# Torque and Rotation (12) (ELCA)

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Notes for:

## Object RaceEdit

• Which object would win in a race?
• Solid Sphere
• Solid Sphere
• Hollow Sphere
• Point Mass – all mass is considered to be located at 1 point.
• Extended Object – a definite, finite size and shape.

## Rotational Motion versus Transitional MotionEdit

• Translation' – motion in which an entire object moves.
• Rotation – motion in which an object spins.

Rotational (rolling) motion can be analyzed separately from translational (linear) motion. Axis of rotation is the point around which an object is free to rotate.

## TorqueEdit

• Torque – is the ability of a force to rotate an object around some axis.

### Force versus TorqueEdit

Force is an action that creates changes in linear motion. An example is a push or pull. Torque is an action that causes objects to rotate, such as a twisting motion.

• FORCE AND TORQUE NOT THE SAME FORCE

Torque is created by a force. Torque depends on where the force is applied and the point about which the object rotates.

### Center of RotationEdit

• Center of Rotation – the point or line about which an object turns.

Force applied far from the center of rotation produces greater torque. Force applied close tot he cent of rotation produces less torque.

### What Type of MotionEdit

The type of motion resulting from an applied force depends on:

1. The location and direction of the force
2. Whether torque is created by the force
3. Is any part of the object fixed in place

### Line of ActionEdit

• Line of Action – an imaginary line that follows the direction of a force and passes through its point of application.

Forces do not have to be perpendicular to an object to cause the object to rotate. If a cat pushes on a door at an angle, the door would still rotate, just not as easily. Torque is created when the line of action does not pass through the center of rotation.

Example: Using a wrench – when you pull on the wrench at a distance from the center of rotation (the bolt), you create torque.

• Force creates torque because it acts to rotate the wrench.

### A force that causes no torqueEdit

If the force goes through the center of rotation, such as when you push on a wrench in its long direction, the bolt is not tightened, and no torque is created. The application of force does not cause the wrench to rotate.

### Direction of ForceEdit

When opening a door, you apply a force to the knob perpendicular to the door. Pushing parallel to the or, toward the hinges, does not open the door.

### Maximum TorqueEdit

For maximum torque, the force should be applied in a direction that creates the greatest lever arm.

### Lever ArmEdit

Lever Arm is the perpendiuclar distance between the line of action of the force and the center of rotation.

### Calculating TorqueEdit

Torque ($\tau$) created by a force is equal to the magnitude of the force ($F$) times the lever arm ($d$).

$\tau = Fd$

• $\tau$ = torque (Nm)
• $F$ = Force (N)
• $d$ = distance (m)

$\tau = Fd(sin \Theta)$

• $\tau$ = torque (Nm)
• $F$ = Force (N)
• $d$ = distance (m)
• $\Theta$ = Angle of Force Applied (º)

#### Word of CautionEdit

When calculating torque, be careful determining the length of the lever arm. If the line of action passes through the center of rotation, the lever arm is zero, and so is the torque, regardless fo the size of the force applied.

#### More than 1 torqueEdit

When > 1 torque acts on an object the torques are combined to determine the net torque. If the torque tend to make an object spin in the same direction, they are added together. If the torques tend to make an object spin in opposite direction, the torques are subtracted.

#### Signage for torquesEdit

Counterclockwise rotation creates positive torque while clockwise rotation creates negative torque.

#### Total TorqueEdit

Total Torque is calculated by adding up each individual torque, keeping up with positive and negative signs. Unit is the N•m, same as for work, but shoudn't be confused with work because each represents different physicals quantities.

#### ExampleEdit

A basketball being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 14 cm from the axis of rotation. Find the net torque acting on the ball.

