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| BETA : Cyclic : Angular Momentum 01 : Angular Momentum versus Linear Momentum : BETA Physics Homework Exercise
Description : Angular momentum versus linear momentum.
Discussion : Angular momentum is the product of linear momentum times the radius. See Figure Angular Momentum versus Linear Momentum
The homework exercise has 3 sets of questions with 3 questions per set.
p, L, omega
L, p, v
m, r, omega
The basic equations are :
p = m v
L = m v r = p r = m omega r
omega = v / r
rearranging
L = m omega r2
p = L / r
v = omega r
m = p / v
r = L / (m * v)
where
m = mass
v = linear velocity
r = radius
p = linear momentum
L = angular momentum
omega = angular velocity (ω)
By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Angular Momentum 02 : Conservation Of Angular Momentum : BETA Physics Homework Exercise
Description : Conservation of angular momentum.
Discussion : The change of angular velocity omega (ω) due to changes in inertia. Angular momentum remains constant.
The homework exercise has 3 sets of questions with 2 questions per set.
L, omegaf
L, omegai
If, Ii
The basic equations are :
L = Ii omegai = If omegaf
omegaf = L / If
rearranging
omegai = L / Ii
If = L / omegaf
Ii = L / omegai
where
L = angular momentum
omegai = initial angular velocity (ωi)
omegaf = final angular velocity (ωf)
Ii = initial inertia
If = final inertia
By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Rotational Energy 01 : Rotational Energy Of A Mass And Wheel Connected By A Rope : BETA Physics Homework Exercise
Description : Rotational energy of a mass and wheel connected by a rope.
Discussion : The gravitational potential energy of the mass is converted to linear kinetic energy of the mass and rotational kinetic energy of the wheel. See Figure Rotational Energy Of A Mass And Wheel
The homework exercise has 3 sets of questions with 4 questions per set.
Ep, El, Er, omega, I
Er, m, r, Ep, h
v, omega, Ep, El, Er
The basic equations are :
Ep = m g h
El = 1/2 m v2
Er = Ep - El
omega = v / r
I = 2 Er / omega2
rearranging
Er = 1/2 I omega2
m = 2 * El / v2
r = v / omega
h = Ep / (m g)
omega = v / r
Solving for v : difficult
v = √(2 m g h / (m + I / r2))
where
v = velocity of mass
m = mass
omega = angular velocity of wheel (ω)
I = rotational inertia
r = radius
g = gravity acceleration
h = height moved by mass
Ep = change in potential energy
El = change in linear kinetic energy
Er = change in rotational energy
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| BETA : Cyclic : Rotational Energy 02 : Rotational Energy Of A Mass And Roller Connected By A Rope : BETA Physics Homework Exercise
Description : Rotational energy of a mass and roller connected by a rope
Discussion : The gravitational potential energy of the mass is converted to linear kinetic energy of the mass and roller, and rotational kinetic energy of the roller. See Figure Rotational Energy Of A Mass And Roller
The homework exercise has 3 sets of questions with 5 questions per set.
Ep, El, Er, omega, I
Er, mm, r, Ep, h
v, omega, Ep, El, Er
The basic equations are :
Ep = mm g h
El = 1/2 (mm + mr) v2
Er = Ep - El
omega = v / r
I = 2 Er / omega2
rearranging
Er = 1/2 I omega2
mm = 2 * El / v2 - mr
r = v / omega
h = Ep / (mm g)
omega = v / r
Solving for v : difficult
v = √(2 mm g h / (mm + mr + I / r2))
where
v = velocity of mass
mm = mass of mass
mr = mass of roller.omega = angular velocity of wheel (ω)
I = rotational inertia
r = radius
g = gravity acceleration
h = height moved by mass
Ep = change in potential energy
El = change in linear kinetic energy
Er = change in rotational energy
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| BETA : Cyclic : Rotational Energy 03 : Rotational Energy Of A Mass Rolling Down A Slope : BETA Physics Homework Exercise
Description : Rotational energy of a mass rolling down a slope.
