| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Forces:}\] | List all forces acting immediately after release. |
| \[\begin{aligned} &\text{(i) }\;\vec F_{\!\text{pivot}} \\ &\text{(ii) }\;\vec W = M g \;(\text{down}) \end{aligned}\] |
\(\vec F_{\!\text{pivot}}\) is the unknown reaction at the hinge; \(\vec W\) acts at the centre (mid-point \(L/2\) from pivot). | |
| 2 | \[\tau_\text{net}=W\left(\dfrac{L}{2}\right)\] | Only weight produces a moment about the pivot because the hinge force passes through the pivot. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\tau_\text{net}=I\alpha\] | Rotational analogue of Newtons second law. |
| 2 | \[Mg\left(\dfrac{L}{2}\right)=\left(\dfrac13 M L^2\right)\alpha\] | Substitute \(\tau_\text{net}\) from Part (a) and the given moment of inertia. |
| 3 | \[\alpha=\dfrac{3g}{2L}\] | Cancel \(M\) and solve algebraically for \(\alpha\). |
| 4 | \[a_t=\alpha\left(\dfrac{L}{2}\right)\] | Tangential acceleration of the centre of mass equals \(\alpha r\) with \(r=L/2\). |
| 5 | \[a_t=\dfrac{3g}{2L}\left(\dfrac{L}{2}\right)=\dfrac{3g}{4}\] | Simplify to obtain \(a_t\). |
| 6 | \[\boxed{\alpha=\dfrac{3g}{2L}},\;\boxed{\;a_t=\dfrac{3g}{4}\;}\] | Box the requested expressions. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\tau=Mg\left(\dfrac{L}{2}\right)\sin\theta\] | General expression when the rod makes angle \(\theta\) with vertical. (Pure algebra; \(\sin\theta\) is geometric.) |
| 2 | \[\alpha=\dfrac{\tau}{I}=\dfrac{3g}{2L}\sin\theta\] | Insert \(I=\tfrac13 ML^2\). Shows \(\alpha\propto\sin\theta\). |
| 3 | \[\alpha_{\text{max}}\text{ at }\theta=90^\circ,\;\alpha=0\text{ when }\theta=0^\circ\] | \(\sin90^\circ=1\) (horizontal start) gives maximum; \(\sin0=0\) (vertical) gives zero. Thus \(\alpha\) decreases monotonically. |
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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?
Young David experimented with slings before tackling Goliath. He found that he could develop an angular speed of \( 8.0 \) \( \text{rev/s} \) in a sling \( 0.60 \) \( \text{m} \) long. If he increased the length to \( 0.90 \) \( \text{m} \), he could revolve the sling only \( 6.0 \) times per second.
A rod may freely rotate about an axis that is perpendicular to the rod and is along the plane of the page. The rod is divided into four sections of equal length of 0.2 m each, and four forces are exerted on the rod, as shown in the figure. Frictional forces are considered negligible. Which of the following describes an additional torque that must be applied in order to keep the rod from rotating?
Pulleys \( X \) and \( Y \) are each attached to a block by a string that wraps around the pulley. Both blocks are released and have the same linear acceleration \( a \). As the blocks fall, the pulleys rotate about their centers. Pulley \( Y \) has a larger radius than Pulley \( X \). How does the angular acceleration \( \alpha_X \) of Pulley \( X \) compare to the angular acceleration \( \alpha_Y \) of Pulley \( Y \)?
A net torque is applied to the edge of a spinning object as it rotates about its internal axis. The table shows the net torque exerted on the object at different instants in time. How can a student use the data table to determine the change in angular momentum of the object from \( 0 \) to \( 6 \) \( \text{s} \)? Justify your selection.
| Time \( (\text{s}) \) | Net Torque \( (\text{N} \cdot \text{m}) \) |
|---|---|
| 0 | 0 |
| 2 | 1.5 |
| 4 | 3.0 |
| 6 | 4.5 |
Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?
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?
Three masses are attached to a \( 1.5 \, \text{m} \) long massless bar. Mass 1 is \( 2 \, \text{kg} \) and is attached to the far left side of the bar. Mass 2 is \( 4 \, \text{kg} \) and is attached to the far right side of the bar. Mass 3 is \( 4 \, \text{kg} \) and is attached to the middle of the bar. At what distance from the far left side of the bar can a string be attached to hold the bar up horizontally?
In lacrosse, a typical throw is made by rotating the stick through an angle of roughly \(90^\circ\), then releasing the ball when the stick is vertical, as shown above. If the \(1 \, \text{meter}\) long stick is at rest when horizontal and the ball leaves the stick with a velocity of \(10 \, \text{m/s}\), what angular acceleration must the stick experience?
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 | |
|---|---|---|---|
| \( (a) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\alpha \) |
| \( (b) \) | \( R\omega \) | \( R\alpha \) | \( R\omega^{2} \) |
| \( (c) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega \) |
| \( (d) \) | \( R\omega \) | \( R\omega^{2} \) | \( R\omega \) |
| \( (e) \) | \( R\omega^{2} \) | \( R\alpha \) | \( R\omega^{2} \) |
\(\alpha = \dfrac{3g}{2L}\)
\(a_t = \dfrac{3g}{4}\)
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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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