• $F_1$ = 11 N
• $d_1$ = 7.0 cm = 0.07 m
• $F_2$ = 15 N
• $d_2$ = 14 cm = 0.14 m

$\tau_\text{net} = -(11 \text{N} \times 0.07 \text{m}) + -(11 \text{N} \times 0.14 \text{m}) = (-2.1 \text{Nm}) + (-0.77 \text{Nm}) = -2.9 \text{Nm}$

A student pushers with a minimum force of 50.0 N on the middle of a door to open it.

1. What minimum force must be applied at the edge of the door in order for the door to open?
• 25 Nm
2. What minimum force must be applied to the hinged side of the door in order for the door to be opened?
• 0 Nm

### Same Force, Different Torques?Edit

Torque is always calculated around a particular center of rotation. The same force may cause different torques when an object is allowed to rotate around different points.

## Center of MassEdit

An object is free to rotate as it travels through the air, if gravity is the only force acting on it, will rotate around its center of mass.

• Center of Mass – Point at which all the mass of the body can be considered to be considered.

### ParabolaEdit

The shape of an object and the way its mass is distributed affect the way the object moves and balances. A thrown ball follows a parabolic motion, what about an irregularly shaped object. AN irregularly shaped object will also spin in flight, but the general shape of its path will also be a parabola. One specific point of the object will move in a perfect parabola. This point is located on the axis about which the object spins.

### ExampleEdit

Hammer rotates about its center of mass. The center of mass moves along the path of a projectile. Regularly shaped objects (sphere or a cube) rotate around its geometric center.

### Spinning AxesEdit

There are three different axes about which an object will naturally spin. The point where these axes intersect is the center of mass. The center of mass is the average position of all the particles that make up the object's mass.

### Complete MotionEdit

The complete motion of a thrown object is the combination of both translational (linear) and rotational motion. The center of mass moves as if the object were a point mass, with all of its mass concentrated at that point for purposes of analyzing its motion.

### Finding the Center of MassEdit

For symmetrical shape objects, the center of mass is at the geometric center of the object. For irregularly shaped objects, the center of mass can be found by spinning the object and finding the intersection of the three spin aces. There may not be substance an object's center of mass. the center of mass of a doughnut is in the doughnut "hole", likewise for a coffee cup, boomerang, or empty box.

### Center of GravityEdit

• Center of Gravity – The average position of an object's weight.

The center of gravity is also the point at which we consider the force of gravity to act on an object. If the acceleration due to gravity is the same at every point in an object, the center of mass and center of gravity are at the same location. The two terms can be used interchangeably.

### Balance and the Center of MassEdit

To balance an object on your finger, you must place you finger directly under the object's center of gravity. The object balances because the torque caused by the force of the object's weight is equal on each side. For an object to remain upright, its center of gravity must be above its area of support. The area of support includes the entire region surrounded by the actual supports. The larder the area of support, the less likely an object is to topple over. An object will topple over if its center of mass is not above the area of support.

## Rotational EquilibriumEdit

Rotational Equilibrium

### Newton's First Law and Rotational MotionEdit

A spinning object will keep spinning at a constant angular speed unless acted on by a net torque. A torque is needed to change the angular speed or axis of a rotating object.

### Rotational InertiaEdit

Inertia is the name for an object's resistance to a change in its motion. Rotational Inertia' is the term used to describe an object's resistance to a change in its rotational motion about some axis.

### Mass and Rotational InertiaEdit

The more mass an object has, the greater its linear inertia and the harder it is to change its motion. Same for rotational inertia – objects with greater mass usually have greater rotational inertia than objects with less mass.

For example, bicycle wheels are hollow with spokes rather than solid disks to reduce the wheels' mass and make them easier to spin. But, rotational inertia depends not only on total mass, but also on the way its mass is distributed.

### Mass DistributionEdit

When mass is concentrated near the object's axis of rotation, it is easy to spin. When mass is far from the axis of rotation it is harder to spin because the mass has a greater distance to move with each rotation. (see Figure 9.9)

### Moment of InertiaEdit

A solid object contains mass distributed at different distances from the center of rotation. Rotational inertia depends on the square of the radius, the distribution of mass makes a big difference for solid objects.

