y = 2 sin (2x) and y = 0 on the interval, [0, π/2]
Given the symmetry of the curve it is obvious that x-bar is at π/4.
First find the area.
Using the table of integrals:
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25. Moment of Inertia M r CR F
Suppose a particle of mass M is attached to a pivot by a mass-less rod of length r. As the particle travels around the circle, we know that the distance it travels is equal to the angle the rod sweeps out measured in radians multiplied by the radius r.
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We know that the linear acceleration of the mass is the product of r and alpha, the angular acceleration.
To accelerate the mass we must apply a force F to the particle, then F = ma.
On the other hand, since we are dealing with a rotation system we would like to work with torque, instead of force so we multiply both sides of the equation by r, then T = Fr = mra.
Finally we use the above equation to convert from linear acceleration to angular acceleration; T = mra = mr(αr) = mr2α
The quantity mr2 is called the moment of inertia of the particle.
Notice the similarity to N2L. Recall that mass in N2L is also referred to as the inertial mass. Moment of Inertia in rotational motion is the analogue to mass in linear motion. It depends not only on the mass but also on the distribution of mass.
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26. Moment of Inertia con’t
Imagine rotating the object above around the x-axis. Note that the strip indicated in the figure remains at a constant distance from the x-axis . Therefore, this strip has the same moment of inertia as a particle of the same mass lying at the same distance above the x-axis..
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If we assume that the object has been cut our of some material of uniform density ρ, the mass of the strip is simply ρw(y)dy
And the moment of inertia for the object is given by
Where a and b are the minimum and maximum values of the y variable.
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27. Calculating Moment of Inertia
Find the moment of inertia of a circular disk of radius R, mass M and uniform density that rotates on a axis passing through its center. Divide the disk into rings of mass dm with radius r and thickness dr. Let dA represent the area of the ring
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28. Calculating Moment of Inertia
The dm is to the total mass m as dA is to the total area A
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Therefore dm = (M/A)(dA)
dA = 2πrdr and A = πr2
Substituting the above into the dm equation gives:
Integrating from 0 to R gives
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29. Moment of Inertia con’t
Typical shapes of uniform density moments of Inertia are readily available in engineering references. Table 12.2 in the text gives a short list of common moments of Inertia.
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30. Parallel Axis Theorem AP Physics C
The moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about any axis parallel to that axis through the center of mass is given by
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Iparallel = ICM +Md2
The expression added to the center of mass moment of inertia will be recognized as the moment of inertia of a point mass - the moment of inertia about a parallel axis is the center of mass moment plus the moment of inertia of the entire object treated as a point mass at the center of mass.
To solve many problems about rotational motion, it is important to know the moment of inertia of each object involved. Calculating the moments of inertia of various objects, even highly symmetrical ones, may be a lengthy and tedious process. While it is important to be able to calculate moments of inertia from the definition (I = mr2 ), in most cases it is useful simply to recall the moment of inertia of a particular type of object. The moments of inertia of frequently occurring shapes (such as a uniform rod, a uniform or a hollow cylinder, a uniform or a hollow sphere) are well known and readily available from any mechanics text, including your textbook. However, one must take into account that an object has not one but an infinite number of moments of inertia. One of the distinctions between the moment of inertia and mass (the latter being the measure of translational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moments of inertia that you can find in the textbooks are usually calculated with respect to an axis passing through the center of mass of the object. However, in many problems the axis of rotation does not pass through the center of mass. Does that mean that one has to go through the lengthy process of finding the moment of inertia from scratch? It turns out that in many cases, calculating the moment of inertia can be done rather easily if one uses the parallel-axis theorem. Mathematically, it can be expressed as , where is the moment of inertia about an axis passing through the center of mass, is the total mass of the object, and is the moment of inertia about another axis, parallel to the one for which is calculated and located a distance from the center of mass.
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31. ||-axis Theorem Proof
The first term of the expansion is the moment of inertia around the center of mass.
The last term is d2M
The middle term is the center of mass when x’ is zero and the term is zero
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32. Rotational Kinetic Energy
An objects rotational kinetic energy is the sum of the kinetic energies of each of the particles making up the object.
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Substituting rω for v we get
Factoring out the ½ and ω we get the expression . Note that the summation of mr2 is the moment of inertia of the object.
The rotational kinetic Energy is ½ the product of the moment of inertia and the angular velocity squared.
If the rotational axis is not through the center of mass the object may move up and down, changing the gravitational potential energy and the mechanical energy becomes:
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33. Rotational Dynamics Torque is the rotational equivalent of force.
