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Experiment 8 Rotational Equilibrium of Torques

Questions

What are the definitions of torque, moment of inertia and angular acceleration? How does the moment of inertia relate to probability distributions? How do torques add and what is necessary to establish rotational equilibrium? Concepts Part 1 Torque A torque is a ‘turning force’. It results when a force is applied to an object some distance from an axis of rotation. Fundamentally, torque is the multiplication of the force and the distance. A single torque applied to an object will cause it to rotate and have an angular acceleration. As with static equilibrium, two or more torques applied at different points on an object may cancel each other out. Then, the (initially stationary) rotatable object does not rotate. Mathematically, the torque due to a force about a given axis is defined as the vector cross product of the force and a radius vector which extends from the rotation axis to the point where the force acts. or (1) The vector cross product is a third vector that points perpendicular to the plane containing the original two vectors. The length of this third vector is the product of the magnitude of each vector and the sine of the angle between these two vectors. Two torques oppose one another when their vectors point in opposite directions. When the object and the applied force are drawn on a page and the axis of rotation is perpendicular to the page, a positive torque is taken to point out of the plane of the page. In Eqn. (1) the factor of sinq could be written next to the either the radius or the Force. This suggests two equivalent methods for calculating a torque. Consider a ladder leaning against a wall as in Figure 2. The ladder can pivot about its base and its weight, W, exerts a torque on the ladder. The value of the torque depends on the angle of elevation, e, of the ladder. Notice that the angle between r and W is q as in Eqn (1). Method 1: The position vector r points from the pivot to where the force is applied. Then, r sinq is always the perpendicular distance from the pivot to a line, which runs through the force vector (dotted in Fig. 1). This perpendicular distance has two names. It is called the moment arm or the lever arm. It is symbolized by r^ which is read as r-perp. Therefore t = r^ F. (2)

! τ ≡ ! r × ! F τ = rFsinθ

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Figure 1. A rectangular ladder leaning against a wall’s corner. Method 2: If the sinq is written next to the Force, then F sinq is the component of the force which is perpendicular to the radius vector. This component is symbolized by F^, and it is read as, (you guessed it) F-perp. In this way, t = r F^. After calculating individual torques, you will apply the condition for rotational equilibrium: the vector sum of all torques exerted by these forces about any chosen axis (perpendicular to the plane of the forces) is zero:

(3)

Method

Figures 2 and 3 depict equipment used in the first part of this week’s experiment (a model crane boom and the Torque Wheel). The torque wheel can rotate about its center. In Fig. 3, the force F will produce a clockwise rotation. Clockwise torques are usually taken to be negative (directed into the plane of the wheel). The magnitude of the torque is given by the value of the force, F multiplied by its moment arm.

Figure 2. Model Crane Boom on Figure 3. The Torque Wheel on

the Pasco Mechanics Board the Pasco Mechanics Board

W

W's line of action

r

ε

θ

r

τ

i i=1

n

∑ = 0

10 20

30 40millimeter scale Force, F

moment arm

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Procedure 1) Logistics Note: As with the static forces in experiment earlier this semester, three

exercises are described in this part. Each person, in a group of three, can take the lead in setting up the apparatus, making measurements and calculating values. However, each lab partner must have a hand-written record of all three sets of data.

2) Measure and record the mass of the rotatable bar and separately the mass of the angle

scales where the plastic bars mount. Do not include the mass of the bar’s magnetic mount.

3) First, mount the rotatable bar horizontally with the pivot at the center of the bar. Apply

unequal weights to opposite ends of the bar and make adjustments until the bar is horizontal and in rotational equilibrium. Test for rotational equilibrium by nudging the bar to see if it returns to its horizontal position. Calculate the net torque on the bar and compare it to zero. Add a very small mass to one mass hanger to determine the minimum additional torque that eliminates rotational equilibrium.

4) Second, set up the model crane boom as pictured in Fig. 2. Measure and record masses, angles and locations of the forces on the bar. Use the sum of torques to calculate the tension in the ‘cable’ that attaches to the middle part of the rotatable bar. Compare this tension to the weight of the hanging mass. How much additional mass must be added to the crane boom’s ‘load’ for the spring scale’s reading to change appreciably?

5) Third, set up rotational equilibrium of the torque wheel using three hanging masses and

three pulleys. Do not place much more than about 100 grams in each mass hanger. Also, avoid a situation where two of the three torques nearly sum to zero and the third torque is almost zero. Again, test for equilibrium by nudging the wheel clockwise and counter- clockwise. Measure the masses, tension in the string and the moment arms for all three torques. Calculate the sum of all torques and compare this to zero. Again, add a small amount of mass to one mass hanger to determine the minimum additional torque that eliminates rotational equilibrium.

6) Spend some time thinking about the sources of error in Part 1. For each of the three

procedures above, calculate the amount of mass added to eliminate equilibrium as a percentage of the smallest force used to obtain equilibrium. Discuss how to do this with your partners, make the necessary measurements and calculations. You can consider error negligible if it is less than 10% of the smallest force in each procedure.