$I = m_1r^2 + m_2r^2 + m_3r^2 + m_4r^2 + \text{...}$

The moment of inertia for object of the same shape is different depending on which axis the object is rotating around.

### Solid Versus HollowEdit

A hoop (hollow object) has a greater moment of inertia than a greater moment of inertia than a solid disk of the same mass because all of the mass of the hoop is at a large radius. For solids, the mass at the center contributes no rotational inertia because its radius is zero.

### Not Rotating Around the CenterEdit

The axis about which an object rotates affects its moment of inertia. If the axis does not pass through its center of mass, it usually has a greater moment of inertia.

### FormulasEdit

There are some simple formulae used to determine the moment of inertia for some shapes. The units for moments of inertia are $kg \times m^2$.

### EquilibriumEdit

For an object to be in equilibrium, it must have a net force of zero, which is called translational equilibrium, and it must have a net torque of zero, which is rotational equilibrium.

$\Sigma F=0$

$\Sigma \tau =0$

### Rotational EquilibriumEdit

Rotational Equilibrium is when the net torque applied is zero. An object such as a see-saw that is balanced and not rotating. An object rotating at a constant speed, like a wheel.

Rotational Equilibrium is used to determine unknown forces. ANy object that is not moving is in rotational equilibrium and transitional equilibrium.

#### Example ProblemsEdit

1) A ten meter bridge that weights 500 N and is supported at both ends. A person who weighs 750 N stands 2 m from one end of the bridge. What are the forces, $F_A$ and $F_B$, holding up the bridge at either end.

• $w_p$ = 750 N
• $d_\text{p}$ = 2 m
• $d_\text{b}$ = 10 m
• $w_b$ = 500 N
• $F_A + F_B = 1250 N$

Rotational Equilibrium equal zero at any point.

From $F_A$, the left end of the bridge, the torque is zero because its line of action passes through the center of rotation.

$(-750 \text{N})(2 \text{m}) + (-500 \text{N})(5 \text{m}) = (F_B)(10 \text{m}) = 0$

$-1500 \text{Nm} - 2500 \text{Nm} + F_B(10 \text{m}) = 0$

$F_B$ = 400 N

$F_A + 400 \text{N} = 1250 \text{N}$

$F_A = 850 \text{N}$

2) A uniform 5.00 m long horizontal beam that weighs 315 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53º with the horizontal, and a 545 N person is standing 1.50 m from the pin. Find the force in the cable, $F_T$, and the force exerted on the beam by the wall, $R$, if the beam is in equilibrium.

• $d_b$ = 5.0 m
• $w_b$ = 315 N
• $\Theta$ = 53º
• $w_p$ = 545 N
• $d_p$ = 1.50 m
• $F_T$ = ?
• $R$ = ?

$(F_T \times 5 \text{m} \times (\sin{53})) - (1.5 \text{m} \times 545 \text{N}) - (315 \text{N})(2.5 \text{m}) = 0$

$(F_T \times 5 \text{m} \times (\sin{53})) = (1.5 \text{m} \times 545 \text{N}) + (315 \text{N})(2.5 \text{m})$

$(F_T \times 3.99 \text{m} = 1605 \text{Nm}$

$F_T = 402 \text{N}$

$F_x = R_x - F_T(\cos{\Theta}) = 0$

$R_x = 402 \text{N} (\cos{53})$

$R_x = 241.9 \text{N}$

$F_y = R_y - F_T (\sin{\Theta}) - F_\text{g,p} - F_\text{g,b}= 0$

$R_y = F_\text{g,p} + F_\text{g,b} - F_T(\sin{\Theta})$

$R_y = 545 \text{N} + 315 \text{N} - 402 \text{N} \times (\sin{53})$

$R_y = 539 \text{N}$

$R = \sqrt{R_x^2 + R_y^2}$

$R = 591 \text{N}$

## Newton's Second Law for RotationEdit

### Redefining Newton's Second LawEdit

When a net force acts on a n object, the resulting acceleration of the object depends on the object's mass. When an object experiences a net torque, the resulting change in the rotational motion depends on the object's moment of inertia.

