AP Physics C: Mechanics flashcards

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Unit 1: Kinematics

1. Topic 1.1 · Spot the mistakeA runner jogs once around a 400 m track in 80 s. A student says the average velocity is 5 m/s, the same as the average speed. What is wrong?

   Velocity is a vector built from displacement. After one full lap the displacement is zero, so the average velocity is zero. Average speed is distance over time, $400/80=5$ m/s. Speed is a scalar and velocity is a vector, so they can differ.

   Review: 1.1 Scalars and Vectors

2. Topic 1.1 · Work it outVector $\overrightarrow{A}=3\hat{i}-4\hat{j}$ m and $\overrightarrow{B}=-\hat{i}+6\hat{j}$ m. Find $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ in unit vector notation and the magnitude of $\overrightarrow{C}$.

   Add components: $\overrightarrow{C}=(3-1)\hat{i}+(-4+6)\hat{j}=2\hat{i}+2\hat{j}$ m. The magnitude is $\sqrt{2^2+2^2}=2\sqrt{2}\approx 2.8$ m, at ${45}^{\circ }$ above the $+x$ axis.

   Review: 1.1 Scalars and Vectors

3. Topic 1.2 · Work it outA particle's position is $x(t)=2t^3-6t$ (meters, $t$ in seconds). Find its velocity and acceleration at $t=2$ s, and say whether its speed is increasing or decreasing.

   $v=dx/dt=6t^2-6=18$ m/s and $a=dv/dt=12t=24$ m/s². Velocity and acceleration are both positive, so they point the same way and the speed is increasing.

   Review: 1.2 Displacement, Velocity, and Acceleration

4. Topic 1.2 · Spot the mistakeA ball is thrown straight up. At the top of its path its velocity is zero. A student says its acceleration is zero there too. What is wrong?

   Acceleration is the rate of change of velocity, and the velocity is still changing at the top, from upward to downward. The acceleration is 10 m/s² downward during the whole flight, including the instant at the top.

   Review: 1.2 Displacement, Velocity, and Acceleration

5. Topic 1.3 · Work it outA cart's velocity rises steadily from 0 to 8 m/s in 4 s, stays at 8 m/s for 3 s, then falls steadily to 0 in 2 s. Find its displacement and its average velocity over the 9 s.

   Displacement is the area under the velocity-time graph: $\frac{1}{2}(4)(8)+(3)(8)+\frac{1}{2}(2)(8)=16+24+8=48$ m. The average velocity is $48/9\approx 5.3$ m/s.

   Review: 1.3 Representing Motion

6. Topic 1.3 · Work it outA ball is thrown straight up at 20 m/s. Use $g=10$ m/s². Find the maximum height above the launch point and the time to return to it.

   At the top $v=0$, so $v^2=v_0^2-2g\mathrm{\Delta }y$ gives $\mathrm{\Delta }y=400/20=20$ m. Time up is $v_0/g=2$ s and the trip is symmetric, so the total time is 4 s.

   Review: 1.3 Representing Motion

7. Topic 1.4 · Work it outA boat moves at 4 m/s relative to still water and points straight across a 100 m wide river that flows at 3 m/s. Find its speed relative to the shore and how far downstream it lands.

   Relative to the shore the velocities add as vectors, perpendicular here: $\sqrt{4^2+3^2}=5$ m/s. The crossing takes $100/4=25$ s, and in that time the current carries it $3\times 25=75$ m downstream.

   Review: 1.4 Reference Frames and Relative Motion

8. Topic 1.4 · Say it in contextA ball falls near a person standing on the ground and near a passenger in a car moving at constant velocity. Do the two observers agree on the ball's velocity? On its acceleration?

   They disagree on the velocity, because each measures it relative to their own frame. They agree on the acceleration, because the frames differ only by a constant velocity. Acceleration is the same in every inertial frame.

   Review: 1.4 Reference Frames and Relative Motion

9. Topic 1.5 · Work it outA ball is kicked horizontally at 6 m/s from the edge of a 20 m high cliff. Use $g=10$ m/s². Find the time to land, the horizontal distance traveled, and the speed on impact.