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Tangential force is equal to mass times tangential acceleration
Multiplying both sides by r gives torque on the left
Given a mass with forces applied, the net torque is the summation of the moments of each particle which is the Moment of Inertia.
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34. Rotation About a Fixed Axis
Many applications in rotational dynamics involve objects connected via ropes to pulleys.
Generally assume that the rope does not slip on the cylinder/pulley.
The ropes motion must match the motion at the rim of the pulley. The tension in the rope is the same at the pulley and at the object being lowered. The acceleration of the object must be the same as the tangential acceleration of the pulley.
The velocity of the object is the same as the tangential velocity of the pulley.
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35. Bucket Problem
A 2.0 kg bucket is attached to a mass-less string that is wrapped around a 1.0 kg, 4.0 cm diameter cylinder, as shown. The cylinder rotates on an axel through the center. The bucket is released from rest 1.0 m above the floor, How long does it take to reach the floor
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36. Bucket Problem con’t
The net force on the bucket is the upward tension of the rope minus the downward force of gravity
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The cylinder is under the downward tension T which causes a torque with arm, R.
The net torque is also equal to the angular acceleration times the rotational moment of inertia.
Combining the two shows the angular acceleration to be equal to the product of the net torque divided by the moment of inertia. The moment of inertia for the cylinder is (1/2)MR2.
Since the tangential acceleration, α, is CCW, it is positive. The acceleration of the bucket, ay is downward and is negative. Since they are equal one is equal to the opposite of the other.
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37. Bucket Problem Con’t
Given that the acceleration of the bucket is negative 2T/M, we can solve for the tension on the rope, T.
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Substituting this into the Bucket equation gives
WE know everything except ay which we can solve for.
The time of fall can be found from the kinematic equation
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38. Static Equilibrium
Statics is a major course for engineers. The analysis of structures to insure they do not move or are in total static equilibrium.
Essential conditions for Static Equilibrium is that the net force is zero and the net torque is zero.
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39. Statics Problem
A 3.0 m ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of μs, that prevents the ladder from slipping?
This is a static situation so the sum of forces in the x and y directions are equal to zero.
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The net torque around the bottom must be equal to zero
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40. Statics Problem
From the previous slide we have three equations and three unknowns.
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Since fs equals n2 we get Mg/2tan60 which is equal to μsn1 and n1 equals mg therefore
Dividing by mg gives:
This is the minimum value of μs which will keep the ladder from slipping, evaluating gives:
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41. Balance and Stability θc h t/2
The critical angle is the angle of tilt so that the center of gravity is over the pivot point. In other words there is not torque.
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It is obvious that the combination of a high center of gravity and a narrow base of support leads to a object with a tendency to tip over.
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42. Rolling Motion
Rolling motion is a combination of translation and rotation.
The path of a point on the rim of the object traces out a figure called a cycloid. The length of one period of the cycloid is 2piR or the circumference of the circle.
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43. Rolling Motion con’t
For a given particle, i, the location vector for i is the sum of the location vector of the center of mass of the object, rcm, and the vector locating i, relative to the center of mass, cm.
Taking the derivative with respect to t give the velocity equation shown. This says that the velocity of I is the sum of the translational velocity and the rotational velocity.
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44. Rolling Motion con’t
Note that when the point is rotating in the lower half of the circle it has negative x and y components. At the bottom the velocity has the same magnitude as the translational velocity but opposite sign and the velocity of the point is instantaneously zero.
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45. Rolling Motion con’t
At the top of the circle the points rotational velocity is the same as the translational and the net velocity is twice the translational velocity.
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46. Rolling Kinetic Energy
Since the velocity of a point on an object is the sum of the translational and rotational velocities, it makes sense that the kinetic energy is the sum of the rotational and translational kinetic energies.
Point P can be viewed as the pivot point for the whole object since it is instantaneously at rest.
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We know that the rotational kinetic energy is one-half the product of the moment of inertia and the angular velocity.
Using the parallel axis theorem we can restate this in terms of the center of mass, cm.
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47. Great Downhill Race
The figure shows a contest in which a sphere, cylinder, a circular hoop and a particle all with the same mass and starting height run a race down hill. Who will win?
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48. Downhill Race con’t
Using the Conservation of energy equation is this case the sum of the rotational kinetic energy and translational kinetic energy equals the loss of potential energy over height h.
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Recall from table 12-2 in your text that the rotational inertias about the center of mass of standard figures differ by a fractional factor, c.
Substituting into the Conservation of Energy equation we get:
Solving for the velocity at the bottom of the ramp, we see that velocity is independent of the Mass and Radius
Using the values on table 12-2 we see that the particle wins and the hoop is last.
