Physics book - Ch 9 Conceptual Book – Ch 11 Unit 8 - Rotation Physics book - Ch 9 Conceptual Book – Ch 11

Part 1 - Torque Physics Book Sections 9.1-9.3 Conceptual Book Sections 11.1-11.3

Rotation Rotation is “spinning” around an axis Rotation is different than revolution Rotation: turning around an internal axis – the Earth’s rotation gives us the days and nights. Revolution: circling an external axis – the Earth’s revolution around the sun gives us the years (and with the tilt on its axis the seasons).

Rotational Equilibrium Previously we have studied translation equilibrium for an object acted on by forces that act at a single point. Equilibrium if 𝐹 =0. (net force = 0) When a body is acted on by forces that DO NOT have a common line of action, it may be in translational but not rotational equilibrium. It may not move up, down, left, or right, but it may rotate. So, we must consider both the point of application and the magnitude in studying rotational equilibrium.

Line of Action and Axis of Rotation The line of action of a force is an imaginary line extended indefinitely along the vector in both directions. When lines of action at not the same, then rotation may occur about a point called the axis of rotation.

The Moment Arm The distance from the axis of rotation to the line of action of a force is called the moment arm (r). The moment arm determines the effectiveness of a force in causing rotational motion. The bigger the moment arm (the farther away from the axis) the easier it is to cause rotation.

Torque Torque is the tendency to produce a change in rotational motion. The “twisting force” that causes rotation 𝑇𝑜𝑟𝑞𝑢𝑒=𝑓𝑜𝑟𝑐𝑒 × 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚 𝜏=𝐹𝑟 If 𝜃≠90°: 𝜏=𝑟𝐹 sin 𝜃

Torque 𝑟 (the moment arm) should be measured perpendicular to the line of action of the force F. 𝒓 𝐬𝐢𝐧 𝜽 adjusts the measurement for us. Units for torque: Newton-meters (𝑁∙𝑚) Torque is a vector! Counter-clockwise rotation - torque is positive. Clockwise rotation – torque is negative.

Example #1 - Torque A force of 250 N is exerted on a cable wrapped clockwise around a drum that has a diameter of 120 mm. What is the torque produced about the center of the drum? Since r is perpendicular to the line of action of the force (𝜃=0°), the moment arm is equal to the radius of the drum. 𝜏=𝐹𝑟= 250 𝑁 0.06 𝑚 =15.0 𝑁∙𝑚 CW motion, so 𝜏=−15.0 𝑁∙𝑚

Example #2 - Torque A mechanic exerts a 17.0 N force at the end of a 25 cm wrench. If this pull is directed leftward at an angle of 37° from the handle, what is the torque produced? 𝜏=𝑟𝐹 sin 𝜃 =.25 17.0 sin 37° ≈ 2.56 𝑁∙𝑚 Motion is CCW, so 𝜏=+2.56 𝑁∙𝑚

Resultant Torque As with forces, we can add torques together to find resultant torques. Be sure to give each torque its correct sign before adding. The resultant will determine whether there will ultimately be clockwise or counter-clockwise rotation.

Resultant Torque Example Two wheels of diameter 60 cm and 20 cm are fastened together and turn on the same axis. A 150 N weight is attached to the right side of the smaller wheel, and a 200 N weight is attached to the left side of the bigger wheel. What is the resultant torque about the central axis? 𝜏 =− .10 150 + .30 200 =45 𝑁∙𝑚 Answer: 45 N∙m CCW

Equilibrium In order for an object to be in translational equilibrium, we must have 𝐹=0 (in both x and y directions). This means that 𝐹 𝑢𝑝 = 𝐹 𝑑𝑜𝑤𝑛 In order for an object to be in rotational equilibrium, we must have 𝜏 =0. This means that 𝜏 𝑐𝑤 = 𝜏 𝑐𝑐𝑤 Both conditions must be true for total equilibrium (object is not accelerating and not spinning faster or slower)

