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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A centrifuge rotor rotating at 9200 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 N·m. If the mass of the rotor is 3.10 kg and it can be approximated as a solid cylinder of radius 0.0710 m, through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is give by \frac{1}{2}mr^2 .
An object travels along a path shown above, with changing velocity as indicated by vectors A and B. Which vector best represents the net acceleration of the object from time t_A to t_B?
A gun can fire a bullet to height h when fired straight up. If the same gun is pointed at an angle of 45° from the vertical, what is the new maximum height of the projectile?
An object is thrown upward at 65 m/s from the top of a 800 m tall building.
A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.
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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_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 Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients 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. |
k | Spring 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 |
Variable | SI 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)} |
Variable | Derived 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)} |
\omega (Angular Velocity) | \text{radians per second (rad/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.
Start with the given measurement: \text{5 km}
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}}
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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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