(Omit 7.6, Rolling Motion)

The fundamental idea of Newtonian dynamics is that "things happen for a reason;" to be more specific, there is no need to explain rest or motion in a straight line at a constant speed. However, acceleration must be caused; that is, acceleration doesn't just happen, it must be explained, and the reason acceleration occurs is that there must be a net force to cause it.

The same is true for rotational motion, but we can be a little more specific in this case. Angular acceleration must be caused, and the cause of angular acceleration is that there must be a net force. However, the force has to be oriented in a certain way to cause angular acceleration. This leads us to the concept of the torque of a force. (In other countries, the term "moment of a force" is used instead of "torque of a force;" the two terms are interchangeable and mean exactly the same thing.)

Torque

The torque of a force relative to a certain axis is a measure of the effectiveness of the force in producing rotation of an object about the axis. The torque depends on the magnitude of the force, the direction of the force, and the distance between the rotation axis and the point at which the force is applied.

examples: wrench, screwdriver, crow bar, wine press, etc.

Example: Determine the net torque of the tension forces about the axle of the pulley.

Chapter 7: Rotational MotionTuesday, September 17, 2013

10:00 PM

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Example: Determine the net torque about (a) point A, and (b) point B.

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Centre of Mass (Centre of Gravity)

For continuous mass distributions, one can use a similar formula involving integration.

Example: Determine the centre of mass for the system of three identical coins.

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Example: (a) Determine the centre of mass of the two bars considered as one unit.(b) Determine the net torque of the bars about the far left end.

Strategy: Pretend that the mass of each bar is concentrated at its geometrical centre.

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Centripetal and Tangential Acceleration

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Example: A disk of radius 5.0 cm starts from rest and rotates with a constant angular acceleration of 0.6 rad/s2. After 4.3 s, determine(a) the speed of a point on the edge of the disk.(b) the centripetal and tangential components of acceleration.

Newton's second law relates the total force acting on an object to its acceleration.

The rotational analogue of Newton's second law relates the net torque acting on an object to its angular acceleration.

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Rotational Dynamics; Moment of Inertia

The heart of Newtonian mechanics is Newton's second law of motion, which tells us how forces affect motion; specifically, Newton's second law tells us how much acceleration you get when a given net force acts on an object of mass m:

What is the analogue for rotational motion? Consider the following argument for a point particle:

It appears that mr2 plays the role for angular motion that m plays for linear motion; that is, mr2 is a kind of rotational inertia, or resistance to angular acceleration, in the same way that mass a measure of the resistance to acceleration. The quantity mr2 is called the moment of inertia for a point particle of mass m. For a collection of point masses, the total moment of inertia about a specified axis is

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For a continuous mass distribution (like a baseball bat, or rotating parts of machinery) one needs calculus to determine the moment of inertia. A table of moments of inertia for common objects is found on Page 216 of the textbook. It's worthwhile having a look at the table and getting a feel for it.

Examples: implements attached to baseball bats for warmups; sledge hammer

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Example: A pulley with a frictionless bearing and mass 1.3 kg and radius 0.35 m has a rope going over it as shown in the diagram. Determine the angular acceleration of the pulley.

The rope touches the pulley in such a way that the force exerted by the rope on the pulley is at right angles to the "radius vector" that starts at the centre of the pulley and touches the edge of the pulley at the point where the rope touches the pulley. Thus, the net torque on the pulley is

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Example: A wheel has a moment of inertia of 0.55 kg m2 and an initial angular speed of 0.85 rev/s. Friction in the wheel bearings causes the wheel to stop turning after 14.5 s. Determine the magnitude of the frictional torque.

Note that the sign of the angular acceleration is positive, which makes sense because the counterclockwise torque is greater than the clockwise torque, so the net torque is ccw.

First determine the angular acceleration of the wheel as it slows down, assuming that it is constant. Then use the relationship between torque and angular acceleration to determine the magnitude of the torque.

Also note that because we need only the magnitude of the torque, we can ignore the signs of the torque and angular acceleration.

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Example: A 1.2 kg block and a 1.8 kg block are attached to the ends of a rope that hangs over a solid pulley of diameter 0.42 m and mass 1.3 kg. The axle of the pulley is frictionless and the rim of the pulley has enough friction so that the rope rotates the pulley without slipping. Determine the accelerations of the blocks, the tensions in the rope, and the angular acceleration of the pulley.

As usual, the first step is to draw free-body diagrams. However, the new wrinkle in this problem is that we'll also draw a free-body diagram for the pulley. Then we'll apply Newton's second law to each block, and we'll also apply the rotational analogue of Newton's second law to the pulley.

In applying the rotational analogue of Newton's second law to the pulley, we will get an equation involving the angular acceleration. It will be helpful to connect this with the acceleration of the rope, which is the same as the tangentialacceleration of the rim of the pulley, because we assume that the rope turns the pulley without slipping. That is,

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Solve equations (1) and (2) for the tensions, and then substitute the results into equation (3), then solve for the acceleration.

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Are these results reasonable? Is there a way to check them?

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