Rotational Motion

6-1 Angular Position, Velocity, & Acceleration

Angular Displacement (  ) Measured in radians (rad) this unit has no dimensions Counterclockwise is positive 6-1 Angular Position, Velocity, & Acceleration Angular Quantities

Angular Velocity (  ) Defined in the same way as velocity Average velocity is displacement divided by time SI unit Instantaneous Angular Velocity 6-1 Angular Position, Velocity, & Acceleration

Angular Acceleration (  ) Defined like acceleration Angular acceleration is change in angular velocity divide by time SI unit Instantaneous Angular Acceleration 6-1 Angular Position, Velocity, & Acceleration

6.2 Rotational Kinematics

All the equations used in kinematics (for conditions of constant acceleration) Are exactly the same, but with the new quantities 6.2 Rotational Kinematics

6.3 Connections Between Linear & Rotational

For an object moving in a circular path At any given time, the object is moving with a linear velocity tangent to the arc The speed would be circumference divide by the time for one revolution Since  would be 2  /t 6.3 Connections Between Linear & Rotational

Similarly we can determine that So in a situation where a spinning object causes linear motion Circumference of the tire Will equal the displacement 6.3 Connections Between Linear & Rotational

6.4 Rotational K & the Moment of Inertia

In a spinning object, each particle has kinetic energy Lets assume that we have a rod of uniform mass rotating about its end 6.4 Rotational K & the Moment of Inertia v  r

Now, just consider a piece of the rod at the very end To calculate the kinetic energy of the mass Converting To rotational quantities 6.4 Rotational K & the Moment of Inertia v  r

So K depends on angular velocity And it depends on the distribution of mass This mass distribution is called Moment of Inertia 6.4 Rotational K & the Moment of Inertia v  r

Moment of Inertia varies with shape, mass, and axis of rotation (You will need these to solve problems) 6.4 Rotational K & the Moment of Inertia

6.5 Conservation of Energy

We are adding a new type of Kinetic Energy into our existing Energy Equation Adding Rotational Kinetic Energy Where K r is defined as 6.5 Conservation of Energy

So if a solid ball is rolling down a slope The equation would become Using the moment of inertia of a solid sphere And the relationship between v and  6.5 Conservation of Energy

Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m? 6.5 Conservation of Energy

Example: A yo-yo is released from rest and allowed to drop as the top end of the string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm. Assume it acts as a solid disc rotating around its center. What is the angular speed of the yo-yo after it has dropped 0.50m? What quantities remain in the equation below? Expanded equation? Moment of inertia? Solve (remember v=r  ) Cancel and insert values 6.5 Conservation of Energy

6.6 Torque

Rotational Dynamics – the causes of rotational motion How do we make an object spin Apply a force away from the pivot 6.6 Torque

The ability to spin increases with force and the distance from the pivot point If the force is parallel to the distance, no rotation occurs 6.6 Torque

It is the perpendicular component of force that causes rotation This quantity is called Torque (  ) And is measure in Nm 6.6 Torque

As with angular quantities Counterclockwise torque is positive Clockwise torques are defined as negative 6.6 Torque r F r F

6.7 Torque and Angular Acceleration

If an unbalanced torque is applied to an object Consider a point where the force is applied The acceleration of that point is 6.7 Torque and Angular Acceleration

Since We can write this equation as Combining we get Now, math tricks, multiply by r/r The of the fraction is torque, and the bottom is inertia Usually written as 6.7 Torque and Angular Acceleration

Applying Newton’s second law to rotational motion we get 6.7 Torque and Angular Acceleration

6.8 Static Equilibrium

Static Equilibrium How do we calculate the forces needed to support a bridge? The bridge is at static (not moving) equilibrium 6.8 Static Equilibrium

Using a diagram of the bridge Two conditions for equilibrium 1. The sum of the forces equals zero 6.8 Static Equilibrium P1P1 P2P2 W

2.Sum of the torques equals zero For torque we need a pivot point – the actual point does not matter because the object is not rotating Use a pivot point that eliminates a variable 6.8 Static Equilibrium

The equation becomes 6.8 Static Equilibrium P1P1 P2P2 W

Try the problem with some variable The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon? 6.8 Static Equilibrium P1P1 P2P2 W

The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon? First Condition of Equilibrium 6.8 Static Equilibrium P1P1 P2P2 W

The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon? Second Condition of Equilibrium 6.8 Static Equilibrium P1P1 P2P2 W

The bridge is 100 m long with two pylons that are 10 m from each end. The mass of the bridge is 10,000 kg. What is the force on each pylon? Combine equations 6.8 Static Equilibrium P1P1 P2P2 W

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) 6.8 Static Equilibrium

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Diagram? 6.8 Static Equilibrium Wall ladder Dude  =60 o

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Force Diagram? 6.8 Static Equilibrium Wall ladder Dude  =60 o N1N1 f WLWL WDWD N2N2

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Force Equations? 6.8 Static Equilibrium Wall ladder  =60 o N1N1 f WLWL WDWD N2N2

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Pivot Point for Torque? 6.8 Static Equilibrium Wall ladder  =60 o N1N1 f WLWL WDWD N2N2

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Distances? 6.8 Static Equilibrium Wall ladder  =60 o N1N1 f WLWL WDWD N2N2 20m10m r

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Angles? 6.8 Static Equilibrium Wall ladder  =60 o N1N1 f WLWL WDWD N2N2 20m10m r  =60 o  =30 o

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Torque Equation? 6.8 Static Equilibrium Wall ladder N1N1 f WLWL WDWD N2N2 20m10m r  =60 o  =30 o

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Values? 6.8 Static Equilibrium Wall ladder N1N1 f WLWL WDWD N2N2 20m10m r  =60 o  =30 o

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Value for N 1 ? 6.8 Static Equilibrium Wall ladder N1N1 f WLWL WDWD N2N2 20m10m r  =60 o  =30 o

A 75 kg man climbs a ladder that is at 60 o to the horizontal. The ladder has mass of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder is How far up the ladder can the man climb before it begins to slip? (Assume that there is no friction at the top of the ladder) Value for f ? 6.8 Static Equilibrium Wall ladder N1N1 f WLWL WDWD N2N2 20m10m r  =60 o  =30 o

6.9 Center of Mass and Balance

An object is balanced, if the net torque is zero If the rod above is uniform, then when the pivot is in the middle, half the mass is on one side and half on the other This balance point is called the center of mass 6.9 Center of Mass and Balance Wall

6.10 Conservation of Angular Momentum

When an object spins, it has angular momentum Defined as The conservation of angular momentum is treated just like the conservation of linear momentum 6.10 Conservation of Angular Momentum v Conservation of L

So the equation for conservation becomes 6.10 Conservation of Angular Momentum

Example: A small mass, m, attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v 0 =2.4m/s in a circle of radius r 0 =0.80m. The string is then pulled slowly through the hole so that the radius is reduced to r=0.48m. What is the speed, v, of the mass now? 6.10 Conservation of Angular Momentum

Example: A small mass, m, attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v 0 =2.4m/s in a circle of radius r 0 =0.80m. The string is then pulled slowly through the hole so that the radius is reduced to r=0.48m. What is the speed, v, of the mass now? Using conservation of momentum (only one object) Mass on the end of a string so Masses cancel and we can convert to v Enter numbers and solve 6.10 Conservation of Angular Momentum

Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the center of the record. What is the new velocity of the record player? 6.10 Conservation of Angular Momentum

Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the center of the record. What is the new velocity of the record player? The record player is a solid disk, the bug is a point mass Convert rev/min to rad/s Substitute and solve 6.10 Conservation of Angular Momentum v