   Vertical: $20=\frac{1}{2}(10)t^2$ gives $t=2$ s. Horizontal: $6\times 2=12$ m. On impact $v_y=10\times 2=20$ m/s and $v_x=6$ m/s, so the speed is $\sqrt{6^2+{20}^2}\approx 20.9$ m/s.

   Review: 1.5 Motion in Two or Three Dimensions

10. Topic 1.5 · Spot the mistakeOne ball is dropped from rest while another is launched horizontally from the same height at the same moment. A student says the launched ball lands later because it travels farther. What is wrong?

    Motion in one dimension does not change motion in a perpendicular dimension. Both balls start with zero vertical velocity and accelerate downward at $g$, so they land together. The launched ball just covers more ground in that time.

    Review: 1.5 Motion in Two or Three Dimensions

Unit 2: Force and Translational Dynamics

1. Topic 2.1 · Work it outMasses of 2 kg at $x=0$, 3 kg at $x=4$ m and 5 kg at $x=6$ m lie on a line. Where is their center of mass?

   $x_{cm}=\frac{\mathrm{\Sigma }{m}_i{x}_i}{\mathrm{\Sigma }{m}_i}=\frac{2(0)+3(4)+5(6)}{2+3+5}=\frac{42}{10}=4.2$ m. It lies closer to the heavier masses.

   Review: 2.1 Systems and Center of Mass

2. Topic 2.1 · Work it outA thin rod of length $L$ has linear mass density $\lambda =kx$, where $x$ is measured from one end. Find its total mass and the position of its center of mass.

   $M={\int }_0^{L}kxdx=\frac{k{L}^2}{2}$. Then $x_{cm}=\frac{1}{M}{\int }_0^{L}x(kx)dx=\frac{k{L}^3/3}{k{L}^2/2}=\frac{2L}{3}$. It sits past the middle, toward the denser end.

   Review: 2.1 Systems and Center of Mass

3. Topic 2.2 · Spot the mistakeA block slides down a rough incline. A student's free-body diagram includes an arrow labeled "force of motion" down the slope. What is wrong, and which forces belong on the diagram?

   A force must come from an interaction with another object, and nothing exerts a "force of motion". Show only forces on the block: gravity (straight down), the normal force (perpendicular to the surface) and kinetic friction (up the slope). Draw each as one arrow from the dot.

   Review: 2.2 Forces and Free-Body Diagrams

4. Topic 2.3 · Spot the mistakeA heavy truck collides with a small car. A student says the truck exerts a larger force on the car than the car exerts on the truck, because the truck is heavier. What is wrong?

   By Newton's third law the two forces are equal in magnitude and opposite in direction, and they act on different objects. The car has the larger acceleration because the same size force acts on a smaller mass.

   Review: 2.3 Newton’s Third Law

5. Topic 2.3 · Work it outIn an ideal Atwood machine a 3 kg mass and a 1 kg mass hang from an ideal string over an ideal pulley. Use $g=10$ m/s². Find the acceleration and the tension.

   Treat the two masses and the string as one system: $(3-1)(10)=(3+1)a$, so $a=5$ m/s². For the 1 kg mass, $T-10=1(5)$, so $T=15$ N. Check with the 3 kg mass: $30-15=3(5)$. Tension is the same throughout an ideal string.

   Review: 2.3 Newton’s Third Law

6. Topic 2.4 · Say it in contextA crate slides across a floor at constant velocity while a 40 N horizontal force pushes it. What is the friction force, and what does the constant velocity tell you?

   Constant velocity means the net force is zero, so friction is 40 N opposite the push. The forces are balanced, which does not mean there is no force or no motion.

   Review: 2.4 Newton’s First Law

7. Topic 2.4 · Say it in contextA puck slides on frictionless ice at 3 m/s east. A steady 2 N force then pushes it north. What happens to its eastward velocity?

   Nothing. The forces are unbalanced only in the north direction, so velocity changes only northward. The puck keeps its 3 m/s eastward component and gains northward velocity.

   Review: 2.4 Newton’s First Law

8. Topic 2.5 · Work it outA 2 kg object starts from rest at the origin. A net force $F(t)=6t$ N (t in seconds) acts along $+x$. Find its velocity and position at $t=2$ s.

