Rotational Kinematics. Angular Position, Velocity, and Acceleration.
EQUILIBRIUM ROTATIONAL KINETIC ENERGY ANGULAR...
Transcript of EQUILIBRIUM ROTATIONAL KINETIC ENERGY ANGULAR...
EQUILIBRIUM
ROTATIONAL KINETIC ENERGY
ANGULAR MOMENTUM
EquilibriumTwo Conditions for Equilibrium
1) Sum of Forces must be zero
Tacos ! (m= 10 kg)
Determine the Force on Each Chain
EquilibriumTwo Conditions for Equilibrium
2) Sum of Torques must be zero
Tacos ! (m=5kg)
Determine the Tension on the Chain
T=?
Length 1.50m
50o
X
mg
T
EquilibriumTwo Conditions for Equilibrium
2) Sum of Torques must be zero
m= 4.0 kg
Pizza! (m=2.0kg)
2.0m1.3 m45o
Determine the Tension on the Chain X
mg
T
mg
EquilibriumTwo Conditions for Equilibrium
1) Sum of Forces must be zero2) Sum Torques must be zero
Length 3.0m
Tacos ! (m= 10 kg)
X=0m X=2.0m
Determine the Force on Each Chain
mg
T1 T2
X
EquilibriumTwo Conditions for Equilibrium
1) Sum of Forces must be zero2) Sum Torques must be zero
L= 5.0,, m=15kg
m= 50kg, x=3.0m
F1 =? F2 =?
mg
T1 T2
X
mg
Energy in Rotational MotionObjects in motion have Kinetic Energy
Rotating Objects have Kinetic Energy
Energy of single particle:
𝐾𝐸 =1
2𝑚𝑣2 → 𝑣 = 𝜔𝑟 →
1
2𝑚 𝜔𝑟 2
𝐾𝐸 =1
2𝑚𝑟2 𝜔2 → 𝑚𝑟2 = 𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 =
𝑖
𝑚𝑖𝑟𝑖2
𝐾𝐸 =1
2𝐼𝜔2
Energy in Rotational Motion
A force of 10N acts for 2.0m on 20 kg Solid Disc with a radius of 0.2m. What is the final angular speed of the wheel as it spins on its axis?
F = 10Nd=2m
m=20kgr=0.20m
𝑊 = ∆𝐾𝐸
𝐼 =1
2𝑚𝑟2
+
Energy in Rotational Motion• Rolling Objects have both Translational Kinetic Energy and Rotational
Kinetic Energy
• Total Energy =KEt+KEr = 1
2𝑚𝑣2+
1
2𝐼𝜔2
• Find the total energy of a 4.0kg bowling ball moving at 8 m/s. (𝐼 =2
5𝑚𝑟2)
Conservation of Energy
m=1.5 kgr=0.20m
Basketball
1.5m
V=?
GPE
KE + RKE
𝐺𝑃𝐸 = 𝐾𝐸𝑇 + 𝐾𝐸𝑅 → simplify
Gravitational Potential Converted to both translational and rotational kinetic energy.
Conservation of Energy
• Hollow Sphere
• Hoop
• Disc,
• Solid Sphere
Which is which?
How can you tell?
Conservation of Energy• Which has more energy? Hoop, disc? (m=2kg, r =0.40m, vi=6m/s)
• How high up the ramp will each go?
Angular Momentum• Momentum for a point mass rotating around an axis.
• 𝐿 = 𝑟 × 𝑚𝑣 → 𝑣 = 𝜔𝑟 → 𝐿 = 𝑟 ×𝑚 𝜔𝑟 = 𝑚𝑟2𝜔 → 𝑚𝑟2 = 𝐼
• 𝐿 = 𝑚𝑟2𝜔
•𝐿 = 𝐼𝜔• 𝑝 = 𝑚𝑣 ↔ 𝐿 = 𝐼𝜔
Right Hand Rule
Angular Momentum
• Determine the angular momentum of a 2.0kg bike wheel, spinning at 200rpm. (r=0.35m)
Determine the angular momentum of a 5.0kg dog, on the outside edge of a merry go round, moving at 2.0 m/s at 1.2m from the center.
Conservation of Angular Momentum• 𝐿1 = 𝐿2 → 𝐼𝜔 = 𝐼𝜔
Conservation of Angular Momentum
𝐿1 = 𝐿2 → 𝐼𝜔 = 𝐼𝜔
A figure skater with an Initial Moment of Inertia of 3.8 kg*m2 isspinning at 30 rpm, stands and accelerates to 90 rpm. What is her new moment of Inertia?
Conservation of Angular Momentum• 𝐿1 = 𝐿2 → 𝐼𝜔 = 𝐼𝜔
• 𝑚𝑣1𝑟1 = 𝑚𝑣2𝑟2
Mass of Blocks = 2kgR1=0.4mV1 = 5m/s
R2 = 0.15V2=?
Conservation of Angular Momentum
M=0.80kgL=0.60m
M=1.2kgr=0.30mω1=9.0rad/sω2=?
𝐿1 = 𝐿2 → 𝐼𝜔 + 𝐼𝜔 = 𝐼𝜔 + 𝐼𝜔
A B