Ex: How far away from the fulcrum should the blue mass be placed so that the see saw balances? 𝜏 𝑐𝑤 = 𝜏 𝑐𝑐𝑤 3 9.8 2 =5 9.8 𝑟→𝑟=1.2 𝑚

You Try: How far away from the fulcrum should the 3N weight be placed to balance the apparatus? 𝜏 𝑐𝑤 = 𝜏 𝑐𝑐𝑤 3𝑟=5(0.2)→𝑟=0.33 𝑚

Application: Parallel Force Problems Ex: A 60-kg painter stands 2.0 m from one end of a 6.0 m plank that is supported at each end by a scaffold. How much of the painter’s weight must each end of the scaffold support? Ignore the weight of the plank. Problems like this are common IRL. Engineers must be able to make calculations like this in order to ensure the safety of public in buildings, on bridges, etc. Parallel force because we are looking at all vertical forces, but they are on different lines of action.

Ex #1: A 60-kg painter stands 2. 0 m from one end of a 6 Ex #1: A 60-kg painter stands 2.0 m from one end of a 6.0 m plank that is supported at each end by a scaffold. How much of the painter’s weight must each end of the scaffold support? Ignore the weight of the plank. Begin by drawing a force diagram. In this problem, you can choose the point of rotation. Usually an end is simplest so that there will be no torque at one end. First use 𝜏 𝑐𝑤 = 𝜏 𝑐𝑐𝑤 : 𝐹 𝑤 𝑟 𝑤 = 𝐹 2 𝑟 2 60 9.8 2.0 = 𝐹 2 6.0 𝐹 2 =196 𝑁 Next use 𝐹 =0: 𝐹 1 + 𝐹 2 − 𝐹 𝑤 =0 𝐹 1 +196− 60 9.8 =0 𝐹 1 =392 𝑁

You Try Ex #2: A bricklayer with a mass of 75 kg stands on an 8.0 m scaffold 3.0 m from the left end. He has a pile of bricks, which have a mass of 20 kg, 3.0 m from the other end. How much weight much each end support? Ignore the weight of the scaffold. Answers: left end: 𝐹 1 =533 𝑁 right end: 𝐹 2 =398 𝑁

Center of Gravity The center of gravity of any object is that point at which all of its weight can be considered to be concentrated. 𝐶𝐺= 𝑇𝑜𝑡𝑎𝑙 𝑇𝑜𝑟𝑞𝑢𝑒 𝑇𝑜𝑡𝑎𝑙 𝑊𝑒𝑖𝑔ℎ𝑡 For uniform objects, the center of gravity is in the middle or center. For non-uniform objects, the center of gravity can be toward one end or the other. To do problems involving non-uniform objects, the problem will have to give you the center of gravity.

Center of Gravity Problem (Uniform) Ex #3: A 50-m uniform steel beam (weight = 200 N) has the following weights on it: 175 N at the left end, 150N at 10 m from the left, 75 N at 20 m from the left, 25 N at 10 m from the right, and 50 N at the right end. Where is the center of gravity for the whole system? Make a table of weights, moment arms, and torques. Don’t forget the weight of the beam in the middle! Table and answer on next slide.

Center of Gravity Problem (Uniform) Answer: 𝐶𝐺= 11500 675 ≈17 𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 Force (Weight) Moment Arm Torque 200 N 25 m 5000 Nm 175 N 0 m 0 Nm 150 N 10 m 1500 Nm 75 N 20 m 25 N 50-10 = 40 m 1000 Nm 50 N 50 m 2500 Nm Total: 675 N Total: 11,500 Nm

You Try: Non-Uniform CG Ex #4: The empty body of a sports car (length = 6 m) weighs 1200 N and has a center of gravity that is 3.5 m from the front end. The following additions are made to the car: an 800 N engine is added at 1 m from the front end, front seats (300 N) are added at 2.5 m from the front end, and back seats (250 N) are added at 2 m from the back end. What is the car’s new center of gravity? Answer: 2.6 meters from the front end.