Torque vs Work - Revision history

Bbreer6: /* References */

2019-07-25T17:33:17Z

References

← Older revision		Revision as of 13:33, 25 July 2019
Line 213:		Line 213:
	[http://physics.stackexchange.com/ Physics Stack Exchange]<div></div>		[http://physics.stackexchange.com/ Physics Stack Exchange]<div></div>
	[https://en.wikipedia.org/wiki/Torque Wikipedia Torque]<div></div>		[https://en.wikipedia.org/wiki/Torque Wikipedia Torque]<div></div>


	~~jderemer3~~

	[[Category: Angular Momentum]]		[[Category: Angular Momentum]]

Bbreer6: /* Difficult Torque */

2019-07-25T17:32:57Z

Difficult Torque

← Older revision		Revision as of 13:32, 25 July 2019
Line 161:		Line 161:
	<math> \vec{r} = <12, -2, 6> m </math>		<math> \vec{r} = <12, -2, 6> m </math>

	<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>		<math> \vec{\tau} = \vec{r}_{cm} \times \vec{F} </math>

	<math> \vec{\tau} = <12, -2, 6> m X <2, 1, 7> N </math>		<math> \vec{\tau} = <12, -2, 6> m \times <2, 1, 7> N </math>

	<math> \vec{\tau} = (12 m * 2 N)(iXi) + (12 m * 1 N)(iXj) + (12 m * 7 N)(iXk) + (-2 m * 2 N)(jXi) + (-2 m * 1 N)(jXj) + (-2 m * 7 N)(jXk) + (6 m * 2 N)(kXi) + (6 m * 1 N)(kXj) + (6 m * 7 N)(kXk) </math>		<math> \vec{\tau} = (12 m * 2 N)(iXi) + (12 m * 1 N)(iXj) + (12 m * 7 N)(iXk) + (-2 m * 2 N)(jXi) + (-2 m * 1 N)(jXj) + (-2 m * 7 N)(jXk) + (6 m * 2 N)(kXi) + (6 m * 1 N)(kXj) + (6 m * 7 N)(kXk) </math>
Line 169:		Line 169:
	<math> \vec{\tau} = (24 Nm)(0) + (12 Nm)(k) + (84 Nm)(-j) + (-4 Nm)(-k) + (-2 Nm)(0) + (-14 Nm)(i) + (12 Nm)(j) + (6 Nm)(-i) + (42 Nm)(0) </math>		<math> \vec{\tau} = (24 Nm)(0) + (12 Nm)(k) + (84 Nm)(-j) + (-4 Nm)(-k) + (-2 Nm)(0) + (-14 Nm)(i) + (12 Nm)(j) + (6 Nm)(-i) + (42 Nm)(0) </math>

	<math> \vec{\tau} = <-20, -72, 16> N*m </math>		<math> \vec{\tau} = <-20, -72, 16> N \cdot m </math>

	For these types of problems, it is best to split the problem into smaller pieces, identify the lever arms where torques are applied, and work your way back up.		For these types of problems, it is best to split the problem into smaller pieces, identify the lever arms where torques are applied, and work your way back up.

Bbreer6: /* Middling Work */

2019-07-25T17:32:25Z

Middling Work

← Older revision		Revision as of 13:32, 25 July 2019
Line 127:		Line 127:
	Solution:		Solution:

	<math> W = \vec{F}*\vec{r} </math>		<math> W = \vec{F} \cdot \vec{r} </math>

	<math> W_{1} = <30, 0, 0> N * <15, 0, 0> m </math>		<math> W_{1} = <30, 0, 0> N \cdot <15, 0, 0> m </math>

	<math> W_{1} = (30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>		<math> W_{1} = (30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>
Line 137:		Line 137:
	<math> W_{1} = 450 Nm </math>		<math> W_{1} = 450 Nm </math>

	<math> W_{2} = <-30, 0, 0> N * <15, 0, 0> m </math>		<math> W_{2} = <-30, 0, 0> N \cdot <15, 0, 0> m </math>

	<math> W_{2} = (-30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>		<math> W_{2} = (-30 N)(15 m) + (0 N)(0 m) + (0 N)(0 m) </math>
Line 149:		Line 149:
	<math> W_{net} = 450 Nm - 450 Nm </math>		<math> W_{net} = 450 Nm - 450 Nm </math>

