Torque

Physics 2

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point.

For example, consider tightening a bolt with a wrench.

Which of the 3 forces shown will tighten the bolt?

Pivot Point FA

FB

FC

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point.

For example, consider tightening a bolt with a wrench.

Which of the 3 forces shown will tighten the bolt?

Pivot Point FA

FB

FC

Force B will tend to rotate the bolt clockwise, which will tighten it.

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point.

For example, consider tightening a bolt with a wrench.

Which of the 3 forces shown will tighten the bolt?

Pivot Point FA

FB

FC

Force B will tend to rotate the bolt clockwise, which will tighten it.

Notice that force A will tend to rotate the bolt counter-clockwise, loosening it.

What does force C do?

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point.

For example, consider tightening a bolt with a wrench.

Which of the 3 forces shown will tighten the bolt?

Pivot Point FA

FB

FC

Force B will tend to rotate the bolt clockwise, which will tighten it.

Notice that force A will tend to rotate the bolt counter-clockwise, loosening it.

What does force C do?

Force C doesn’t cause any rotation at all – there is no torque generated by force C. Why not?

Torque

Torque is what causes angular acceleration (just like a force causes linear acceleration)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point.

For example, consider tightening a bolt with a wrench.

Which of the 3 forces shown will tighten the bolt?

Pivot Point FA

FB

FC

Force B will tend to rotate the bolt clockwise, which will tighten it.

Notice that force A will tend to rotate the bolt counter-clockwise, loosening it.

What does force C do?

Force C doesn’t cause any rotation at all – there is no torque generated by force C. Why not?

Force C points directly at the pivot point – no torque is created in this case.

Torque

Direction is perpendicular to both radius and force.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). )sin(Fr

Fr

Torque

Direction is perpendicular to both radius and force.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). )sin(Fr

Fr

Pivot Point FA

r

θ

Take a look at the diagram – r and θ are shown for force A.

Torque

Direction is perpendicular to both radius and force.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). )sin(Fr

Fr

Pivot Point FA

r

θ

Take a look at the diagram – r and θ are shown for force A.

There are 2 ways to interpret the formula.

•If you group the Fsin(θ) together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular (can you see why from the formula?)

Torque

Direction is perpendicular to both radius and force.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). )sin(Fr

Fr

Pivot Point FA

r

θ

Take a look at the diagram – r and θ are shown for force A.

There are 2 ways to interpret the formula.

•If you group the Fsin(θ) together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular (can you see why from the formula?)

•The other option is to group the rsin(θ) together and call it the “lever arm” for the force. Think of this as the shortest distance from the pivot point to where the force is applied. This is the effective radius of the force. Again, to get maximum torque the angle should be 90°.

Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°

Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°

mN6.30)50sin()N20()m2(1

The direction of this torque is into the page (use the right-hand-rule).

Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°

mN6.30)50sin()N20()m2(1

The direction of this torque is into the page (use the right-hand-rule).

mN0.26)120sin()N30()m1(2

The direction of this torque is out of the page (use the right-hand-rule).

Note that the angle could also be 60° and we get the same answer.

Torque

We mentioned earlier that torques produce angular accelerations. We have a formula for this relationship:

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

This is really just Newton’s 2nd law applied to rotational motion. The moment of inertia, I, takes the place of the mass, and we use angular acceleration instead of linear.

I

Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°

Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°We will need to find the moment of inertia for this rod.

From our table of shapes (p.299 in textbook) we find a formula:

2231

231

mkg67.6)m2)(kg5(I

MLI

Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°We will need to find the moment of inertia for this rod.

From our table of shapes (p.299 in textbook) we find a formula:

2231

231

mkg67.6)m2)(kg5(I

MLI

Now we can apply our torque formula. We’ll define positive torque to mean “out of the page”. Alternately you can think of this as “counter-clockwise”. We can put a little arrow symbol to indicate this in the diagram.

With this choice torque 1 is negative, and torque 2 is positive.

+

Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown.

F1 is applied at the right end, and F2 is at the center.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

F1=20N

F2=30N

120°

50°We will need to find the moment of inertia for this rod.

From our table of shapes (p.299 in textbook) we find a formula:

2231

231

mkg67.6)m2)(kg5(I

MLI

srad

2

69.0

mkg67.6mN0.26mN6.30

I

Now we can apply our torque formula. We’ll define positive torque to mean “out of the page”. Alternately you can think of this as “counter-clockwise”. We can put a little arrow symbol to indicate this in the diagram.

With this choice torque 1 is negative, and torque 2 is positive.

+

The negative sign means that the angular acceleration vector points into the page, and the object’s will accelerate in the clockwise direction.