9/16/2018 Physics 253

Chapter 10: Rotational Motional About a Fixed Axis Review of Angular Quantities & Motion and Torque (10-1,10-2,10-3, 10-5) Solving Problems in Rotational Dynamics (10-6, 10-7) Determining Moments of Inertia, Conservation of Angular Momentum, and Rotational Kinetic Energy (10-8,10-9, 10-10) Rotational + Translational Motion (10-11, 10-12) 9/16/2018 Physics 253

Review of Quiz 3 Points Score Distribution Average: 36 Total: 50 High: 50 Low: 6 Score Distribution 00-10: 1 11-20: 3 21-30: 22 31-40: 42 41-50: 24 Average: 36 9/16/2018 Physics 253

Problem 1 Problem 1 (5 points): How much work must be supplied to vertically lift a 10.0-kg box 0.50 meters? a) 2N b) -98.0N c) 4.9 J d) 49 J Answer: The force and displacement are in the same direction so W=(F)(d)=(mg)(d)= (10.0kg)(9.8m/s2)(0.50m)=49J 9/16/2018 Physics 253

Problem 2 Problem 2 (5 points): The earth’s gravitational potential energy is defined to be zero at: a) the earth’s surface b) infinite distance c) near the moon d) the earth’s center Answer: 9/16/2018 Physics 253

Problem 3 Problem 3 (5 points): What is the center of mass for m1 = 5.0 kg, x1 = 2.0 m, m2 = 12 kg, x2 = 6.0 m? a) 82m b) 4.8 m c) 8.5m d)0.021m Answer: 9/16/2018 Physics 253

Problem 4 Problem 4 (5 points): A spring compressed by 0.10 m stores 30J of energy. What is the spring constant? a) 6000N/m b) 600 N/m c) 300N/m d) 0.03N/m Answer: 9/16/2018 Physics 253

Problem 5 Problem 5 (5 points): Two astronauts in outer space and initially at rest “push off” and move away from one another. The first astronaut has a mass of 100.0kg and a velocity of +2.5m/s. If the second astronaut has a velocity of -4.0m/s, what is her mass? Answer: 9/16/2018 Physics 253

Problem 6 Problem 6 (10 points): A motorcyclist is trying to leap across a canyon as shown in the figure. When he leaves the cliff the cycle has a speed of 38.0 m/s. Using conservation of energy find the speed with which the cycle strikes the ground on the other side. Answer: 9/16/2018 Physics 253

Problem 7 Problem 7 (15 points): ): A 5x104 kg spaceship is traveling at a speed of 1.1x104 m/s. The engine exerts a force of 4x105N parallel to the displacement and fires until the displacement is 2.5x106 m. (No forces act on the vessel except that generated by its engine.) Determine (a) the initial kinetic energy, b) the work done by the engine and c) using the Work-Energy Theorem the final velocity of the spaceship. Answer: 9/16/2018 Physics 253

Torque = (Lever Arm) x (Magnitude of the force) Review Torque Torque = (Lever Arm) x (Magnitude of the force) 9/16/2018 Physics 253

Rotational Dynamics: Torque and Rotational Inertia Just as force is proportional to acceleration, torque seems to be proportional to angular acceleration: Note that the constant of proportionality for linear motion is mass. Just on dimensional grounds the analogous constant must be different for rotational motion. Turns out we can get there starting with the 2nd Law. 9/16/2018 Physics 253

Consider a particle of mass m rotating on a circle of radius r at the end of a massless string or rod and subject to a force F. The constant of proportionality is mR2 and represents the rotational inertia of the particle and is often called the moment of inertia. 9/16/2018 Physics 253

9/16/2018 Physics 253

Comments on Moment of Inertia Serves the same role for rotational motion as mass does for linear motion But since I=SmiRi2 is a sum over many objects it depends on mass distribution If of equal mass, a larger cylinder will have a greater moment of inertia than a smaller one. Something intuitively true. When mass is far from the axis, it also hard to rotate something, again something familiar. For rotational motion the mass of a body cannot be considered as concentrated at the center of mass. Still it can be extended to the center of mass 9/16/2018 Physics 253

Example 1: Moments of Inertia Calculations Two small weights of mass 5.0 and 7.0kg are mounted 4.0m apart on a massless rod. Calculate the moment of inertia I About an axis halfway between the weights About an axis 0.50 meter to left of the 5.0kg mass Calculate the force on the 7.0kg mass needed to achieve an acceleration of 1 rad/sec2 9/16/2018 Physics 253

9/16/2018 Physics 253

Characteristics of the Moment of Inertia Different for different axes. Masses close to the axis contribute little, but masses distant contribute much. Calculations can be difficult because the mass distributions are not uniform. They can be worked out taking the sum to the limit and using calculus: Experimentally done by measuring a for a known t. 9/16/2018 Physics 253

Example 2: A Heavy Pulley A 15.0 N force is applied to a cord around a 4.00kg pulley at a radius of 33.0cm. The pulley accelerates uniformly from rest to 30.0 rad/s in 3.00s. If there is a frictional torque at the axel of 1.10mN what is the pulley’s moment of inertia? 9/16/2018 Physics 253

9/16/2018 Physics 253

9/16/2018 Physics 253

9/16/2018 Physics 253

Example 3: A Pulley and Bucket Lets take the heavy pulley problem a bit further and hang a bucket of weight 15.0N (m=1.53kg) from it. Calculate The angular acceleration of the pulley, a, and the linear acceleration of the bucket, a. If the pulley and bucket start at rest calculate the angular velocity of the pulley, w, and the linear velocity of the bucket, v, at 3.00s. 9/16/2018 Physics 253

Our strategy will be to analyze the rotational motion of the pulley analyze the linear motion of the bucket Connect the two and solve for accelerations Use accelerations to find the final velocity 9/16/2018 Physics 253

9/16/2018 Physics 253

Example 4: A Rotating Rod A uniform rod of mass M and length L can pivot freely up or down. The rod is held horizontally and then released. Assuming the force of gravity acts at the center of mass, at the moment of release determine The angular acceleration of the rod The linear acceleration at the tip of the rod. 9/16/2018 Physics 253

9/16/2018 Physics 253

Solving Problems with Rotational Motion Draw the free-body diagram! Show all forces and WHERE they act. Put gravity at the CG or CM. Identify the axis of rotation and calculate torques about it. CCW is positive CW is negative. Apply the 2nd Laws Rotational motion: St=Ia Translational motion: SF=ma Solve the set of equations for any missing info. Smell test! 9/16/2018 Physics 253

Ok we’ve defined torque To Summarize Ok we’ve defined torque This led to the analog of Newton’s 2nd Law for angular motion: Where the moment of inertia serves the role of mass Next we’ll learn how to calculate I and explore angular momentums and rotational energy. 9/16/2018 Physics 253