	<math> W_{net} = 0 N*m </math>		<math> W_{net} = 0 N \cdot m </math>

	===Difficult Torque===		===Difficult Torque===

Bbreer6: /* Simple Torque */

2019-07-25T17:29:54Z

Simple Torque

← Older revision		Revision as of 13:29, 25 July 2019
Line 109:		Line 109:
	Solution:		Solution:

	<math> \vec{\tau} = \vec{r}_{cm} X \vec{F} </math>		<math> \vec{\tau} = \vec{r}_{cm} \times \vec{F} </math>

	<math> \vec{\tau} = <0, 4, 0> m X <-2, 0, 0> N </math>		<math> \vec{\tau} = <0, 4, 0> m \times <-2, 0, 0> N </math>

	<math> \vec{\tau} = (0 m * -2 N)(iXi) + (0 m * 0 N)(iXj) + (0 m * 0 N)(iXk) + (4 m * -2 N)(jXi) + (4 m * 0 N)(jXj) + (4 m * 0 N)(jXk) + (0 m * -2 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>		<math> \vec{\tau} = (0 m * -2 N)(iXi) + (0 m * 0 N)(iXj) + (0 m * 0 N)(iXk) + (4 m * -2 N)(jXi) + (4 m * 0 N)(jXj) + (4 m * 0 N)(jXk) + (0 m * -2 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>
Line 118:		Line 118:

	<math> \vec{\tau} = <0, 0, 8> N*m </math>		<math> \vec{\tau} = <0, 0, 8> N*m </math>


	===Middling Work===		===Middling Work===

Bbreer6: /* Nuances in Torque */

2019-07-25T17:29:33Z

Nuances in Torque

← Older revision		Revision as of 13:29, 25 July 2019
Line 89:		Line 89:
	Using these values, let's take the cross product of position cross force to find the torque cause by the force on the wheel.		Using these values, let's take the cross product of position cross force to find the torque cause by the force on the wheel.

	Torque: <math> \vec{\tau} = \vec{r}_{cm} X \vec{F} = <2, 0, 0> m X <-15, 0, 0> N </math>		Torque: <math> \vec{\tau} = \vec{r}_{cm} \times \vec{F} = <2, 0, 0> m \times <-15, 0, 0> N </math>

	<math> \vec{\tau} = (2 m * -15 N)(iXi) + (2 m * 0 N)(iXj) + (2 m * 0 N)(iXk) + (0 m * -15 N)(jXi) + (0 m * 0 N)(jXj) + (0 m * 0 N)(jXk) + (0 m * 0 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>		<math> \vec{\tau} = (2 m * -15 N)(iXi) + (2 m * 0 N)(iXj) + (2 m * 0 N)(iXk) + (0 m * -15 N)(jXi) + (0 m * 0 N)(jXj) + (0 m * 0 N)(jXk) + (0 m * 0 N)(kXi) + (0 m * 0 N)(kXj) + (0 m * 0 N)(kXk) </math>
Line 98:		Line 98:

	So, even though a force was applied at a distance relative to the center of mass of the wheel, there was no net torque because the force was acting straight through the center of mass of the wheel. This is important to remember as any force acting through an objects center of mass may be disregarded as having any effect on the torque and therefore does not change the angular momentum of the object.		So, even though a force was applied at a distance relative to the center of mass of the wheel, there was no net torque because the force was acting straight through the center of mass of the wheel. This is important to remember as any force acting through an objects center of mass may be disregarded as having any effect on the torque and therefore does not change the angular momentum of the object.


	==Example Problems==		==Example Problems==

Bbreer6: /* Mathematical Model: The Concepts */

2019-07-25T17:29:05Z

Mathematical Model: The Concepts

← Older revision		Revision as of 13:29, 25 July 2019
Line 47:		Line 47:
	As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.		As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.

	<math> W = \~~delta~~ K = K_{final} - K_{initial} </math>		<math> W = \Delta K = K_{final} - K_{initial} </math>

	<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>		<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>

Bbreer6: /* Mathematical Model: The Concepts */

2019-07-25T17:28:46Z

Mathematical Model: The Concepts

← Older revision		Revision as of 13:28, 25 July 2019
Line 19:		Line 19:
	The first of these concepts is '''torque''', which is a measurement of the moment that a force causes in a system. A moment is also known as a system's tendency of a force to cause rotation.		The first of these concepts is '''torque''', which is a measurement of the moment that a force causes in a system. A moment is also known as a system's tendency of a force to cause rotation.

