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Using the energy conservation principle: the kinetic energy at release will be converted to potential energy at the highest point. Calculate height h above the wheel center.

Alternatively, you can use both rotational and linear kinematics to determine the max height as shown below.

Step Derivation/Formula Reasoning
1 r = \frac{D}{2} Calculate the radius r of the wheel by dividing the diameter D by 2. The given diameter D is 24 cm, so r = 24 \, \text{cm} / 2 = 12 \, \text{cm} = 0.12 \, \text{m} .
2 {\omega_f }^2 ={\omega_i }^2 + 2\alpha \Delta \theta Use the angular kinematic equation to find the final angular velocity \omega_f . Assume initial angular velocity \omega_i is zero since the wheel starts from rest and angular acceleration \alpha is given.
3 \omega_f = \sqrt{2 \alpha \theta} Rearrange equation in step 2 and calculate the final angular velocity \omega_f .
4 v = r \omega_f To find the height ( a linear distance) we need to find the linear speed v of the ball at the point of release. Use the equation v = r \omega .
5 v = 0.12 \times \sqrt{2 \times 470 \times \frac{3\pi}{2}} Substitute values and calculate v = 7.99 \, m/s .
6 {v_f}^2 = {v_0}^2 + 2a\Delta x Linear kinematic equation to solve for the change in vertical height, given that the final speed is 0 at the max height.
7 \Delta x  = h_{max} = \frac{-{v_0}^2}{2g} Solve for \Delta x which is the maximum height.
8 h = \frac{-(0.12 \times \sqrt{2 \times 470 \times \frac{3\pi}{2}})^2}{2 \times -9.81} Calculate h by substituting v from step 7 and solve.
9 h_{max} = 3.25 \, m  This is the final step where the height h is computed with all values substituted and simplified.

This table gives a detailed breakdown of calculating the height that a steel ball reaches above the center of a wheel, considering the kinematic equations for rotational and linear motions and energy conservation principles.

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h_{max} = 3.25 \, m

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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