Here is the solution to the problem:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\tau = r \cdot F\) | Calculate the torque (\(\tau\)) applied to the sphere. Torque is the product of the force applied (F) and the distance (r) from the axis of rotation. |
| 2 | \(\tau = 0.30 \, \text{m} \cdot 2.0 \, \text{N}\) | Substitute the given values for \(r\) and \(F\) into the torque formula. |
| 3 | \(\tau = 0.60 \, \text{N} \cdot \text{m}\) | Calculate the torque. |
| 4 | \(\tau = I \cdot \alpha\) | Relate torque (\(\tau\)) to angular acceleration (\(\alpha\)) using the moment of inertia (\(I\)). |
| 5 | \(0.60 \, \text{N} \cdot \text{m} = 0.06 \, \text{kg} \cdot \text{m}^2 \cdot \alpha\) | Substitute the known values of torque and moment of inertia into the formula to solve for \(\alpha\). |
| 6 | \(\alpha = 10 \, \text{rad/s}^2\) | Solve for the angular acceleration by dividing the torque by the moment of inertia. |
| 7 | \(\omega_f = \omega_i + \alpha \cdot t\) | Use the angular kinematics equation that relates initial angular velocity (\(\omega_i\)), angular acceleration (\(\alpha\)), and time (\(t\)) to final angular velocity (\(\omega_f\)). Since we are bringing the sphere to rest, \(\omega_f = 0\). |
| 8 | \(0 = 20 \, \text{rad/s} + (-10 \, \text{rad/s}^2) \cdot t\) | Substitute \(\omega_i = 20 \, \text{rad/s}\) and \(\alpha = -10 \, \text{rad/s}^2\) (negative because the sphere is slowing down). |
| 9 | \(-20 \, \text{rad/s} = -10 \, \text{rad/s}^2 \cdot t\) | Rearrange the equation to solve for \(t\). |
| 10 | \(t = 2 \, \text{s}\) | Solve for \(t\) by dividing both sides by \(-10 \, \text{rad/s}^2\). |
| 11 | \(\boxed{2 \, \text{s}}\) | The time required to stop the sphere is 2 seconds, which matches option (b). |
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What condition(s) are necessary for static equilibrium?
The moment of inertia of a solid cylinder about its axis is given by \( 0.5MR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
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?
A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
A solid ball and a cylinder roll down an inclined plane. Which reaches the bottom first? Hint the rotational inertia of a sphere about its center is \(I = \frac{2}{5}mR^{2}\) and the rotational inertia of a cylinder about its center is \(I = \frac{1}{2}mR^{2}\).
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of \( 5.0 \) \( \text{rev/s} \) in \( 8.0 \) \( \text{s} \). 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 \( 12.0 \) \( \text{s} \). 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 disk of radius 35 cm rotates at a constant angular velocity of 10 rad/s. How fast does a point on the rim of the disk travel (in m/s)?
A solid sphere \( \left( I = \frac{2}{5}MR^2 \right) \) and a solid cylinder \( \left( I = \frac{1}{2}MR^2 \right) \), both uniform and of the same mass and radius, roll without slipping at the same forward speed. It is correct to say that the total kinetic energy of the solid sphere is
Suppose just two external forces act on a stationary, rigid object and the two forces are equal in magnitude and opposite in direction. Under what condition does the object start to rotate?
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]
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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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