10-27-99
Sections 8.4 - 8.6
We've looked at the rotational equivalents of displacement, velocity, and acceleration; now we'll extend the parallel between straight-line motion and rotational motion by investigating the rotational equivalent of force, which is torque.
To get something to move in a straight-line, or to deflect an object traveling in a straight line, it is necessary to apply a force. Similarly, to start something spinning, or to alter the rotation of a spinning object, a torque must be applied.
A torque is a force exerted at a distance from the axis of rotation; the easiest way to think of torque is to consider a door. When you open a door, where do you push? If you exert a force at the hinge, the door will not move; the easiest way to open a door is to exert a force on the side of the door opposite the hinge, and to push or pull with a force perpendicular to the door. This maximizes the torque you exert.
I will state the equation for torque in a slightly different way than the book does. Note that the symbol for torque is the Greek letter tau. Torque is the product of the distance from the point of rotation to where the force is applied x the force x the sine of the angle between the line you measure distance along and the line of the force:
In a given situation, there are usually three ways to determine the torque arising from a particular force. Consider the example of the torque exerted by a rope tied to the end of a hinged rod, as shown in the diagram.
The first thing to notice is that the torque is a counter-clockwise torque, as it tends to make the rod spin in a counter-clockwise direction. The rod does not spin because the rope's torque is balanced by a clockwise torque coming from the weight of the rod itself. We'll look at that in more detail later; for now, consider just the torque exerted by the rope.
There are three equivalent ways to determine this torque, as shown in the diagram below.
Method 1 - In method one, simply measure r from the hinge along the rod to where the force is applied, multiply by the force, and then multiply by the sine of the angle between the rod (the line you measure r along) and the force.
Method 2 - For method two, set up a right-angled triangle, so that there is a 90° angle between the line you measure the distance along and the line of the force. This is the way the textbook does it; done in this way, the line you measure distance along is called the lever arm. If we give the lever arm the symbol l, from the right-angled triangle it is clear that
Using this to calculate the torque gives:
Method 3 - In this method, split the force into components, perpendicular to the rod and parallel to the rod. The component parallel to the rod is along a line passing through the hinge, so it is not trying to make the rod spin clockwise or counter-clockwise; it produces zero torque. The perpendicular component (F sinq) gives plenty of torque, the size of which is given by:
Any force that is along a line which passes through the axis of rotation produces no torque. Note that torque is a vector quantity, and, like angular displacement, angular velocity, and angular acceleration, is in a direction perpendicular to the plane of rotation. The same right-hand rule used for angular velocity, etc., can be applied to torque; for convenience, though, we'll probably just talk about the directions as clockwise and counter-clockwise.
We've looked at the rotational equivalents of several straight-line motion variables, so let's extend the parallel a little more by discussing the rotational equivalent of mass, which is something called the moment of inertia.
Mass is a measure of how difficult it is to get something to move in a straight line, or to change an object's straight-line motion. The more mass something has, the harder it is to start it moving, or to stop it once it starts. Similarly, the moment of inertia of an object is a measure of how difficult it is to start it spinning, or to alter an object's spinning motion. The moment of inertia depends on the mass of an object, but it also depends on how that mass is distributed relative to the axis of rotation: an object where the mass is concentrated close to the axis of rotation is easier to spin than an object of identical mass with the mass concentrated far from the axis of rotation.
The moment of inertia of an object depends on where the axis of rotation is. The moment of inertia can be found by breaking up the object into little pieces, multiplying the mass of each little piece by the square of the distance it is from the axis of rotation, and adding all these products up:
Fortunately, for common objects rotating about typical axes of rotation, these sums have been worked out, so we don't have to do it ourselves. A table of some of these moments of inertia can be found on page 223 in the textbook. Note that for an object where the mass is all concentrated at the same distance from the axis of rotation, such as a small ball being swung in a circle on a string, the moment of inertia is simply MR2 . For objects where the mass is distributed at different distances from the axis of rotation, there is some multiplying factor in front of the MR2.
You can figure out the rotational equivalent of any straight-line motion equation by substituting the corresponding rotational variables for the straight-line motion variables (angular displacement for displacement, angular velocity for velocity, angular acceleration for acceleration, torque for force, and moment of inertia for mass). Try this for Newton's second law:
Replace force by torque, m by I, and acceleration by angular acceleration and you get:
We've dealt with this kind of problem before, but we've never accounted for the pulley. Now we will. There are two masses, one sitting on a table, attached to the second mass which is hanging down over a pulley. When you let the system go, the hanging mass is pulled down by gravity, accelerating the mass sitting on the table. When you looked at this situation previously, you treated the pulley as being massless and frictionless. We'll still treat it as frictionless, but now let's work with a real pulley.
A 111 N block sits on a table; the coefficient of kinetic friction between the block and the table is 0.300. This block is attached to a 258 N block by a rope that passes over a pulley; the second block hangs down below the pulley. The pulley is a solid disk with a mass of 1.25 kg and an unknown radius. The rope passes over the pulley on the outer edge. What is the acceleration of the blocks?
As usual, the first place to start is with a free-body diagram of each block and the pulley. Note that because the pulley has an angular acceleration, the tensions in the two parts of the rope have to be different, so there are different tension forces acting on the two blocks.
Then, as usual, the next step is to apply Newton's second law and write down the force and/or torque equations. For block 1, the force equations look like this:
For block 2, the force equation is:
The pulley is rotating, not moving in a straight line, so do the sum of the torques:
Just a quick note about positive directions...you know that the system will accelerate so that block 2 accelerates down, so make down the positive direction for block 2. Block 1 accelerates right, so make right the positive direction for block 1, and for the pulley, which will have a clockwise angular acceleration, make clockwise the positive direction.
We have three equations, with a bunch of unknowns, the two tensions, the moment of inertia, the acceleration, and the angular acceleration. The moment of inertia is easy to calculate, because we know what the pulley looks like (a solid disk) and we have the mass and radius:
The next step is to make the connection between the angular acceleration of the pulley and the acceleration of the two blocks. Assume the rope does not slip on the pulley, so a point on the pulley which is in contact with the rope has a tangential acceleration equal to the acceleration of any point on the rope, which is equal to the acceleration of the blocks. Recalling the relationship between the angular acceleration and the tangential acceleration gives:
Plugging this, and the expression for the moment of inertia, into the torque equation gives:
All the factors of r, the radius of the pulley, cancel out, leaving:
Substituting the expressions derived above for the two tensions gives:
This can be solved for the acceleration:
Accounting for the mass of the pulley just gives an extra term in the denominator. Plugging in the numbers and solving for the acceleration gives:
Neglecting the mass of the pulley gives a = 5.97 m/s2.