	<math> \vec{\tau} = \vec{r}~~_cm X~~ \vec{F} </math>		<math> \vec{\tau} = \vec{r_{cm}} \times \vec{F} </math>

	Torque can also be represented through rotational kinematics, whereby the rotational acceleration and the moment of inertia are multiplied together to equal torque:		Torque can also be represented through rotational kinematics, whereby the rotational acceleration and the moment of inertia are multiplied together to equal torque:
Line 30:		Line 30:
	The second concept is '''work''', which is the measurement of a force acting over a distance on a given system.		The second concept is '''work''', which is the measurement of a force acting over a distance on a given system.

	<math> W = \vec{F}*\vec{r} </math>		<math> W = \vec{F} \cdot \vec{r} </math>

	<i>where '''W''' is the ~~scaler~~ dot product of the force vector and change in position vector</i>		<i>where '''W''' is the scalar dot product of the force vector and change in position vector</i>


	Both of these concepts are measured using the units of <math> Newton * meters </math> , however torque is a vector value while work is a ~~scaler~~ value. This is a key point to keep in mind.		Both of these concepts are measured using the units of <math> Newton * meters </math> , however torque is a vector value while work is a scalar value. This is a key point to keep in mind.


Line 47:		Line 47:
	As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.		As it happens, work on a system is also equivalent to the change in kinetic energy of a system and therefore can also be expressed in joules. Positive work will add kinetic energy to a system whereas negative work will subtract kinetic energy from a system.

	<math> W = ∆K = K_{final} - K_{initial} </math>		<math> W = \delta K = K_{final} - K_{initial} </math>

	<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>		<i>where '''W''' is the work done and '''K''' is the kinetic energy of the system.</i>

Bbreer6: /* Example Problems */

2019-07-25T05:03:55Z

Example Problems

← Older revision		Revision as of 01:03, 25 July 2019
Line 101:		Line 101:

	==Example Problems==		==Example Problems==

	~~Be sure to show all steps in your solution and include diagrams whenever possible~~

	===Simple Torque===		===Simple Torque===
Line 125:		Line 123:
	===Middling Work===		===Middling Work===

	In the system below, determine the net work that is accomplished by the two forces.		In the system below, determine the net work that is accomplished by the two forces:

	[[File:Wvt007.jpg\|300px\|middle]]		[[File:Wvt007.jpg\|300px\|middle]]
Line 175:		Line 173:
	<math> \vec{\tau} = <-20, -72, 16> N*m </math>		<math> \vec{\tau} = <-20, -72, 16> N*m </math>

			For these types of problems, it is best to split the problem into smaller pieces, identify the lever arms where torques are applied, and work your way back up.

	==Connectedness==		==Connectedness==

Bbreer6: /* Mathematical Model: The Concepts */

2019-07-15T03:18:05Z

Mathematical Model: The Concepts

← Older revision		Revision as of 23:18, 14 July 2019
Line 52:		Line 52:


	So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a ~~scaler~~ measurement, while torque is a vector representation. ~~They~~ must be ~~measure~~ individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.		So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a scalar measurement, while torque is a vector representation. The two quantities must be measured individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.

	===Nuances in Work===		===Nuances in Work===

Bbreer6: /* Mathematical Model: The Concepts */

2019-07-15T03:06:12Z

Mathematical Model: The Concepts

← Older revision		Revision as of 23:06, 14 July 2019
Line 20:		Line 20:

	<math> \vec{\tau} = \vec{r}_cm X \vec{F} </math>		<math> \vec{\tau} = \vec{r}_cm X \vec{F} </math>

			Torque can also be represented through rotational kinematics, whereby the rotational acceleration and the moment of inertia are multiplied together to equal torque:
			<math> \vec{\tau} = \vec{I} \cdot \vec{\alpha}</math>

	[[File:Wvt001.jpg\|300px\|thumb\|right\|Work on an Object]]		[[File:Wvt001.jpg\|300px\|thumb\|right\|Work on an Object]]
Line 50:		Line 53:

	So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a scaler measurement, while torque is a vector representation. They must be measure individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.		So, work can be expressed in terms of joules, which means that torque can be expressed in terms of joules too, right? Well, it's not quite as straightforward as changing units. Remember, work is a scaler measurement, while torque is a vector representation. They must be measure individually because while work effects the linear momentum of a system, torque effects the angular momentum of a system. To put it in common terms, work relates to the movement of a system from one position to another while torque relates to the twisting of the system itself. So in fact, they represent very different concepts.


	===Nuances in Work===		===Nuances in Work===