0 attempts
0% avg
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 | [katex]I = \frac{2}{5} m r^2[/katex] | The moment of inertia (I) for a solid sphere about an axis through its center is [katex]\frac{2}{5} m r^2[/katex], where [katex]m[/katex] is the mass and [katex]r[/katex] is the radius of the sphere. |
| 2 | [katex]r = \frac{0.72 \, \text{m}}{2} = 0.36 \, \text{m}[/katex] | Convert the diameter of the sphere to radius. This helps in computing the moment of inertia and further calculations. |
| 3 | [katex] \tau = I \alpha[/katex] | Torque ([katex]\tau[/katex]) is related to the angular acceleration ([katex]\alpha[/katex]) by the formula, where [katex]I[/katex] is the moment of inertia. |
| 4 | [katex]\theta = \omega_f t – \frac{1}{2} \alpha t^2[/katex] | The angular displacement ([katex] \theta [/katex]) can be expressed in terms of final angular velocity ([katex] \omega_f [/katex]), angular acceleration ([katex] \alpha [/katex]), and time ([katex] t [/katex]). Here, starting from rest simplifies to [katex] \theta = \frac{1}{2} \alpha t^2 [/katex]. |
| 5 | [katex]\theta = 160 \times 2\pi \text{ rad} = 1005.31 \text{ rad}[/katex] | Convert the number of revolutions to radians (since [katex]1[/katex] revolution = [katex]2\pi[/katex] radians). |
| 6 | [katex]1005.31 \text{ rad} = \frac{1}{2} \alpha (15.0 \text{ s})^2[/katex] | Use the total revolutions in radians and solve for angular acceleration [katex]\alpha[/katex] using the time elapsed. |
| 7 | [katex]\alpha = \frac{2 \times 1005.31 \text{ rad}}{(15.0 \text{ s})^2} = 8.937 \text{ rad/s}^2[/katex] | Calculate the angular acceleration [katex]\alpha[/katex]. |
| 8 | [katex]\tau = I \alpha \implies 10.8 \text{ Nm} = \frac{2}{5} m (0.36 \text{ m})^2 \times 8.937 \text{ rad/s}^2[/katex] | Plug values of moment of inertia and angular acceleration into the torque equation to solve for mass [katex]m[/katex]. |
| 9 | [katex]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}[/katex] | Final step: solve the equation for mass, providing the solution to the problem. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
Four forces are exerted on a disk of radius \( R \) that is free to spin about its center, as shown above. The magnitudes are proportional to the length of the force vectors, where \( F_1 = F_4 \), \( F_2 = F_3 \), and \( F_1 = 2F_2 \). Which two forces combine to exert zero net torque on the disk?
A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?
A \( 50 \, \text{kg} \) person is sitting on a seesaw \( 1.2 \, \text{m} \) from the balance point. On the other side, a \( 70 \, \text{kg} \) person is balanced. How far from the balance point is the second person sitting?
A \( 4 \)-\( \text{kg} \) ball and a \( 1 \)-\( \text{kg} \) ball are positioned a distance \( L \) apart on a bar of negligible mass. How far from the \( 4 \)-\( \text{kg} \) mass should the fulcrum be placed to balance the bar?
Why are doorknobs located on the side of the door opposite the hinges?
A disk is initially rotating counterclockwise around a fixed axis with angular speed \( \omega_0 \). At time \( t = 0 \), the two forces shown in the figure above are exerted on the disk. If counterclockwise is positive, which of the following could show the angular velocity of the disk as a function of time?
The downward motion of an elevator is controlled by a cable that unwinds from a cylinder of radius \( 0.20 \) \( \text{m} \). What is the angular velocity of the cylinder when the downward speed of the elevator is \( 1.2 \) \( \text{m/s} \)?
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 disk increases from 2 complete revolutions in 2 seconds to 5 complete revolutions in 2 seconds. What is its average angular acceleration?
A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope.
What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?
Note: [katex] I_\text{disk} = \frac{1}{2}mr^2 [/katex]
23.3 kg
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
