Torque/Rotational Energy Mock Exam. Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK.

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1 AP Physics C Spring, 2017 Torque/Rotational Energy Mock Exam Name: Answer Key Mr. Leonard Instructions: (105 points) Answer the following questions. SHOW ALL OF YOUR WORK. (22 pts ) 1. Two masses are attached to opposite ends of a string as shown in the figure below. The two boxes have masses m 1 = 5.0 kg and m 2 = 10.0 kg. The pulley has a mass of 4.0 kg and a radius of 10 cm. 22 pts m 1 m 2 (a) (6 pts) Draw a force diagram for both masses as well as the pulley. m 1 m 2 Solution: T 2 T 1 T m 1 1 m 2 T 2 (b) (6 pts) Apply Newton s Second Laws to to the two boxes and the rotational version of Newton s Second Law to the pulley. Write the resulting expressions for each. Solution: Because the pulley has mass, a torque is required to rotate it. As a result the tensions on the two sides of the pulley must be different. This difference in tension is required to create the torque necessary to rotate the pulley. I will denote the tension in the rope connected to m 1 as T 1, and the tension in the rope connected to m 2 as T pts

2 AP Physics C/Torque/Rotational Energy Mock Exam Page 2 of 9 Name: Answer Key We use the translational Newton s Law ( F = ma) for m 1 and m 2 : F1 = m 1 a T 1 = m 1 a F2 = m 2 a F g,2 T 2 = m 2 a And we use the rotational Newton s Law ( τ = I α) for the pulley: τ = I α ( ) 1 T 2 R T 1 R = 2 M R 2 α T 2 T 1 = 1 M (R α) 2 T 2 T 1 = 1 2 M a Notice that we use relationship a = R α to express the equation above in terms of a. This was not necessary for part (b), but will set us up for part (c), where we are asked to solve for the acceleration. Note that you always must eliminate α and R from the rotational version of Newton s Second Lab before you can solve for the acceleration. (c) (4 pts) Use the equations you found in part (b) to solve for the acceleration of the two boxes. Solution: Recall that we can eliminate tension from these equations by adding them together. Remember the tension must cancel due to Newton s Third Law, and, if your tensions do not cancel when you add these equations together, there is a mistake in your application of Newton s Second Law. Adding the equations we found in part (b) together, and solving for a gives: T 1 + (F g,2 T 2 ) + ( T 2 T 1 ) = (m 1 + m ) M a m 2 g = (m 1 + m ) M a a = m 2 g m 1 + m M = 5.76 m s 2 (d) (6 pts) Calculate the tension at both ends of the rope. Solution: To find the tension, we simply plug the result we got from part (c) into the equations we found in part (b). T 1 = m 1 a = 28.8 N F g,2 T 2 = m 2 a = 40.4 N 10 pts

3 AP Physics C/Torque/Rotational Energy Mock Exam Page 3 of 9 Name: Answer Key (23 pts ) 2. A large rope is wound around two identical cylinders. Each cylinder has a mass M and a radius R. One of the cylinders is suspended from the ceiling, while the other cylinder is free to fall. The cylinders unwind as they fall so that the rope does not slip against the cylinder. 23 pts M, R α 1 M, R α 2 a Hint: The acceleration of the falling cylinder is related to α 1 and α 2 according to a = R (α 1 + α 2 ). (a) (8 pts) Write the rotational equations of motion for both cylinders. Express the equations in terms of quantities T, M, R, α 1, and α 2. Using the resulting equations, determine how the quantities α 1 and α 2 are related. Solution: The rotational equation of motion is τ = I α. For both cylinders the torque is τ = T R. Since tension is the same at opposite ends of a rope, and the radius of the two pulleys is the same, the torque (τ = T R) will be the same for both pulleys. Moreover, because M and R are the same for both pulleys, their rotational inertia I are the same. Looking at the rotational equation of motion for pulley one gives: τ 1 = I α 1 T R = I α 1 α 1 = T R I Similarly, the rotational equation of motion for pulley two gives: τ 2 = I α 2 T R = I α 2 α 2 = T R I Since T, R and I are the same for both pulleys, α 1 = α 2. Because α 1 is equal to α 1, I will refer to both as simply α for the remainder of the problem. Additionally, we are told that a = R(α 1 + α 2 ). Because α 1 = α 2 = α, this becomes a = 2Rα. (b) (5 pts) Write the translational equation of motion of the falling cylinder. Solution: The translational equation of motion is F = m a. There are two forces acting on the falling pulley: gravity and tension. If we say that down is the positive direction of motion, we get: F = M a F g T = M a 13 pts

