## Homework Helpers: Physics

## 4 Rotational and Circular Motion

### Lesson 4–2: Torque

What causes a rotational motion, as described in our last lesson? Newton”s first law reminds us that an object at rest remains at rest, and an object in motion remains in motion, unless acted upon by an unbalanced force. Applying this law to rotational motion we come to realize that if an object is rotating, no external force is required to keep it rotating. However, if a rotating object was initially at rest, an unbalanced force must have been involved at some point to get it to start rotating. Keep in mind that not every force will cause a rotational motion. It is possible for you to throw a football without spinning it, causing only translational motion. Or you can throw a football with what they call a “spiral,” giving it both translational and rotational motion. In both cases, you applied an unbalanced force, but you only would have applied **torque** in the later case.

**Torque ( τ)**

A measure of the ability of a force to rotate an object around an axis.

Torque =

distance of force from axis × force × sine of angle between the force and lever*τ* = dF sin *θ*

Do you remember playing on a seesaw when you were a child? Did you encounter situations where you were on one side of the seesaw and a child with a much different weight was on the other? Can you recall what you had to do to compensate for the fact that you both were different weights? By trial and error, you may have found that you could compensate for the different weights by having the lighter child move further back on one side of the seesaw, away from the center, and having the heavier child move up, closer to the center. By shifting your positions on the seesaw, you may have been able to achieve **rotational equilibrium**, allowing you to keep the seesaw horizontal and motionless. You did this by making sure that the net **torque** on the seesaw was zero.

Torque is measured in N · m. If the net torque on an object such as a seesaw is zero, it will keep on doing what it is doing (maintain constant angular velocity), and the object is said to be at rotational equilibrium. When the net torque on an object that is free to rotate around an axis or pivot point is not equal to zero, there will be an angular acceleration. It is the net torque that we apply that allows us to open doors and turn screws and bolts. It was the net torque applied by your weight that allowed the seesaw to move down when you played at the park.

__Figure 4.3__ illustrates how to measure the distance between the pivot point (axis of rotation) and the applied force.

The reason why you needed the lighter child to move toward the end of the seesaw, away from the pivot point, is to maximize the torque on that side of the seesaw by increasing the value of d. Moving the heavier child inward, toward the pivot point, decreases the torque on that side of the seesaw by decreasing the value of d.

*Figure 4.3*

At rotational equilibrium, d_{1} F_{1} sin *θ*_{1} = d_{2} F_{2} sin *θ*_{2}

There are many similar real-life examples that involve torque. For example, if you want to open a very heavy swinging door, you should apply the force as far away from the pivot point (hinges) as possible. If you want to lift a heavy object with a lever, you want to apply the force as far away from the pivot point (fulcrum) as you can. If you are trying to remove a stubborn lug nut from your tire, you want to apply the force at the end of the handle of the tire iron. In each of these cases you may also maximize the torque by applying the force perpendicular to the surface or lever arm.

**Example 1**

A woman applies a force of 185 N perpendicular to the handle of a wrench at a distance of 85 cm from center of the nut (axis of rotation) she is turning. Calculate the magnitude of the applied torque.

Suppose the same woman applied a force with the same magnitude on the same wrench at the same distance from the axis of rotation, but she didn”t apply the force perpendicular to the (lever arm) wrench? How would that change our answer?

**Example 2**

A woman applies a force of 185 N at an angle of 60.0° to the handle of a wrench at a distance of 85 cm from center of the nut (axis of rotation) she is turning. Calculate the magnitude of the applied torque.

So, the woman would be exerting the same effort, but only some of her effort would translate to torque. Where would the rest of her effort be directed? Into trying to push the nut away from her, parallel to the direction of the handle.

Torque problems become more complex as multiple forces are applied. Sometimes multiple forces work together, and sometimes they oppose each other. To account for this, torques that would cause clockwise rotation are usually considered negative, while torques that would cause counterclockwise rotation are considered positive.

**Example 3**

A child with a mass of 35.0 kg sits on the right side of a horizontal seesaw, at a distance of 1.2 m from the pivot point. How far to the left of the pivot point should a 25.0 kg child sit in order to achieve rotational equilibrium?

**Convert:**

We will use Newton”s second law and the mass of the children to find their weights, which will be equal to the forces that they will exert on the seesaw. Because the seesaw is horizontal, the force of each weight will be applied perpendicular to it.

F_{1} = FW_{1} = *m*g = (35.0 kg)(9.81 m/s^{2}) = 343.35 N = 343 N

F_{2} = FW_{2} = *m*g = (25.0 kg)(9.81 m/s^{2}) = 245.25 N = 245 N

**Find:** d_{2}

**Isolate:**

We start with rotational equilibrium: *τ*_{net} = *τ*_{1} + *τ*_{2} = 0 N · m

Subtracting *τ*_{1} from both sides, we get: *τ*_{2} = *−τ*_{1}

Which is equivalent to: F_{2} d_{2} sin *θ*_{2} = –(F_{1} d_{1} sin *θ*_{1})

So, the second child should sit 1.68 m to the left of the pivot point. The negative sign in our answer indicates that this child should be on the opposite side of the pivot point as the first child.

Lesson 4–2 Review

__1.__ ______________ is the ability of an applied force to produce rotational motion.

__2.__ How far from a pivot point should a 25.0 N force be applied perpendicularly to the surface of a lever to achieve a torque of 17.5 N × m?

__3.__ A force of 18.5 N is applied at 35.0° to the surface of a seesaw at a distance of 0.875 m from the pivot point. Find the torque produced by this force.

__4.__ Imagine that you need to change a flat tire on your car. What is the advantage of working with a longer wrench when trying to remove the lug nuts?