   $a=F/m=3t$. Then $v={\int }_0^{2}3tdt=1.5t^2=6$ m/s and $x={\int }_0^{2}1.5t^{2}dt=0.5t^3=4$ m.

   Review: 2.5 Newton’s Second Law

9. Topic 2.5 · Spot the mistakeA ball is rising after being tossed upward. A student says there must be a net upward force, since the ball is moving up. What is wrong?

   Net force sets the acceleration, not the direction of motion. After release the only force is gravity, pointing down, so the net force is downward and the rising ball is slowing down.

   Review: 2.5 Newton’s Second Law

10. Topic 2.6 · Work it outEarth's gravitational field strength at its surface is 10 N/kg. Find the field strength one Earth radius above the surface, and the weight of a 60 kg person there.

    One radius above the surface is $r=2R$, and $g\propto 1/r^2$, so the field is $10/4=2.5$ N/kg. The weight is $60\times 2.5=150$ N, a quarter of the surface value.

    Review: 2.6 Gravitational Force

11. Topic 2.6 · Say it in contextA 50 kg student stands on a scale in an elevator accelerating upward at 2 m/s². Use $g=10$ m/s². What does the scale read? What does it read if the cable breaks and the elevator falls freely?

    The scale reads the normal force: $N-mg=ma$, so $N=50(12)=600$ N. In free fall gravity is the only force, so $N=0$ and the student seems weightless, though gravity still acts.

    Review: 2.6 Gravitational Force

12. Topic 2.6 · Say it in contextInside a planet of uniform density, how does the gravitational field strength at half the planet's radius compare with the surface value? What is the field inside a thin hollow shell?

    Only the mass closer to the center than the object contributes to the net force, because the pulls of the outer layers cancel. The field grows in proportion to the distance from the center, so it is half the surface value at $R/2$. Inside a thin hollow shell the shell's net gravitational field is zero.

    Review: 2.6 Gravitational Force

13. Topic 2.7 · Work it outA 5 kg block rests on a floor with ${\mu }_s=0.6$ and ${\mu }_k=0.4$. Use $g=10$ m/s². Does a 25 N horizontal push move it, and what is friction? What happens with a 35 N push?

    $N=50$ N, so maximum static friction is $0.6\times 50=30$ N. At 25 N the block stays put and static friction equals 25 N. At 35 N it slips, friction drops to $0.4\times 50=20$ N, and $a=15/5=3$ m/s².

    Review: 2.7 Kinetic and Static Friction

14. Topic 2.7 · Spot the mistakeA student says friction always equals $\mu N$ and gets bigger when the surfaces touch over a larger area. What is wrong?

    Only kinetic friction is ${\mu }_{k}N$. Static friction takes whatever value prevents slipping, up to a maximum of ${\mu }_{s}N$. Friction does not depend on the area of contact.

    Review: 2.7 Kinetic and Static Friction

15. Topic 2.8 · Work it outA 0.5 kg block hangs at rest from a spring that stretches 0.1 m. Use $g=10$ m/s². Find $k$. Then find the equivalent spring constant of two such springs joined end to end, and of two side by side.

    At rest $kx=mg$, so $k=5/0.1=50$ N/m. In series $1/k_{eq}=1/50+1/50$, so $k_{eq}=25$ N/m, smaller than either spring. In parallel $k_{eq}=50+50=100$ N/m.

    Review: 2.8 Spring Forces

16. Topic 2.9 · Work it outA 2 kg object falls from rest with air resistance $F_r=-kv$, where $k=4$ kg/s. Use $g=10$ m/s². Find its terminal speed and its speed after 0.5 s, given $v=v_t(1-e^{-kt/m})$.

    At terminal speed the net force is zero, so $mg=k{v}_t$ and $v_t=20/4=5$ m/s. With $kt/m=1$, $v=5(1-e^{-1})\approx 3.2$ m/s. The speed approaches 5 m/s but never exceeds it.

    Review: 2.9 Resistive Forces

17. Topic 2.9 · Say it in contextA puck on frictionless ice is launched at $v_0$ and feels only a drag force $F=-kv$. Write $v(t)$, and decide whether the puck ever stops and whether its total distance is finite.