4 AP Physics C/Torque/Rotational Energy Mock Exam Page 4 of 9 Name: Answer Key (c) (5 pts) Calculate the acceleration of the falling pulley. Express your answer as a fraction of g. Solution: We still need to do some work with our rotational equation of motion. Generally, in rope and pulley problems, we eliminate R from the equation and rewrite everything in terms of a. Normally we accomplish this by substituting in a formula for I (in this case I = 1/2 M R 2 ), as well as using a relationship between a and α (in this case a = 2R α). Substituting I = 1 2 M R2 into the rotational equation of motion gives: T R = I( α ) 1 T R = 2 M R 2 α T = 1 2 M R α Substituting a = 2R α into the expression above gives: T = 1 2 M ( a 2 T = 1 4 M a Now, we can plug the expression we just found into translation equation we found in part (b). ) F g T = M a M g 1 4 M a = M a M g = 5 4 M a g = 5 4 a a = 4 5 g (d) (5 pts) Write an expression for the tension in the rope. Express your answer in terms of M, and g. Solution: Finally, we find the tension by plugging our answer from part (c) into any of the equations we found above. For example, using the translational equation of motion to solve for T gives: F g T = M a M g T = M 4 5 g T = 1 5 M g 10 pts

5 AP Physics C/Torque/Rotational Energy Mock Exam Page 5 of 9 Name: Answer Key (18 pts ) 3. A single blade of an airplane propeller has a total mass of 80 kg and length of 2.0 meter. The mass per unit length of blade is given by: λ(x) = ( 60 kg ) m 2 (1.0 m 2 x 2 ) where x is the distance measured from the center of the blade. 18 pts 1.0 m (a) (10 pts) Calculate the moment of inertia of a single propeller blade rotating about its center. Solution: Recall that the formula for the rotational inertia is: I = dm x 2 Where dm is the mass of a tiny slice of the object. In this case, because we are given λ (the mass per length), dm = λ dx. Plugging this into our formula for I gives: I = = dm x 2 λ(x) x 2 dx Recall that, when we determine the bounds of the integral, we call the axis of rotation x = 0. Since the propeller extends 1.0 meter to either side of the axis, the bounds of the integral are ±1 meter. Plugging the equation for λ(x) into the integral above, and taking the integral gives: I = = λ(x) x 2 dx (1 x 2 ) x 2 dx = 60 x 2 x 4 dx 1 [ ] x 3 = 60 3 x [ 1 = ( ) 4 = = 16 kg m 2 ( )] (b) (8 pts) An actual airplane propeller consists of four propeller blades each of which rotates about the end of the blade. Calculate the moment of inertia of the entire airplane propeller. 18 pts

6 AP Physics C/Torque/Rotational Energy Mock Exam Page 6 of 9 Name: Answer Key Solution: If we can find the rotational inertia of a single propeller blade rotated about its end, we can simply multiply this by four to find the total rotational inertia of the propeller. To find the rotational inertia of the blade rotated about its end, we use the parallel axis theorem, which states: I = I c.m. + M d 2, where I c.m. is the rotational inertia of the object rotated about its center (which is what we found in part (a)) and d is the distance between the center of the object and the axis around which it is rotated ( in this case d = 1 m). Putting all of this together gives: I = I c.m. + M d 2 = 16 + (80 kg)(1 m) 2 = 96 kg m 2 Finally, since there are four propeller blades, the total rotational inertia is: I total = 4(96 kg m 2 ) = 384 kg m 2. (22 pts ) 4. A simple yo-yo consists of a string wound around the outer edge of a solid cylinder with a mass of 1.2 kg. The yo-yo is released from rest while the end of the string is held in place causing the yo-yo to unwind as it falls. The string does not slip against the surface of the yo-yo as its falls. (a) (4 pts) If the string is 1.2 m long, and the yo-yo has zero gravitational potential energy when the string is fully unwound, what is the initial potential energy of the yo-yo? Solution: The initial potential energy is gravitational, which is given by m g h. Since the yo-yo is 1.2 m above its lowest point, h = 1.2 m. Putting all of this together gives: P E g = m g h = J. 22 pts (b) (10 pts) Write an expression for the kinetic energy of the yo-yo in terms of M and v. Solution: As the yo-yo falls, it unravels. As a result, we have both rotational kinetic energy as well as translational kinetic energy. So, 14 pts