Inertia is a measure of resistance to acceleration.

Rotating motion requires an equation that relates torque and angular acceleration.

### Moment of InertiaEdit

Because of the $r$ (radius) in $mr^2$, mass that is farther from the axis of rotation has much more rotational inertia than mass close to axis.

#### ExampleEdit

A 1 kg mass on a 2 m rod has 4 times the rotational inertia as a 1 kg mass on a 1 m rod. Therefore, it takes 4 times as much torque to produce the same angular acceleration.

### Torques and NewtonEdit

When a torque is applied to an object, it spins in the direction of the applied torque. Angular speed increases at a rate directly proportional to the net torque and inversely proportional to its moment of inertia. The greater the torque, the greater the angular acceleration.

$\alpha = \frac{\tau}{I}$

• $\alpha$ – rotational acceleration $\frac{\text{rad}}{s^2}$
• $\tau$ – torque $N /times m$
• $I$ – Moment of Inertia $kg \times m^2$

#### ExampleEdit

A bicycle wheel has a moment of inertia of 0.1 $kg \times m^2$. When the brakes are applied, a force of 100 N is applied at the rim of the wheel, at a radius of 0.35 M. What is the angular acceleration?

• $\alpha$$? \frac{\text{rad}}{s^2}$
• $\tau$$? N /times m$
• $I$$0.1 kg \times m^2$

$\tau = Fr$

$\tau = 1000 \text{N} \times 0.3 \text{m} = 35 \text{Nm}$

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{35 \text{Nm}}{0.1 kg \times m^2} = 350 \frac{rad}{s^2}$

### Rotational Motion of Newton's Second LawEdit

Rotational motion has a new set of variables, but the relationships are similar. Force causes linear acceleration, torque causes angular acceleration. Mass is a measure of linear inertia, moment of inertia (I) is a measure of rotational inertia.

### Angular MomentumEdit

Because a rotating ovject has inertia, it also possesses momentum associated with its rotation. This si called angular momentum.

$L = l \omega$

• $L$ = Angular Momentum $kg \times \frac{m^2}{s}$
• $I$ = Moment of Interia $kg \times m^2$
• $\omega$ = Angular Speed $\frac{rad}{s}$

### Conservation of Angular MomentumEdit

When the net external torque acting on an object(s) is zero, angular momentum of the object does not change. Angular momentum can be conserved.

Example:

Ice Skater in a spin – arms outstretched, large moment of inertia. Arms brought it to the body, smaller moment of inertia. Because angular momentum is constant, angular speed increases to compensate for the smaller moment of inertia.

#### Example ProblemEdit

A 0.11 kg mouse rides on the edge of a Lazy Susan that has a mass of 1.3 kg and a radius of 2.5 m. If the Lazy Susan begins with an angular speed of 3.0 $\frac{rad}{s}$, what is its angular speed after the mouse walks from the edge to a point 0.15 m from the center?

• $m_\text{m,i} = 0.11 kg$
• $m_\text{LS} = 1.3 kg$
• $r_\text{LS} = 2.5 m$
• $r_\text{m,f} = 0.15 m$
• $\omega_\text{i} = 3.0 \frac{rad}{s}$
• $I\text{LS} = \frac{1}{2}mr^2 = \frac{1}{2} (1.3 kg) ((.25m)^2) = 0.0406 kgm^2$
• $I\text{m,i} = mr^2 = (0.11 kg) ((.25 m)^2) = .0069 kgm^2$
• $I\text{m,f} = mr^2 = (0.11 kg) ((.15 m)^2) = .0025 kgm^2$