Discussion : The gravitational potential energy of the mass is converted to linear kinetic energy and rotational kinetic energy. See Figure Rotational Energy Of A Mass Rolling Down A slope
The homework exercise has 3 sets of questions with 4 questions per set.
Ep, El, Er, omega, I
Er, m, r, Ep, h
v, omega, Ep, El, Er
The basic equations are :
Ep = m g h
El = 1/2 m v2
Er = Ep - El
omega = v / r
I = 2 Er / omega2
rearranging
Er = 1/2 I omega2
m = 2 * El / v2
r = v / omega
h = Ep / (m g)
omega = v / r
Solving for v : difficult
v = √(2 m g h / (m + I / r2))
where
v = velocity of mass
m = mass
omega = angular velocity of wheel (ω)
I = rotational inertia
r = radius
g = gravity acceleration
h = height moved by mass
Ep = change in potential energy
El = change in linear kinetic energy
Er = change in rotational energy
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| BETA : Cyclic : Torque And Inertia 01 : Torque And Forces Acting On A Bicycle : BETA Physics Homework Exercise
Description : Torque and forces acting on a bicycle pedal, sprockets and wheel.
Discussion : Torque is the product of force times radius. The downwards force on a bicycle pedal produces a torque in the front socket, which produces a force in the chain, which in turn produces a torque on the back sprocket, and which produces a force between the wheel and the ground. See Figure Torque Acting On A Bicycle Pedal Sprockets And Wheel
The homework exercise has 2 sets of questions with 4 questions per set.
Tf, Fc, Tb, Fw
Tb, Fc, Tf, Fp
The basic equations are :
Tf = Fp rp
Fc = Tf / rf
Tb = Fc rb
Fw = Tb / rw
rearranging
Tb = Fw rw
Fc = Tb / rb
Tf = Fc rf
Fp = Tf / rp
where
Fp = pedal force
Fc = chain force
Fw = wheel force
rp = pedal radius
rf = front sprocket radius
rb = back sprocket radius
rw = wheel radius
Tf = front sprocket torque
Tb = back sprocket torque
By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Torque And Inertia 02 : Moment Of Inertia Of Cylinder And Sphere : BETA Physics Homework Exercise
Description : Moment of inertia of simple shapes; hollow cylinder, solid cylinder, hollow sphere, solid sphere.
Discussion : The moment of inertia is the integral of the mass times the radius. The formula for the hollow cylinder and the hollow sphere applies only when the wall thickness of the cylinder or sphere is very thin.
The homework exercise has 3 sets of questions with 4 questions per set.
Ihc, Isc, Ihs, Iss
m, Ihc, Isc, Iss
r, Ihc, Isc, Ihs
The basic equations are :
Ihc = m r2
Isc = 1/2 m r2
Ihs = 2/3 m r2
Iss = 2/5 m r2
rearranging
m = Ihc / r2 = 2 Isc / r2 = 3/2 Ihs / r2 = 5/2 Iss / r2
r = √(Ihc / m) = √(2 Isc / m) = √(3/2 Ihs / m) = √(5/2 Iss / m)
where
Ihc = hollow cylinder moment of inertia
Isc = solid cylinder moment of inertia
Ihs = hollow sphere moment of inertia
Iss = solid sphere moment of inertia
m = mass
r = radius
By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Torque And Inertia 03 : Torque Inertia And Angular Acceleration : BETA Physics Homework Exercise
Description : Torque, inertia and angular acceleration.
Discussion : Torque, inertia and angular acceleration are the rotaional equivalents of force, mass and acceleration.
The homework exercise has 3 sets of questions with 3 questions per set.
