Rotational Motion 6 1 Angular Position Velocity Acceleration
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Rotational Motion
6 -1 Angular Position, Velocity, & Acceleration
Angular Displacement (q) Angular Quantities Measured in radians (rad) this unit has no dimensions Counterclockwise is positive 6 -1 Angular Position, Velocity, & Acceleration
Angular Velocity (w) 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 (a) 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 w would be 2 p/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 w v r 6. 4 Rotational K & the Moment of Inertia
Now, just consider a piece of the rod at the very end To calculate the kinetic energy of the mass w v Converting To rotational r quantities 6. 4 Rotational K & the Moment of Inertia
So K depends on angular velocity And it depends on the distribution of mass This mass distribution is called Moment of Inertia w v r 6. 4 Rotational K & the Moment of Inertia
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 Kr 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 w 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. 056 kg 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. 50 m? 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. 056 kg 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. 50 m? What quantities remain in the equation below? Expanded equation? Moment of inertia? Solve (remember v=rw) 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 (t) And is measure in Nm 6. 6 Torque
As with angular quantities Counterclockwise torque is positive Clockwise torques are defined as negative r r F F 6. 6 Torque
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 P 1 P 2 W Two conditions for equilibrium 1. The sum of the forces equals zero 6. 8 Static Equilibrium
2. Sum of the torques equals zero 3. For torque we need a pivot point – the actual point does not matter because the object is not rotating 4. Use a pivot point that eliminates a variable 6. 8 Static Equilibrium
The equation becomes P 1 P 2 W 6. 8 Static Equilibrium
Try the problem with some variable P 1 P 2 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? 6. 8 Static Equilibrium
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? P 1 P 2 W First Condition of Equilibrium 6. 8 Static Equilibrium
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? P 1 P 2 W Second Condition of Equilibrium 6. 8 Static Equilibrium
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? P 1 P 2 W Combine equations 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 0. 20. 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 0. 20. 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? Dude Wall ladder q=60 o 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 0. 20. 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) N 2 Force Diagram? Dude Wall ladder N 1 WD WL q=60 o f 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 0. 20. 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? N 2 Wall ladder N 1 WD WL q=60 o f 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 0. 20. 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? N 2 Wall ladder N 1 WD WL q=60 o f 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 0. 20. 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) N 2 Distances? 20 m 10 m ladder N 1 WD WL q=60 o f 6. 8 Static Equilibrium r Wall
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 0. 20. 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) N 2 Angles? 20 m 10 m q=30 o ladder N 1 q=30 o WD q=30 o WL o q=60 f o 6. 8 Static Equilibrium r Wall
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 0. 20. 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) N 2 Torque Equation? 20 m 10 m q=30 o ladder N 1 q=30 o WD WL q=60 o f 6. 8 Static Equilibrium r Wall
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 0. 20. 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) N 2 Values? 20 m 10 m q=30 o ladder N 1 q=30 o WD WL q=60 o f 6. 8 Static Equilibrium r Wall
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 0. 20. 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) N 2 Value for N 1? 20 m 10 m q=30 o ladder N 1 q=30 o WD WL q=60 o f 6. 8 Static Equilibrium r Wall
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 0. 20. 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) N 2 Value for f? 20 m 10 m q=30 o ladder N 1 q=30 o WD WL q=60 o f 6. 8 Static Equilibrium r Wall
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 Wall 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
6. 10 Conservation of Angular Momentum
When an object spins, it has angular momentum v Defined as The conservation of angular momentum is treated just like the conservation of linear momentum Conservation of L 6. 10 Conservation of Angular Momentum
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. 4 m/s in a circle of radius r 0=0. 80 m. The string is then pulled slowly through the hole so that the radius is reduced to r=0. 48 m. 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. 4 m/s in a circle of radius r 0=0. 80 m. The string is then pulled slowly through the hole so that the radius is reduced to r=0. 48 m. 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 15 cm. A bug with a mass of 0. 25 kg lands 10 cm 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 15 cm. A bug with a mass of 0. 25 kg lands 10 cm from the center of the record. What is the new velocity of the record player? v 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
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