### Linear vs Rotational Equations

In Circular motion speed review, we covered rotational kinematics. Here we saw that all linear variables has a “rotational counterpart.” The chart below should help you give an idea on how to turn any linear concept into the rotational counterpart.

Linear Variables | Rotational Variables |
---|---|

∆x (displacement in meters ) | ∆θ (angular displacement in radians) |

∆v (velocity in m/s) | ∆ω (angular velocity in rad/s) |

a (acceleration in m/s^{2}) | 𝛂 (angular acceleration in rad/s^{2}) |

m (mass aka inertia) | I (rotational inertia in kg·m^{2}) |

Formulas | Replace linear variable with rotational counterparts |

F = ma | τ = I𝜶 |

KE = 1/2mv^{2} | KE _{rotational}= 1/2Iω^{2} |

p = mv → Linear momentum | L = Iω → Angular momentum |

∆p = Ft → Impulse | ∆L = τ(t) → Angular impulse |

### Newton’s Second Law

The equation for torque is similar to Newton’s Second Law, Σ F = ma.

Torque = Σ **τ = ***I***𝜶**, this tells us the SUM of all torques acting on a mass will cause rotation.

Just as a net linear force causes a linear acceleration, a net rotational force (Torque) causes a rotational acceleration (𝜶).

*Note: Angular acceleration and rotational acceleration are used interchangeably. They are both written in rad/s^{2} and mean the same thing.

Another equation for Torque is **T = Fr(sinϴ)**

- where “F” is the applied force.
- “r” is the distance between the pivot point and the place where the force is applied. We call this the “lever arm”
- Note the
**(sinϴ)**tells us to use ONLY the component of force applied**perpendicularly**to the lever arm.

In some cases, you will have to combine both equations to solve the problem:

*I*𝜶 = Fr(sinϴ)

#### Rotational Inertia

Note that rotational inertia is also referred to as a *moment of inertia.* It’s very similar to how we use mass, BUT it also involves the use of the radius of the object.

You do NOT need to memorize the equations below for the AP exam. And they will not be given on the formula sheet.

You DO need to know the rotational inertia for a point mass:

**I _{point mass} = mr^{2}**

*AP Physics C students might be asked to derive the formula for the rotational inertia of a disk/rod. This involves calculus, so be sure to know how to do this.

Here some other rotational inertia formulas for other common objects.

#### Direction of Torque

Generally we stick to counter clockwise is positive and clockwise is negative.

An easy way to figure out direction. Take your pencil and push your finger in the direction of the force applied in the problem. See which way your pencil spins.

#### Parallel Axis Theorem

*This is for AP Physics C students* *or higher*.

The parallel axis theorem shows how to find the rotational inertia of a mass when the axis of rotation is **not **the center of mass. For example, a disk rotating 3/4 away from the center.

The parallel axis theorem is given as *I* = *I*_{cm} + md^{2}

*I*_{cm}is the rotational inertia of the center of mass.- d is the distance away from the center of mass
- m is the mass of the object

A common question type is when an object with velocity v sticks to another object (like a disk) and causes an angular acceleration. Another question is a roll of paper that unravels along the floor (thus does not rotates through the center).

### 3. Types of torque problems

Here I’ll list the most common types of force problems you will see. It will be worthwhile studying these types of questions and understanding how to solve them. For more practice, there are 60 questions below to help you master torque!

#### Static Equilibrium Questions (**τ**_{net} = 0 → 𝜶 = 0)

_{net}= 0 → 𝜶 = 0

- fulcrum and lever problems
- Tension of a slanted cable required to hold a horizontal platform
- Ladder problems, involving max weight on a ladder before slipping, required friction, etc,

#### Dynamic Torque Questions (**τ**_{net} ≠ 0 → 𝜶 ≠ 0))

_{net}≠ 0 → 𝜶 ≠ 0

- Finding the torque a pulley exerts on a system
- Finding the acceleration of a pulley system. Or finding the either tension in a pulley system
- Frictional torque acting on a system slows it down
- Finding angular acceleration of rotating platforms (like a merry-go-round) given torque

#### Rolling

- Objects, like spheres, will
*roll,*not slide down ramps. - Static friction causes a torque and rotation around the objects center.
- You will also need to apply conservation of energy and utilize rotational kinetic energy

### Solving Torque Problems

Torque problems can be solved just like regular force problems. If you need a quick refresher on how to solve linear force problems skim through Forces Speed Review.