$L_i = L_f$

$(l_\text{mi} + l\text{LS}) \omega_i = (l_\text{mf} + l\text{LS}) \omega_f$

$\frac{(l_\text{mi} + l\text{LS}) \omega_i} {(l_\text{mf} + l\text{LS})} = \omega_f$

$\omega_f = \frac{0.0069 kgm^2 + 0.0406 kgm^2 \times 3.0 \frac{rad}{s}}{0.0025 kgm^2 + 0.0406 kgm^2} = 3.3 \frac{rad}{s}$

### Conservation of Mechanical EnergyEdit

Mechanical energy includes translation kinetic energy and potential energy. Now that we have knowledge about rolling objects, ME also includes rotational kinetic energy.

$KE_\text{rot} = \frac{1}{2}l\omega^2$

• $KE_\text{rot}$ = Rotational Kinetic Energy '"UNIQ852cd633137f588-math-00000062-QINU"'
• $I$ = Moment of Inertia '"UNIQ852cd633137f588-math-00000064-QINU"'
• $\omega$ = Angular Speed '"UNIQ852cd633137f588-math-00000066-QINU"'

$ME = KE_\text{trans} + KE_\text{rot} + PE_\text{g}$

• $ME$ = Mechanical Energy
• $KE_\text{trans}$ = Kinetic Translation Energy
• $KE_\text{rot}$ = Kinetic Rotational Energy
• $PE_\text{g}$ = Potential Energy Due to Gravity

$ME = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + mgh$

$ME_i = ME_f$

• $ME_i$ = Initial Mechanical Energy
• $ME_f$ = Final Mechanical Energy

$mgh_i + \frac{1}{2}mv^2_i + \frac{1}{2}I\omega^2_i = mgh_f + \frac{1}{2}mv^2_f + \frac{1}{2}!\omega^2_f$

#### ExampleEdit

A solid ball with a mass of 4.10 kg and a radius of 0.050 m starts from rest at a height of 2.00 m and rolls down a 30.0º slope. What is the translational speed of the ball when it leaves the incline?

• $m = 4.10 \text{kg}$
• $r = 0.050 \text{m}$
• $h_i = 2.00 \text{m}$
• $v_i = \text{?}$
• $I_i =$
• $\omega_i =$
• $h_f =$
• $v_f =$
• $I_f =$
• $\omega_f =$

$mgh_i + = \frac{1}{2}mv_f^2 \times \frac{1}{2} (\frac{2}{5} \times m \times r^2) \times \frac{v}{r}^2_f$

You can eliminate the value of mass and radius and a single $\frac{1}{2}$

$gh_i = \frac{1}{2}v_f^2 \times (\frac{1}{5} ) \times {v}^2_f$

Fill in the blank and Move things around.

$(9.8 \frac{m}{s^2} \times 2.00 \text{m}) = \frac{1}{2}v_f^2 \times (\frac{1}{5} ) \times {v}^2_f$

$(9.8 \frac{m}{s^2} \times 2.00 \text{m}) = \frac{7}{10}v_f^2$

$\frac{10}{7}(9.8 \frac{m}{s^2} \times 2.00 \text{m}) = v_f^2$

Solve

$v_f^2 = \frac{10}{7}(9.8 \frac{m}{s^2} \times 2.00 \text{m}) = 28 \frac{m^2}{s^2}$

$\sqrt{v_f^2} = \sqrt{28 \frac{m^2}{s^2}}$

$v_f = 5.3 \frac{m}{s}$

Mechanical Advantage (MA) is the advantage that using a machine give us.

$MA = \frac{\text{output force}}{\text{inputforce}} = \frac{F_\text{out}}{F\text{in}} = \frac{d_\text{in}}{d_\text{out}}$

### EfficiencyEdit

The efficiency of a machine is a measure of how much useful work out you get from a machine versus hav much energy (work) you put into the machine.

$\text{eff} = \frac{W_\text{out}}{W_\text{in}}$