To solve this problem, we need to calculate the mass of the sphere given the diameter, torque, number of revolutions, and time. We will use the rotational dynamics principles and formulas.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(I = \frac{2}{5} m r^2\) | The moment of inertia (I) for a solid sphere about an axis through its center is \(\frac{2}{5} m r^2\), where \(m\) is the mass and \(r\) is the radius of the sphere. |
| 2 | \(r = \frac{0.72 \, \text{m}}{2} = 0.36 \, \text{m}\) | Convert the diameter of the sphere to radius. This helps in computing the moment of inertia and further calculations. |
| 3 | \( \tau = I \alpha\) | Torque (\(\tau\)) is related to the angular acceleration (\(\alpha\)) by the formula, where \(I\) is the moment of inertia. |
| 4 | \(\theta = \omega_f t – \frac{1}{2} \alpha t^2\) | The angular displacement (\( \theta \)) can be expressed in terms of final angular velocity (\( \omega_f \)), angular acceleration (\( \alpha \)), and time (\( t \)). Here, starting from rest simplifies to \( \theta = \frac{1}{2} \alpha t^2 \). |
| 5 | \(\theta = 160 \times 2\pi \text{ rad} = 1005.31 \text{ rad}\) | Convert the number of revolutions to radians (since \(1\) revolution = \(2\pi\) radians). |
| 6 | \(1005.31 \text{ rad} = \frac{1}{2} \alpha (15.0 \text{ s})^2\) | Use the total revolutions in radians and solve for angular acceleration \(\alpha\) using the time elapsed. |
| 7 | \(\alpha = \frac{2 \times 1005.31 \text{ rad}}{(15.0 \text{ s})^2} = 8.937 \text{ rad/s}^2\) | Calculate the angular acceleration \(\alpha\). |
| 8 | \(\tau = I \alpha \implies 10.8 \text{ Nm} = \frac{2}{5} m (0.36 \text{ m})^2 \times 8.937 \text{ rad/s}^2\) | Plug values of moment of inertia and angular acceleration into the torque equation to solve for mass \(m\). |
| 9 | \(m = \frac{10.8 \text{ Nm}}{\frac{2}{5} \times (0.36 \text{ m})^2 \times 8.937 \text{ rad/s}^2} \approx 23.3 \text{ kg}\) | Final step: solve the equation for mass, providing the solution to the problem. |
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The angular velocity of a rotating disk of radius \(20 \, \text{cm}\) increases from \(1 \, \text{rad/s}\) to \(3 \, \text{rad/s}\) in \(0.5 \, \text{s}\). What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
A meter stick of mass 200 grams is balanced at the 40-cm mark when a 100-gram mass is suspended from the 10-cm mark. What is the distance from the pivot point to the center of mass of the meter stick? Give your answer in centimeters.
A uniform rod of length \( L \) is pivoted at one end \(45^{\circ}\) below the horizontal and released from rest. The rod swings freely downward. Which of the following best describes the angular acceleration of the rod as it swings from the initial position to the vertical position?
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.
A centrifuge rotor rotating at \( 9200 \) \( \text{rpm} \) is shut off and is eventually brought uniformly to rest by a frictional torque of \( 1.20 \) \( \text{N} \cdot \text{m} \). If the mass of the rotor is \( 3.10 \) \( \text{kg} \) and it can be approximated as a solid cylinder of radius \( 0.0710 \) \( \text{m} \), through how many revolutions will the rotor turn before coming to rest? The moment of inertia of a cylinder is given by \( \frac{1}{2} m r^2 \).
A uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
At time \( t = 0 \), a disk starts from rest and begins spinning about its center with a constant angular acceleration of magnitude \( \alpha \). At time \( t_f \), the disk has angular speed \( \omega_f \). Which of the following expressions correctly compares the final angular displacement \( \theta_f \) of the disk at time \( t_f \) to the angular displacement \( \theta_{1/2} \) at time \( \frac{t_f}{2} \)?
When is the angular momentum of a system constant?
A 6.0-cm-diameter gear rotates with angular velocity \( \omega = \left(20-\frac {1}{2} t^2 \right) \, \text {rad/s} \), where \(t\) is in seconds. At \(t = 4.0 \, \text{s}\), what are
23.3 kg
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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