**Quick Recap So far**

- Instead of up and down, we use clockwise and counter-clockwise as directions.
- Generally, we say that clockwise is the negative direction
- Since toque is a force we can use it in combination with linear forces. This basically means we can have a system of 3 equations: 1 for torque (rotational direction), 1 for horizontal forces (linear-x-direction), and 1 for vertical direction (linear-y-direction)
- Most often friction between the axis of rotation and mass (such as a disk) is negligible. But when friction is
**not**negligible it produces a counter-torque, which causes an angular*deceleration*.

#### Problem Solving Framework for Torque Problems

Follow this outline to solve **any/all** torque problem:

- Draw an FBD/diagram of the situation. You will want to also show the lever arm in the diagram.
- Identify the pivot point. This is simply the point around which the object rotates. For example, a rod of uniform mass tossed up will rotate about its midpoint. You measure the lever arm (r)
*from*the pivot point. - Find the net Torque (
**τ**)of the object. Remember that_{net}**τ**is simply the SUM of all torques around the pivot point. Note that static equilibrium = balanced torque → τ_{net}_{net}= 0.- You can use the formula
**T = Fr(sinϴ)**to find each toruqe then add/subtract them based on the direction.

- You can use the formula
- Lastly, set the net toque you found (
**τ**) equal to_{net}and solve your equation for the unknown value. Note in some cases the system is in static equilibrium, which means that there’s no angular acceleration. Thus*I*𝜶= 0, because*I*𝜶**𝜶**= 0. You still want to make an equation for**τ**in any case._{net}

### Rotational Energy and Momentum

Energy and Momentum are others ways to solve toque problems!

As with linear energy and momentum, there is also rotational energy and momentum. The formulas are identical (just different variables as outlined in the table 1 above).

This will be covered in a separate post.

### 10 Challenging Problems to Help You Master Torque

Since you are at the end of the AP Physics 1 curriculum, these questions were made with the AP exam in mind. There are NO numbers involved. Just variables and pure derivation will be used. All these questions are ranked > 8/10 in difficulty.

##### (1) A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?

##### (2) A uniform solid cylinder of mass M and radius R is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it, as shown in the figure below. The string is then pulled with a constant force F, causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle θ, what is the kinetic energy of the cylinder (in terms of F and θ)?

##### (3) A seesaw is balanced on a fulcrum, with a boy of mass M1 sitting on one end and a girl of mass M2 sitting on the other end. The seesaw is a uniform plank of length L and mass M. The fulcrum is located at the midpoint of the plank.

##### (4) The boy is sitting at a distance d1 from the fulcrum, and the girl is sitting at a distance d2 from the fulcrum, with d1 > d2. The seesaw is level, with the two ends at the same height. What is the minimum mass of the seesaw that will keep it balanced with the two children sitting on it?

##### (5) [Numbers given] A pulley system consists of two blocks of mass 5 kg and 10 kg, connected by a rope of negligible mass that passes over a pulley of radius 0.1 meters and mass 2 kg, as shown in the figure. The pulley is free to rotate about its axis. The system is released from rest, and the block of mass 10 kg starts to move downwards. Assuming that the coefficient of kinetic friction between the pulley and the rope is 0.2, and neglecting air resistance, determine the acceleration of the system, the tension in the rope, and the magnitude and direction of the frictional force exerted on the pulley.

### (7) Five More Problems

##### (1) A 5-meter long ladder is leaning against a wall, with the bottom of the ladder 3 meters from the wall. The ladder is uniform and has a mass of 20 kg. A person of mass 80 kg is standing on the ladder at a distance of 4 meters from the bottom of the ladder. The ladder makes an angle of 60 degrees with the ground. What is the force exerted by the wall on the ladder?

⇨ concepts involved: Torque, linear forces, static torque

⇨ difficulty: medium

##### (2) A meter stick of mass 0.2 kg is pivoted at one end and supported horizontally. A force of 3 N is applied to the free end, perpendicular to the length of the meter stick. What is the net torque about the pivot point?

(a) 0

(b) 1.50

(c) 2.48

(d) 3.00

⇨ concepts involved: net torque

⇨ difficulty: easy

##### (3) A meter stick of mass 200 grams is balanced at the 40-cm mark when a 100-gram mass is suspended from the 10-cm mark. What is the distance in centimeters from the pivot point to the center of mass of the meter stick?

⇨ concepts involved: static equilibrium, toque

⇨ difficulty: medium

##### (4) A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?

⇨ concepts involved: non-static torque, tension

⇨ difficulty: hard

##### (5) A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?

(a) g/L

(b) 3g/4L

(c) (m+1)g/L

(d) (3*m*g)/(2L)

⇨ concepts involved: moment of inertia, torque

⇨ difficulty: medium

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