Torque and Rotational Motion

Torque is force applied at a distance from an axis of rotation, and rotational motion is rotation around a fixed axis.


Before we delve into torque and rotational motion, there are some important things to know:

To convert any angular value to tangential, multiply the angular value by the radius of the rotational motion


l = lever arm, or distance from the axis of rotation at which the force is applied

θ = angular displacement = d/r

ω = angular velocity = change in θ/change in t = v/r

α = angular acceleration = change in ω / change in t = a/r

I = rotational inertia

L = angular momentum


Now that we have those out of the way, we can talk about rotational kinematics.  Remember the first unit?  Now, we are going to use all of those kinematic formulas once again, except  we are going to use rotational  variables.


The same method of solving problems in one-dimensional and two-dimensional kinematics is used when solving rotational kinematics problems.  Using these equations, we can solve a plethora of problems related to rotational kinematics.

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/drot2.gif
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/drot2.gif

Example Problem #1

Example problem #1: There is a rotating disk that starts with ω = 5 rad/sec.  After a force is applied to it over a distance of θ = 8 radians, it ends with ω = 10 rev/sec.  What was the angular acceleration?


Answer: 4.6875 rad/sec^2


Explanation: Because we have the initial angular velocity, final angular velocity, and change in angular displacement and need to solve for angular acceleration, we are going to use equation #3.  Plugging in those given values, we have: 100 = 25 + 2(α)(8).  Solving for α, we get that α = 4.6875 rad/sec^2.

Torque

Torque is the rotational equivalent to force.  Torque can be modeled by the equation τ = Flsinθ.  This means that as force goes up and as the lever arm distance goes up, the torque increases.  This makes sense, as if you try to open a door right next to the hinges with a small amount of force, the door will likely open very slowly, whereas if you open at the handle with a lot of force, it will open much quicker.  When the angle involved between the Force being applied and the lever that isn’t 90 degrees, you use the angle formed between the force and lever arm vectors when their tails are touching.


One common application that requires the use of the torque equation is balancing things on a board on a fulcrum.  When solving problems like these, calculate the net torque on each side of the fulcrum and solve for what you need to add/where you need to add it using that information.


Rotational inertia refers to an object’s resistance to rotation, and there are some problems where you will need to take this into account.  Usually, you will be given the value of rotational inertia, or given one of these formulas.


Angular momentum is another value that you will sometimes have to find, and it is represented by the equation L = Iω.

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Example Problem #2

Example problem #2: A perpendicular force of 10N directed North is applied to a door from a distance of .5m from its axis of rotation, and a perpendicular force of 5N directed South is applied in the other direction, from a distance of .8m from its axis of rotation.  Which way does the door rotate?


Answer: North


Explanation: Using the torque formula, we get that τ (North) = 10 x .5 = 5 Newton meters and that τ (South) = 5 x .8 = 4 Newton meters.  Since the τ (North) > τ (South), the door will rotate towards the North.

Rotational Kinetic Energy

When an object is rolling, it does not only have kinetic energy in the traditional sense, which we call translational kinetic energy, but also has rotational kinetic energy. If an object is spinning but not moving forwards or backwards, it only has rotational kinetic energy.


The formula for Rotational Kinetic Energy is the same as translational kinetic energy, except that the rotational analogs are used instead of their translational counterparts:

E=1/2Iω²


The most common application of something like this is an object rotating down a ramp. If this is the case, the gravitational kinetic energy is transferred to both translational kinetic energy and rotational kinetic energy. The conservation of energy equation will look like this:

mgh=1/2mv² + 1/2Iω².


This equation simplifies very well, see the picture on the right as an example.

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