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49. Angular Momentum
Angular Momentum is the rotational analogue of linear momentum in much the same way torque is the equivalent of force.
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Angular momentum can be defined as a the product of the location vector r and the instantaneous linear momentum vector, p.
Translating the p vector shows that the Angular Momentum vector is upward
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50. Angular Momentum
If the trajectory is circular, the angle is 90 degrees.
Applet
Applet shows the angular momentum vector about a point on the axis of a circle of a particle (m) , moving round a circular path of radius R at v.    
The green vector is the linear momentum vector(p), the blue vector is the position vector (r) and the black vector is the angular momentum vector(L).
The relation between the vectors is L = r x p
The red and magenta vectors are the radial and axial components of the angular momentum vector and have magnitudes of mvR and mvh respectively.
Notice that during the rotation of the particle the axial component of angular momentum remains unchanged and the radial component rotates at the same angular speed as the particle itself.
Experiment by changing the height of the reference point above the plane of the circle, the radius of the circle and the angular velocity of the particle.
Double clicking anywhere in the applet area at any point shifts the origin to that point. Dragging with the mouse changes the orientation of the axes.
The checkbox at the bottom right corner, if checked, causes a half plane containing the z- axis to be displayed.
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51. Angular Momentum
Since torque is the rotational equivalent of force, it seems obvious that torque causes angular acceleration and as a result angular momentum to change.
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For a particle in circular motion with tangential acceleration vt, the angular momentum is along the z axis. Based on our sign convention angular momentum and velocity are positive.
Taking the derivative of the cross product using the product rule gives:
dr/dt is velocity and r X Fnet is the net torque
The velocity and momentum vectors are parallel and the cross product of parallel vectors is zero making the rate of change of Angular Momentum equal to torque.
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52. Conservation of Angular Momentum
Two equal masses are at the ends of a mass-less 500 cm long rod. The rod spins at 2.0 Rev/s about an axis through its midpoint. If the rod lengthens to 160 cm, what is the angular velocity
It has been shown experimentally that in a closed system Angular momentum is conserved, in other words:
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This is a simple example demonstrating the principle.
Substituting into the last equation and solving give s 0.20 rev/sec.
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53. Testing Understanding
The two buckets are spinning at a constant velocity on a frictionless bearing. It starts to rain. What happens? Why?
There is no torque on the buckets so angular momentum is conserved.
Increased mass in buckets increases inertia so angular velocity must decrease.
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54. Problem 1
An 18 cm long bicycle crank arm with a pedal at one end is attached to a 20 cm diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60 rpm to 90 rpm in 10 s.
a. What is the tangential acceleration of the pedal?
b. Want length of chain passes over the top of the sprocket during this interval?
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55. Problem 1 con’t b. Since L = r∆θ, find r∆θ.
a. Since at = rα, find α first. With 60 rpm = 6.28 rad/s and 90 rpm = 9.43 rad/s:
The angular acceleration of the sprocket and chain are the same.
b. Since L = r∆θ, find r∆θ.
The length of chain which has passed over the top of the sprocket is
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56. Problem 2
A 100 g ball and a 200 g ball are connected by a 30 cm mass-less rigid rod. The balls rotate about their center of mass at 120 rpm. What is the speed of the 100g ball?
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57. Problem 3
A 300 g ball and a 600 g ball are connected by a 40 cm mass-less rigid rod, The structure rotates about its center of mass at 100 rpm. What is its rotational kinetic energy?
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First we find the center of Mass
Next we find the inertia
Finally we find the rotational kinetic energy
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58. Problem 4
A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door’s moment of inertia for (a) rotation on its hinges (b) rotation about a vertical axis inside the door, 15 cm from one edge.
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59. Problem 5
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.o kg. What is the magnitude of the torque about his shoulder if he holds the ball (a) straight out to his side, parallel to the floor and (b) straight, but 45 degrees below horizontal.
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60. Problem 5 con’t
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61. Problem 6
Starting from rest, a 12 cm diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disk’s moment of inertia is 2.5x10-5 kgm2. (a) How much torque is applied? (b) How many revolutions does it make before reaching full speed?
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62. Problem 6 con’t a. Using the rotational kinematic equation b.
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63. Problem 7
The two objects in the figure are balanced on the pivot. What is the distance d?
There are three forces acting on the object: the gravitational force (Fg)1 acting through the center of mass of the long rod, the gravitational force (Fg)2 acting through the center of mass of the short rod, and the normal force p on the object applied by the pivot. The translational equilibrium equation (fnet)y is:
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