Rotational Kinetic energy and momentum

Rotational Kinetic Energy

Find ω when beam is at lowest point A 4 m beam with a 30 kg mass is free to rotate on a hinge. It is attached to a wall with a horizontal cable. The cable is then cut, find the angular velocity when the beam is horizontal. . θ = 35o θ = 35o Find ω when beam is at lowest point Answer: ω = 3.6 r/s

Rolling Motion Many rotational motion situations involve rolling objects. Rolling without slipping involves both rotation and translation so you need to account for both rotational and translational kinetic energy.  The point of contact of the object with the surface is the axis of rotation

Example In the past, everything was SLIDING. Now the object is rolling and thus has MORE energy than normal. So let’s assume the ball is like a thin spherical shell and was released from a position 5 m above the ground. Calculate the velocity at the bottom of the incline. If you HAD NOT included the rotational kinetic energy, you see the answer is very much different. 7.7 m/s

Example A bowling ball with a radius of .3 m and mass of 4.6 kg is held by a wire. The wire is swung back so that it is now .5 m off the floor. The string is cut as the bowling ball makes contact with the floor. The resulting distance it travels in 5 seconds is 6 m. What is its rotational kinetic energy and speed? 20.2 J and 15.6 rad/s

Angular Momentum (L)

Example Rank the following from largest to smallest angular momentum.

Example A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially. = 2.71 rad/s

Example a) What is the period of the new motion? Two twin ice skaters separated by 10 meters skate without friction in a circle by holding onto opposite ends of a rope. They move around a circle once every five seconds. By reeling in the rope, they approach each other until they are separated by 2 meters. a) What is the period of the new motion? b) If each skater had a mass of 75 kg, what is the work done by the skaters in pulling closer? TF = T0/25 = 0.2 s W = 7.11x105 J

Example The figure below shows two masses held together by a thread on a rod that is rotating about its center with angular velocity, ω. If the thread breaks, what happens to the system's (a) angular momentum and (b) angular speed. (Increase, decrease or remains the same)

Example A 50 kg figure skater rotates with her arms out at 2 rev/s, this gives her a radius of .7m. She then pulls her arms in which gives her a radius of 0.25 m. Find the final speed in rev/s.

Remember our table??? x (or y) Ɵ Δx ΔƟ Δx=rΔƟ v=Δx/Δt ω=ΔƟ/Δt=2π/T Quantity Linear Rotational Connection Position Displacement Velocity Acceleration 1st kinematic 2nd kinematic 3rd kinematic Centripetal acceleration Inertia Kinetic Energy What causes acceleration Newton’s 2nd Law Momentum x (or y) Ɵ Δx ΔƟ Δx=rΔƟ v=Δx/Δt ω=ΔƟ/Δt=2π/T v=rω a=Δv/Δt α=Δω/Δt a=rα m I KEtrans= ½ mv2 KErot= ½ Iω2 Force Torque τ=r*F Fnet=ma τnet=Iα p=mv L=Iω or mvr