Step | Derivation/Formula | Reasoning |
---|---|---|
1 (a) | v = r \omega | For rolling motion, the linear velocity v at the bottom is related to the angular velocity \omega by the radius r . Here, the radius r is 0.16 m. |
2 (a) | \omega = \frac{v}{r} | Calculate the angular velocity \omega at the bottom using the given linear velocity v = 3.2 \, \text{m/s} . |
3 (a) | \omega = \frac{3.2 \, \text{m/s}}{0.16 \, \text{m}} = 20 \, \text{rad/s} | Substitute the values into the equation to find \omega . |
4 (a) | \alpha = \frac{\omega}{t} | Angular acceleration \alpha is calculated using the angular velocity \omega and the time t it takes to reach that angular velocity. |
5 (a) | v^2 = v_0^2 + 2aL | Using the kinematic equation for linear motion, where a is linear acceleration and L = 1.5 \, \text{m} is the length of the incline. |
6 (a) | 3.2^2 = 0 + 2a \times 1.5 | a = \frac{(3.2)^2}{2 \times 1.5} = \frac{10.24}{3} \approx 3.413 \, \text{m/s}^2 |
7 (a) | a = r\alpha | Relate linear acceleration a to angular acceleration \alpha . |
8 (a) | \alpha = \frac{a}{r} = \frac{3.413 \, \text{m/s}^2}{0.16 \, \text{m}} \approx 21.33 \, \text{rad/s}^2 | Compute angular acceleration \alpha . |
1 (b) | \omega = \alpha t | Use the equation for angular velocity \omega related to angular acceleration \alpha and time t . |
2 (b) | \omega_{\text{rpm}} = 7329 = \omega \frac{60}{2\pi} | Convert \omega from rpm to rad/s for calculation. |
3 (b) | \omega = 7329 \cdot \frac{2\pi}{60} \approx 767.23 \, \text{rad/s} | Find \omega in rad/s. |
4 (b) | t = \frac{\omega}{\alpha} = \frac{767.23}{419} \approx 1.83 \, \text{s} | Compute the time t . |
1 (c) | \Delta \omega = \alpha \Delta t | Angular velocity change \Delta \omega is given, calculate the time \Delta t using angular acceleration \alpha . |
2 (c) | \Delta \omega = 33.3 \, \text{rad/s} – 3.33 \, \text{rad/s} = 29.97 \, \text{rad/s} | Calculate the change in angular velocity \Delta \omega . |
3 (c) | \Delta t = \frac{\Delta \omega}{\alpha} = \frac{29.97}{5.15} \approx 5.82 \, \text{s} | Calculate the time \Delta t needed to reach the final angular velocity. |
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A centrifuge accelerates uniformly from rest to 15,000 rpm in 240 s. Through how many revolutions did it turn in this time?
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during the entire 20-s interval? Assume constant angular acceleration while it is starting and stopping.
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? I_{sphere} = \frac{2}{5}MR^2, I_{cylinder} = \frac{1}{2}MR^2, I_{pipe} = MR^2
A point on the edge of a disk rotates around the center of the disk with an initial angular velocity of 3 rad/s clockwise. The graph shows the point’s angular acceleration as a function of time. The positive direction is considered to be counterclockwise. All frictional forces are considered to be negligible.
How long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
(a) Angular acceleration of the ball is \approx 21.33 \, \text{rad/s}^2 .
(b) Time for the CD player to reach full speed is \approx 1.83 \, \text{s} .
(c) Time to accelerate for the rotating object is \approx 5.82 \, \text{s} .
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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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