T, omegaf, theta
alpha, t, theta
I, omegai, omegaf
The basic equations are :
T = alpha I
omegaf = omegai + alpha t
theta = omegai + 1/2 alpha t2
rearranging
alpha = T / I
I = T / alpha
alpha = (omegaf - omegai) / t
t = (omegaf - omegai) / alpha
other equations
theta = (omegai + omegaf) t / 2
omegai = theta / t - alpha t / 2
where
T = torque
alpha = angular acceleration (α)
I = rotational inertia
t = time
omegai = initial angular velocity (ωi)
omegaf = final angular velocity (ωf)
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| BETA : Cyclic : Torque And Inertia 04 : Acceleration Of A Mass And Wheel Connected By A Rope : BETA Physics Homework Exercise
Description : Acceleration of a mass and wheel connected by a rope.
Discussion : The gravity force on the mass accelerates the wheel. See Figure Acceleration Of A Mass And Wheel
The homework exercise has 3 sets of questions with 4 questions per set.
T, F, a, m
F, T, alpha, I
F, a, T, alpha
The basic equations are :
T = alpha I
F = T / r
a = alpha r
m = F / (g - a)
rearranging
F = m (g -a)
T = F r
alpha = a / r
I = T / alpha
a = g - F / m
solving for F : difficult
a = g - F / m = r alpha = F r2 / I
equating a and rearranging
F = g / (r2 / I + 1 / m )
where
T = torque
F = tension force
a = acceleration of mass
m = mass
alpha = angular acceleration of wheel (α)
I = rotational inertia
r = radius
g = gravitational acceleration
Gravitational acceleration g is a constant = 9.81. By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Torque And Inertia 05 : Acceleration Of A Mass And Roller Connected By A Rope : BETA Physics Homework Exercise
Description : Acceleration of a mass and roller connected by a rope.
Discussion : The gravitational force on the mass accelerates the roller; linear acceleration and angular acceleration. See Figure Acceleration Of A Mass And Roller
The homework exercise has 3 sets of questions with 5 questions per set.
T, Fr, a, Ft, m
Ft, alpha, Fr, T, I
Ft, a, Fr, alpha, T
The basic equations are :
T = alpha I
Fr = T / r
a = alpha r
Ft = Fr + a mr
mm = F / (g - a)
rearranging
Ft = mm (g - a)
a = g - Ft / mm
alpha = a / r
Fr = Ft - a mr
I = T / alpha
alpha = a / r
T = Fr r
T = alpha I
Solving For Ft : Difficult
Ft = g (1 / mm + 1 / mr * (1 - I / (I + mr * r2)))
where
T = torque
Ft = tension force
Fr = resistance force (ground)
a = acceleration of mass
mm = mass of mass
mr = mass of roller
alpha = angular acceleration of wheel (α)
I = rotational inertia
r = radius
g = gravity acceleration
By convention rotational or angular vectors are orientated along the axis of rotation.
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| BETA : Cyclic : Torque And Inertia 06 : Acceleration Of A Mass Rolling Down A Slope : BETA Physics Homework Exercise
Description : Acceleration of a mass rolling down a slope.
Discussion : A rolling mass will accelerate more slowly than a frictionless mass because of the energy required for rotation. Friction force is required for rotation.
The homework exercise has 3 sets of questions with 5 questions per set.
Ft, a, alpha, T, I
T, r, Fr, Ft, theta
m, Fr, T, alpha, a
The basic equations are :
Ft = m g sind(theta)
a = (Ft - Fr) / m
alpha = a / r
T = Fr r
I = T / alpha
rearranging
T = alpha I
alpha = T / I
r = a / alpha
a = alpha r
Fr = T / r
Ft = Fr + m a
Fr = Ft - m a
theta = asind(Ft / (m g)
Solving For Fr : Difficult
Fr = Ft / (m * r2 / I + 1)
where
T = torque
Ft = tangential force
Fr = resistance force (on ground)
a = acceleration of mass
m = mass
r = radius
g = gravity
theta = slope angle ( θ)
alpha = angular acceleration (α)
I = rotational inertia
The angles are in degrees. You are recommended to use the trigonometry degree functions sind() and asind().
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