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# 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 [katex] h [/katex].
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = R\omega[/katex] | The velocity [katex] v [/katex] of the hoop at the bottom is related to the angular velocity [katex] \omega [/katex] and the radius [katex] R [/katex] of the hoop by the no-slip condition. |
| 2 | [katex]KE_{\text{bottom}} = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2[/katex] | Calculate the total kinetic energy at the bottom considering both translational ([katex] \frac{1}{2}m v^2 [/katex]) and rotational ([katex] \frac{1}{2} I \omega^2 [/katex]) kinetic energy. |
| 3 | [katex]I = m R^2[/katex] (for a hoop) | The moment of inertia [katex] I [/katex] of a hoop about its center is [katex] m R^2 [/katex]. |
| 4 | [katex]KE_{\text{bottom}} = \frac{1}{2}m (R\omega)^2 + \frac{1}{2} m R^2 \omega^2 = m R^2 \omega^2[/katex] | Substitute [katex] I [/katex] and [katex] v [/katex] into the kinetic energy expression and simplify. |
| 5 | [katex]PE_{\text{top}} = mgh[/katex] | Calculate the potential energy at the maximum height [katex] h [/katex] using the mass [katex] m [/katex] and gravitational acceleration [katex] g [/katex]. |
| 6 | [katex]KE_{\text{bottom}} = PE_{\text{top}}[/katex] | Apply conservation of mechanical energy, assuming no energy loss to friction or air resistance. |
| 7 | [katex]m R^2 \omega^2 = mgh[/katex] | Set the expressions for kinetic and potential energy equal and simplify. |
| 8 | [katex]R = \sqrt{\frac{gh}{\omega^2}}[/katex] | Solve for [katex] R [/katex]. 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. |
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A point P is at a distance \( R \) from the axis of rotation of a rigid body whose angular velocity and angular acceleration are \( \omega \) and \( \alpha \) respectively. The linear speed, centripetal acceleration, and tangential acceleration of the point can be expressed as:
| Linear speed | Centripetal acceleration | Tangential acceleration | |
|---|---|---|---|
| \( (a) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\alpha \) |
| \( (b) \) | \( R\omega \) | \( R\alpha \) | \( R\omega^{2} \) |
| \( (c) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega \) |
| \( (d) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\omega \) |
| \( (e) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega^{2} \) |
A uniform rod of length \( L \) and mass \( M \) is free to rotate about one end, as shown in the diagram. The free end is released from rest at a horizontal position, as shown. The pivot point is supported by a stand so that only the free end can move. The moment of inertia of a rod about its end is \(\tfrac{1}{3} M L^{2}\).
What is the net torque acting on the pivot supporting a \(10 \, \text{kilogram}\) beam \(2 \, \text{meters}\) long as shown above? Assume that the positive direction is clockwise.
When the speed of a rear-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?
A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 N·m which accelerates it uniformly from rest through a total of 160 revolutions in 15.0 s. What is the mass of the sphere?
Which of the following must be zero if an object is spinning at a constant rate? There may be more than one right answer.
A high-speed drill rotating counterclockwise at \( 2400 \) \( \text{rpm} \) comes to a halt in \( 2.5 \) \( \text{s} \).
Two uniform disks have the same mass but different radii: disk \( 1 \) has a radius \( R \) and disk \( 2 \) has a radius \( 2R \). What is the ratio of the moment of inertia of the first disk to the second disk?
Consider a uniform hoop of radius R and mass M rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy?
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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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