# Part (a): Expression for the radius of the hoop
The solution involves converting the initial kinetic energy into gravitational potential energy at the maximum height h .
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | v = R\omega | The velocity v of the hoop at the bottom is related to the angular velocity \omega and the radius R of the hoop by the no-slip condition. |
| 2 | KE_{\text{bottom}} = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2 | Calculate the total kinetic energy at the bottom considering both translational ( \frac{1}{2}m v^2 ) and rotational ( \frac{1}{2} I \omega^2 ) kinetic energy. |
| 3 | I = m R^2 (for a hoop) | The moment of inertia I of a hoop about its center is m R^2 . |
| 4 | KE_{\text{bottom}} = \frac{1}{2}m (R\omega)^2 + \frac{1}{2} m R^2 \omega^2 = m R^2 \omega^2 | Substitute I and v into the kinetic energy expression and simplify. |
| 5 | PE_{\text{top}} = mgh | Calculate the potential energy at the maximum height h using the mass m and gravitational acceleration g . |
| 6 | KE_{\text{bottom}} = PE_{\text{top}} | Apply conservation of mechanical energy, assuming no energy loss to friction or air resistance. |
| 7 | m R^2 \omega^2 = mgh | Set the expressions for kinetic and potential energy equal and simplify. |
| 8 | R = \sqrt{\frac{gh}{\omega^2}} | Solve for R . This equation provides the radius in terms of the given quantities and constants. |
# Part (b): Direction of friction while the hoop rolls up the ramp
| Step | Explanation |
|---|---|
| Reasoning | The direction of friction must oppose the tendency of slipping. Whether the hoop slides up or down the ramp, it will always try to slide down the ramp. Thus, static friction acts in the direction of motion, up the ramp, to prevent the hoop from sliding back. |
# Part (c): Direction of friction while the hoop rolls down the ramp
| Step | Explanation |
|---|---|
| Reasoning | As explained in part (b), regardless of the direction the hoop travels on the ramp, friction will continue to point up the ramp. As the hoop rotates down the ramp, it want to slip down the ramp, but is countered by static friction that points up the ramp. |
Phy can also check your working. Just snap a picture!
An isolated spherical star of radius R_o , rotates about an axis that passes through its center with an angular velocity of \omega_o . Gravitational forces within the star cause the star’s radius to collapse and decrease to a value r_o <R_o , but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum L of the star immediately after the collapse?
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 .
Two uniform solid balls, one of radius R and mass M, the other of radius 2R and mass 8M, roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
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 rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?
By continuing you (1) agree to our Terms of Sale and Terms of Use 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_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} | |
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} |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