    Separate variables in $mdv/dt=-kv$: $v=v_0{e}^{-kt/m}$. The speed only approaches zero, so it never stops in finite time. The total distance is finite: ${\int }_0^{\infty }vdt=m{v}_0/k$.

    Review: 2.9 Resistive Forces

18. Topic 2.10 · Work it outA roller coaster car goes around a vertical loop of radius 10 m. Use $g=10$ m/s². Find the minimum speed at the top that keeps the car on the track, and the normal force at that speed.

    At the minimum speed gravity alone supplies the centripetal acceleration: $mg=m{v}^2/r$, so $v=\sqrt{gr}=10$ m/s. The normal force is zero at that instant. Any slower and the car would leave the track.

    Review: 2.10 Circular Motion

19. Topic 2.10 · Which one fitsA 1000 kg car rounds a flat curve of radius 50 m at 20 m/s. Use $g=10$ m/s². Which force supplies the centripetal acceleration, and what minimum ${\mu }_s$ keeps the car from sliding?

    Static friction from the road points toward the center. Required: $a_c=v^2/r=400/50=8$ m/s². Friction must reach $m{a}_c$ and is at most ${\mu }_{s}mg$, so ${\mu }_s\ge 8/10=0.8$.

    Review: 2.10 Circular Motion

20. Topic 2.10 · Work it outA satellite circles a planet at radius $R$ with period $T$. A second satellite circles the same planet at radius $4R$. What is its period?

    For circular orbits $T^2=\frac{4{\pi }^2}{GM}{R}^3$, so $T\propto R^{3/2}$. Quadrupling the radius multiplies the period by $4^{3/2}=8$, giving $8T$. The satellite's own mass does not matter.

    Review: 2.10 Circular Motion

Unit 3: Work, Energy, and Power

1. Topic 3.1 · Say it in contextA 2 kg ball moves at 3 m/s relative to the ground. A cyclist moving along with it at 3 m/s measures the ball's kinetic energy. What do the ground observer and the cyclist each find?

   The ground observer finds $\frac{1}{2}(2)(3{)}^2=9$ J. The cyclist sees the ball at rest and finds 0 J. Kinetic energy is a scalar that depends on the observer's frame of reference.

   Review: 3.1 Translational Kinetic Energy

2. Topic 3.1 · Spot the mistakeA student says that when a car's speed doubles its kinetic energy doubles. What is wrong, and what does that mean for stopping distance under the same braking force?

   Kinetic energy is $\frac{1}{2}m{v}^2$, so doubling the speed quadruples it. The brakes must remove four times as much energy, so with the same braking force the stopping distance is four times as long.

   Review: 3.1 Translational Kinetic Energy

3. Topic 3.2 · Work it outA net force $F(x)=3x^2$ N pushes a 4 kg object along $+x$ from $x=0$ to $x=2$ m. The object starts at 1 m/s in the $+x$ direction. Find the work done by the force and the final speed.

   $W={\int }_0^{2}3x^{2}dx=8$ J. Initially $K=\frac{1}{2}(4)(1{)}^2=2$ J, so $K_f=10$ J and $v=\sqrt{2(10)/4}=\sqrt{5}\approx 2.2$ m/s, by the work-energy theorem.

   Review: 3.2 Work

4. Topic 3.2 · Spot the mistakeA person carries a 10 kg box horizontally across a room at constant speed. A student says the person does positive work on the box. What is wrong?

   Work is $Fd\cos \theta$. The upward force on the box is perpendicular to the displacement, so $\theta ={90}^{\circ }$ and the work is zero. A force perpendicular to the motion changes direction, not kinetic energy.

   Review: 3.2 Work

5. Topic 3.2 · Work it outA 2 kg block slides at 6 m/s onto a rough floor with ${\mu }_k=0.5$. Use $g=10$ m/s². How far does it slide before stopping, and how much mechanical energy is dissipated?

   Friction is $0.5\times 20=10$ N. The work-energy theorem gives $-10d=0-\frac{1}{2}(2)(36)$, so $d=3.6$ m. The 36 J of kinetic energy is dissipated as thermal energy.

   Review: 3.2 Work

6. Topic 3.3 · Say it in contextA potential energy graph $U(x)$ has a local minimum at $x=a$ and a local maximum at $x=b$. Classify each equilibrium and give the force at each point.