7 AP Physics C/Torque/Rotational Energy Mock Exam Page 7 of 9 Name: Answer Key KE = KE rot. + KE trans. = 1 2 I ω M v2 = 1 ( ) M R2 ω M v2 = 1 4 M (R ω) M v2 = 1 4 M v M v2 = 3 4 M v2 (c) (4 pts) Use conservation of energy to calculate the translational speed the yo-yo attains after falling 1.2 meters. Solution: Setting the initial potential energy equal to the final kinetic energy, and solving for v gives: E 0 = E f M g h = 3 4 M v 2 g h = 3 4 v2 v 2 = 4 g h 3 4 g h v = 3 = 3.96 m s (d) (4 pts) Suppose a ball is dropped at the same moment the yo-yo is released. Which one reaches the bottom first? (The bottom refers to the bottom of the yo-yo s motion; a point 1.2 m below the point where the object is dropped.) Briefly explain your reasoning. Solution: The ball will reach the bottom first. For the ball, the gravitational potential energy is converted into translational kinetic energy only. Whereas, for the yo-yo, the gravitational potential energy is converted into both kinetic energy and rotational kinetic energy (because the yo-yo has to rotate as it falls). Because the energy is shared between translational and rotational kinetic energies, the yo-yo has a lower translational kinetic energy when it reaches the bottom. This means the yo-yo is traveling slower after falling 1.2 m. Thus, the yo-yo takes longer to reach the bottom of its motion. 8 pts

8 AP Physics C/Torque/Rotational Energy Mock Exam Page 8 of 9 Name: Answer Key (20 pts ) 5. A solid cylinder starts out sliding without rolling down a ramp which is inclined at an angle θ. The cylinder has an initial velocity v 0 and the coefficient of friction between the cylinder and the ramp is µ. Express your answers in terms of v 0, µ, and trigonometric functions. ω 0 = 0rad/s 20 pts v 0 (a) (16 pts) How much time passes before the cylinder rolls without sliding down the ramp? Solution: The first step in answering this question is determining how the velocity and rotational velocity change as the cylinder slides down the ramp. In order to write a formula for v(t), we first need to find the cylinder s acceleration. We can accomplish this by applying Newton s Second Law to the cylinder: θ F = M a F g,x f = m a M g sin θ µ M g cos θ = M a a = g sin θ µ g cos θ Since the acceleration is constant, the velocity is given by: v(t) = v 0 + (g sin θ µ g cos θ) t Similarly, we can find the rotational acceleration of the cylinder by applying the rotational version of Newton s Second Law: τ = I α ( ) 1 f R = 2 M R 2 α f = 1 2 M R α µ M g cos θ = 1 2 M R α 1 2 R α = µ g cos θ α = 2 µ g cos θ R Once again, because the rotational acceleration is constant (and because the initial angular velocity is zero), ω(t) is given by: ( ) 2 µ g cos θ ω(t) = t R We want to know when v(t) = R ω(t). Setting these equations equal, and solving for t gives: 16 pts

9 AP Physics C/Torque/Rotational Energy Mock Exam Page 9 of 9 Name: Answer Key v(t) = R ω(t) ( ) 2 µ g cos θ v 0 + (g sin θ µ g cos θ) t = R t R v 0 + (g sin θ µ g cos θ) t = (2 µ g cos θ)t v 0 = (3µ g cos θ g sin θ) t v 0 t = 3 µ g cos θ g sin θ (b) (4 pts) Beyond a certain angle v increases faster than ω, so that the ball never reaches a point where is rolls without sliding. At what angle does this occur? Solution: As the angle of the ramp increases, the amount of time it takes before the ball begins to roll without sliding increases. This time, which we found a formula for in part (a), will eventually go to infinity when the denominator of that formula goes to zero. This is the critical angle beyond which the ball never rolls without sliding. Solving for when the denominator equals zero gives: 3 µ g cos θ g sin θ = 0 sin θ = 3 µ cos θ tan θ = 3µ θ = tan 1 (3µ) 4 pts