### Rotary motion

In practice most machines involve rotary motion as well as linear motion. This could be such examples as electric motors, gears, pulleys and internal combustion engines. Therefore if we need to calculate how fast a machine will reach full speed (in other words calculate the acceleration of all its components) then we must consider rotary acceleration, and the associated torques, as well as linear acceleration. Fortunately Newton’s second law of motion applies equally well to rotary motion provided we use the correct version of the formula.

Suppose we are looking at the acceleration of a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. We cannot use the conventional form of Newton’s second law, F = ma, because although there is a linear force being applied in the form of the tension in the cord, the acceleration is definitely not in a line. We therefore cannot identify an acceleration a in units of m/s

The mass of an object is really the resistance that the object offers to being moved in a straight line even when there is no friction. It can also be called the inertia of the object; an object with a large mass will accelerate much slower than one with a low mass under the action of an identical force. We are therefore looking for the resistance that an object offers to being rotated in the absence of friction. This must be a combination of mass and shape because a flywheel where most of the mass is concentrated in the rim, supported by slender spokes, will offer much more resistance than a uniform wheel of the same mass. The quantity we are seeking is something known as the

kilograms times metres squared (kg.m

I = mr

We are now in a position to write Newton’s second law in a form suitable for rotary motion

t = Ia

With this equation we can solve problems involving real engineering components that utilize rotary motion and then go on to consider combinations of linear and rotary components. One of the most common rotary motion devices is the flywheel. This is a massive wheel which is mounted on a shaft to deliberately provide a great deal of resistance to angular acceleration. The purpose of the device is to ensure a smooth running speed, especially for something that is being driven by a series of pulses such as those coming from an internal combustion engine.

This general method can be applied to much more complicated shapes such as pulleys which may be regarded as a series of uniform discs on a common axis. Because all the components are all on the same axis, the individual moments of inertia may all be added to give the value for the

single shape. Sometimes the moment of inertia of a rotating object is given directly by the manufacturer but there is also another standard way in which they can quote the value for designers. This is in terms of something called the

I = mk

The first part of this problem is an application of kinematics to rotary motion where we know the time, the initial velocity and the final velocity, and we need to find the acceleration. The equation to use is therefore

w

The start velocity is zero because the flywheel begins from rest. The final angular velocity is

3000 rev/min = 3000 x 2p/60 = 314.2 rad/s

Therefore

314.2 = 0 + 40a

a = 7.854 rad/s

Now that the angular acceleration is known it is possible to calculate the torque or couple which is necessary to produce it, once the moment of inertia is calculated. The flywheel is a uniform disc so the moment of inertia is given by I = mr

Therefore the driving torque is given by

t = Ia = 27 x 7.854 =

Suppose we are looking at the acceleration of a solid disc mounted on a shaft and rotated by a pull cord wrapped around its rim. We cannot use the conventional form of Newton’s second law, F = ma, because although there is a linear force being applied in the form of the tension in the cord, the acceleration is definitely not in a line. We therefore cannot identify an acceleration a in units of m/s

^{2}. Furthermore some of the mass of the disc is close to the axle and not moving from one point to another, only turning on the spot, while some of the mass is moving very quickly at the rim. Now in the section on kinematics we saw how the equations of uniform motion could be adapted for rotary motion by substituting the equivalent rotary quantity instead of the linear term. We found that linear distance travelled s could be replaced by angle of rotation q, while linear acceleration a could be replaced by the angular acceleration a, where a = a/r. It is also quite clear that if we are looking at rotary motion then we should be using the torque t = Fr instead of force. All that remains is to find the quantity that will be the equivalent of mass in rotary motion.The mass of an object is really the resistance that the object offers to being moved in a straight line even when there is no friction. It can also be called the inertia of the object; an object with a large mass will accelerate much slower than one with a low mass under the action of an identical force. We are therefore looking for the resistance that an object offers to being rotated in the absence of friction. This must be a combination of mass and shape because a flywheel where most of the mass is concentrated in the rim, supported by slender spokes, will offer much more resistance than a uniform wheel of the same mass. The quantity we are seeking is something known as the

*mass moment of inertia*or simply as the*moment of inertia, I*, which has units ofkilograms times metres squared (kg.m

^{2}). For a uniform disc, which is the most common shape found in engineering objects such as pulleys, this is given by the formulaI = mr

^{2}/2We are now in a position to write Newton’s second law in a form suitable for rotary motion

t = Ia

With this equation we can solve problems involving real engineering components that utilize rotary motion and then go on to consider combinations of linear and rotary components. One of the most common rotary motion devices is the flywheel. This is a massive wheel which is mounted on a shaft to deliberately provide a great deal of resistance to angular acceleration. The purpose of the device is to ensure a smooth running speed, especially for something that is being driven by a series of pulses such as those coming from an internal combustion engine.

This general method can be applied to much more complicated shapes such as pulleys which may be regarded as a series of uniform discs on a common axis. Because all the components are all on the same axis, the individual moments of inertia may all be added to give the value for the

single shape. Sometimes the moment of inertia of a rotating object is given directly by the manufacturer but there is also another standard way in which they can quote the value for designers. This is in terms of something called the

*radius of gyration, k*. With this the moment of inertia of a body is found by multiplying the mass of the body by the square of the radius of gyration such thatI = mk

^{2}**An electric motor is being used to accelerate a flywheel from rest to a speed of 3000 rev/min in 40 s. The flywheel is a uniform disc of mass 96 kg and radius 0.75 m. Calculate the angular acceleration required and hence find the output torque which must be produced by the motor.**__Example:__The first part of this problem is an application of kinematics to rotary motion where we know the time, the initial velocity and the final velocity, and we need to find the acceleration. The equation to use is therefore

w

_{2}= w_{1}+ atThe start velocity is zero because the flywheel begins from rest. The final angular velocity is

3000 rev/min = 3000 x 2p/60 = 314.2 rad/s

Therefore

314.2 = 0 + 40a

a = 7.854 rad/s

^{2}Now that the angular acceleration is known it is possible to calculate the torque or couple which is necessary to produce it, once the moment of inertia is calculated. The flywheel is a uniform disc so the moment of inertia is given by I = mr

^{2}/2 = 96 x 0.75^{2}/2 = 27 kg.m^{2}Therefore the driving torque is given by

t = Ia = 27 x 7.854 =

**212 N.m**
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