Physics 2Rotational MotionPrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

First some quick geometry review:rx = rWe 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 2 radians in one complete revolution.Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

rx = rDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.Angular Velocity = = Angular Acceleration = = This is the Greek letter omega (not w)This is the Greek letter alpha (looks kinda like a fish)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

rx = rDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.Angular Velocity = = Angular Acceleration = = This is the Greek letter omega (not w)This is the Greek letter alpha (looks kinda like a fish)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBExample: 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).

rx = rDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.Angular Velocity = = Angular Acceleration = = This is the Greek letter omega (not w)This is the Greek letter alpha (looks kinda like a fish)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBExample: 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

rx = rDefinitions of angular velocity and angular acceleration are analogous to what we had for linear motion.Angular Velocity = = Angular Acceleration = = This is the Greek letter omega (not w)This is the Greek letter alpha (looks kinda like a fish)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBExample: 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 secondNow we can use the definition of angular acceleration:Standard units for angular acceleration are radians per second2.

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 ZacconeFor Campus Learning Assistance Services at UCSB

We have already seen one case:x = rThis translates between distance (linear) and angle (rotational)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBWe 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.

We have already seen one case:x = rThis translates between distance (linear) and angle (rotational)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBWe 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 = r linear velocity relates to angular velocityatan = r linear acceleration relates to angular accelerationNotice a pattern here?

We have already seen one case:x = rThis translates between distance (linear) and angle (rotational)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBWe 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 = r linear velocity relates to angular velocityatan = r linear acceleration relates to angular accelerationMultiply the angular quantity by the radius to get the linear quantity.

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)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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)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 ZacconeFor Campus Learning Assistance Services at UCSB

Linear Motion (constant a)Rotational Motion (constant )x=x0+v0t+at2=0+0t+t2v=v0+at=0+tv2=v02+2a(x-x0)2=02+2(-0)

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. Heres a Formula:We know that is angular velocity. Comparing with the formula for linear kinetic energy, what do you think I is?Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

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. Heres 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 ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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 ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownPrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownPrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownAt this point we can substitute the formula for each shape (from table 9.2 on page 279)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownAt this point we can substitute the formula for each shape (from table 9.2 on page 279)Solid SpherePrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownAt this point we can substitute the formula for each shape (from table 9.2 on page 279)Solid SphereHollow SpherePrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Problem 9.48A 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 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?hWe can use conservation of energy for this one. Since they dont 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 with v/r so everything is in terms of the desired unknownAt this point we can substitute the formula for each shape (from table 9.2 on page 279)Solid SphereHollow SphereThe solid sphere is faster because its moment of inertia is smaller.It reaches the bottom first.Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB