1. Physics 2 Rotational Motion Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. First some quick geometry review: θ r x = rθ We need this formula for arc length to see the connection between rotational motion and linear motion. We will also need to be able to convert from revolutions to radians. There are 2π radians in one complete revolution. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. θ r x = rθ Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion. Angular Velocity = ω = Angular Acceleration = α = This is the Greek letter omega (not w) This is the Greek letter alpha (looks kinda like a fish) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. θ r x = rθ Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion. Angular Velocity = ω = Angular Acceleration = α = This is the Greek letter omega (not w) This is the Greek letter alpha (looks kinda like a fish) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm. Find the final angular velocity and the angular acceleration (assume constant).

5. θ r x = rθ Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion. Angular Velocity = ω = Angular Acceleration = α = This is the Greek letter omega (not w) This is the Greek letter alpha (looks kinda like a fish) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm. Find the final angular velocity and the angular acceleration (assume constant). rpm stands for “revolutions per minute” – we can just do a unit conversion: Standard units for angular velocity are radians per second

6. θ r x = rθ Definitions of angular velocity and angular acceleration are analogous to what we had for linear motion. Angular Velocity = ω = Angular Acceleration = α = This is the Greek letter omega (not w) This is the Greek letter alpha (looks kinda like a fish) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A centrifuge starts from rest and spins for 7 seconds until it reaches 1000 rpm. Find the final angular velocity and the angular acceleration (assume constant). rpm stands for “revolutions per minute” – we can just do a unit conversion: Standard units for angular velocity are radians per second Now we can use the definition of angular acceleration: Standard units for angular acceleration are radians per second 2.

7. We already know how to deal with linear motion. We have formulas for kinematics, forces, energy and momentum. We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones! All you have to do is translate the variables. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

8. We have already seen one case: x = rθThis translates between distance (linear) and angle (rotational) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We already know how to deal with linear motion. We have formulas for kinematics, forces, energy and momentum. We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones! All you have to do is translate the variables.

9. We have already seen one case: x = rθThis translates between distance (linear) and angle (rotational) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We already know how to deal with linear motion. We have formulas for kinematics, forces, energy and momentum. We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones! All you have to do is translate the variables. Here are the other variables: v = rω linear velocity relates to angular velocity a tan = rα linear acceleration relates to angular acceleration Notice a pattern here?

10. We have already seen one case: x = rθThis translates between distance (linear) and angle (rotational) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB We already know how to deal with linear motion. We have formulas for kinematics, forces, energy and momentum. We will find similar formulas for rotational motion. Actually, we already know the formulas – they are the same as the linear ones! All you have to do is translate the variables. Here are the other variables: v = rω linear velocity relates to angular velocity a tan = rα linear acceleration relates to angular acceleration Multiply the angular quantity by the radius to get the linear quantity.

11. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

12. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

13. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. We basically have two options on how to proceed. We can switch to angular variables right away, or we can do the corresponding problem in linear variables and translate at the end. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. Switching to angular variables right away: Convert to angular velocity: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. Switching to angular variables right away: Convert to angular velocity: Find angular acceleration: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

16. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. Switching to angular variables right away: Convert to angular velocity: Find angular acceleration: Use a kinematics equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. Switching to angular variables right away: Convert to angular velocity: Find angular acceleration: Use a kinematics equation: Convert to revolutions: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18. We can use this technique to find angular motion formulas from the analogous linear motion formulas we already know. Here are some kinematics formulas. (these are in a table on page 292 of your book) Linear Motion (constant a) Rotational Motion (constant α) x=x 0 +v 0 t+½at 2 θ=θ 0 +ω 0 t+½αt 2 v=v 0 +atω=ω0+αtω=ω0+αt v 2 =v 0 2 +2a(x-x 0 )ω 2 =ω 0 2 +2α(θ-θ 0 ) Here is a kinematics example: A cyclist starts from rest and accelerates to a speed of 15 m/s in a time of 25 sec. Assume the tires have a diameter of 700mm and that the acceleration is constant. Find the total number of revolutions that the wheels make during the 25 second interval. This time do the linear problem first: Find linear acceleration: Use a kinematics equation: Convert to revolutions: (we did some rounding off) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19. Some other topics for linear motion are Energy, Forces and Momentum. All of these have analogues for rotational motion as well. Forces and Momentum will be covered in chapter 10. That leaves us with Energy. Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula: We know that ω is angular velocity. Comparing with the formula for linear kinetic energy, what do you think I is? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20. Some other topics for linear motion are Energy, Forces and Momentum. All of these have analogues for rotational motion as well. Forces and Momentum will be covered in chapter 10. That leaves us with Energy. Any moving object will have Kinetic Energy. This applies to rotating objects. Here’s a Formula: The I in our formula takes the place of m (mass) in the linear formula. We call it Moment of Inertia (or rotational inertia). It plays the same role in rotational motion that mass plays in linear motion (I quantifies how difficult it is to produce an angular acceleration. The value for I will depend on the shape of your object, but the basic rule of thumb is that the farther the mass is from the axis of rotation, the larger the inertia. Page 299 in your book has a table of formulas for different shapes. These are all based on the formula for a point particle. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

21. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

22. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

23. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown At this point we can substitute the formula for each shape (from table 9.2 on page 279) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

27. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown At this point we can substitute the formula for each shape (from table 9.2 on page 279) Solid Sphere Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

28. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown At this point we can substitute the formula for each shape (from table 9.2 on page 279) Solid SphereHollow Sphere Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

29. Problem 9.48 A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle θ above the horizontal. Both spheres start from rest at the same vertical height h. a) How fast is each sphere moving when it reaches the bottom of the hill? b) Which sphere will reach the bottom first, the hollow one or the solid one? h θ We can use conservation of energy for this one. Since they don’t mention it we can ignore rolling friction and just assume that the total energy at the top equals the energy at the bottom. we can replace ω with v/r so everything is in terms of the desired unknown At this point we can substitute the formula for each shape (from table 9.2 on page 279) Solid SphereHollow Sphere The solid sphere is faster because its moment of inertia is smaller. It reaches the bottom first. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB