Rotational Motion Part 3 By: Heather Britton. Rotational Motion Rotational kinetic energy - the...
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Transcript of Rotational Motion Part 3 By: Heather Britton. Rotational Motion Rotational kinetic energy - the...
Rotational Motion Part 3Rotational Motion Part 3
By: Heather BrittonBy: Heather Britton
Rotational MotionRotational Motion
Rotational kinetic energy - the energy an object possesses by rotating
Like other forms of energy it is expressed in Joules (J)
Rotational kinetic energy - the energy an object possesses by rotating
Like other forms of energy it is expressed in Joules (J)
Rotational MotionRotational Motion
Recall that for translational motion KE = (1/2)mv2
v = ωr
Substituting for v we get
KE = (1/2)mω2r2
Recall that for translational motion KE = (1/2)mv2
v = ωr
Substituting for v we get
KE = (1/2)mω2r2
Rotational MotionRotational Motion
Rearranging we get
KE = (1/2)mr2ω2
I = mr2 so our equation for rotational KE becomes
KE = (1/2)Iω2
Rearranging we get
KE = (1/2)mr2ω2
I = mr2 so our equation for rotational KE becomes
KE = (1/2)Iω2
Rotational MotionRotational Motion
In regards to the law of conservation of energy, we now have a new quantity to consider
PEo + KEroto + KEtrano = PE + KErot + KEtran
In regards to the law of conservation of energy, we now have a new quantity to consider
PEo + KEroto + KEtrano = PE + KErot + KEtran
Example 8Example 8
A tennis ball, starting from rest, rolls down a hill (I = (2/3)mr2). It travels down a valley and back up the other side and becomes airborne at a 35o angle. The height difference between the starting point and launch point is 1.8 m. How far down range will the ball travel in the air?
A tennis ball, starting from rest, rolls down a hill (I = (2/3)mr2). It travels down a valley and back up the other side and becomes airborne at a 35o angle. The height difference between the starting point and launch point is 1.8 m. How far down range will the ball travel in the air?
Rotational MotionRotational Motion
Angular momentum - the rotational analog for linear momentum
Recall p = mv
Substituting for angular quantities we get the equation
Angular momentum - the rotational analog for linear momentum
Recall p = mv
Substituting for angular quantities we get the equation
Rotational MotionRotational Motion
L = Iω
ω = rotational velocity measured in rad/s
I = moment of inertia measured in kgm2
L = angular momentum measured in kgm2/s
L = Iω
ω = rotational velocity measured in rad/s
I = moment of inertia measured in kgm2
L = angular momentum measured in kgm2/s
Rotational MotionRotational Motion
Just like impulse F = p/Δt, there is an analog for rotation
See your book for the derivation of the following equation
τ = Iα
Just like impulse F = p/Δt, there is an analog for rotation
See your book for the derivation of the following equation
τ = Iα
Rotational MotionRotational Motion
The law of conservation of angular momentum - the total angular momentum of a rotating body remains constant if the net torque acting on it is zero
The law of conservation of angular momentum - the total angular momentum of a rotating body remains constant if the net torque acting on it is zero
Rotational MotionRotational Motion
Think of a figure skater.....
They spin very fast when they tuck their arms in they spin very fast
When they extend their arms, the rate of rotation slows
Angular momentum is conserved
Think of a figure skater.....
They spin very fast when they tuck their arms in they spin very fast
When they extend their arms, the rate of rotation slows
Angular momentum is conserved
Example 9Example 9
An artificial satellite is placed into an elliptical orbit about Earth. The point of closest approach (perigee) is rp = 8.37 x 106 m from the center of Earth. The greatest distance (apogee) is ra = 2.51 x 107 m. V at the perigee is 8450 m/s. What is the speed at the apogee?
An artificial satellite is placed into an elliptical orbit about Earth. The point of closest approach (perigee) is rp = 8.37 x 106 m from the center of Earth. The greatest distance (apogee) is ra = 2.51 x 107 m. V at the perigee is 8450 m/s. What is the speed at the apogee?