   The minimum at $x=a$ is stable: a small displacement leads to a force back toward $a$. The maximum at $x=b$ is unstable: a small displacement leads to a force away from $b$. At both, $F_x=-dU/dx=0$.

   Review: 3.3 Potential Energy

7. Topic 3.3 · Work it outA particle has $U(x)=x^3-12x$ (joules, $x$ in meters). Find its equilibrium positions and classify each as stable or unstable.

   Set $F_x=-dU/dx=-(3x^2-12)=0$, so $x=\pm 2$ m. The second derivative is $6x$. At $x=2$ it is positive (a minimum, stable). At $x=-2$ it is negative (a maximum, unstable).

   Review: 3.3 Potential Energy

8. Topic 3.3 · Work it outA 0.2 kg block compresses a spring ($k=200$ N/m) by 0.1 m on a frictionless table and is released. Find its speed as it leaves the spring.

   The stored energy is $\frac{1}{2}k(\mathrm{\Delta }x{)}^2=\frac{1}{2}(200)(0.01)=1$ J. All of it becomes kinetic energy, so $\frac{1}{2}(0.2)v^2=1$ and $v=\sqrt{10}\approx 3.2$ m/s.

   Review: 3.3 Potential Energy

9. Topic 3.4 · Which one fitsA block slides down a rough ramp. For the system of block, ramp and Earth, is mechanical energy conserved? Is total energy? What if the system is only the block?

   Mechanical energy is not conserved, because friction dissipates some as thermal energy, but total energy is. For the block alone, gravity and friction do work on it, so its energy changes by the net work, a transfer with its surroundings.

   Review: 3.4 Conservation of Energy

10. Topic 3.4 · Work it outA 0.5 kg ball on a 2 m string is released from rest at a height of 0.8 m above the bottom of its swing. Use $g=10$ m/s². Find its speed and the string tension at the bottom.

    Energy: $mgh=\frac{1}{2}m{v}^2$, so $v=\sqrt{2(10)(0.8)}=4$ m/s. At the bottom $T-mg=m{v}^2/L$, so $T=5+0.5(16)/2=9$ N.

    Review: 3.4 Conservation of Energy

11. Topic 3.5 · Work it outA motor lifts a 50 kg load straight up at a constant 2 m/s. Use $g=10$ m/s². What power does it deliver?

    At constant speed the lifting force equals the weight, 500 N. Power is $P=Fv=500\times 2=1000$ W.

    Review: 3.5 Power

12. Topic 3.5 · Work it outA constant 20 N force, 60 degrees above the horizontal, pulls a crate along the floor at a constant 3 m/s. What instantaneous power does the force deliver?

    Only the component parallel to the velocity delivers power: $P=Fv\cos \theta =20\times 3\times 0.5=30$ W. The vertical component does no work.

    Review: 3.5 Power

Unit 4: Linear Momentum

1. Topic 4.1 · Say it in contextTwo objects have the same kinetic energy, and object A has four times the mass of object B. Compare the magnitudes of their momenta.

   Since $K=p^2/(2m)$, the momentum is $p=\sqrt{2mK}$. With equal $K$, $p$ is proportional to $\sqrt{m}$, so A has twice the momentum of B.

   Review: 4.1 Linear Momentum

2. Topic 4.1 · Say it in contextA 2 kg cart moves east at 3 m/s and a 3 kg cart moves west at 2 m/s. Find the total momentum and the total kinetic energy of the pair.

   Momentum is a vector: taking east as positive, $2(3)+3(-2)=0$. Kinetic energy is a scalar and always adds: $\frac{1}{2}(2)(9)+\frac{1}{2}(3)(4)=9+6=15$ J.

   Review: 4.1 Linear Momentum

3. Topic 4.2 · Work it outA net force on a 2 kg cart starts at 0, rises steadily to 10 N in 2 s, stays at 10 N for 1 s, then falls steadily to 0 in 1 s. The cart starts at rest. Find the impulse and the final speed.

   Impulse is the area under the force-time graph: $\frac{1}{2}(2)(10)+(1)(10)+\frac{1}{2}(1)(10)=25$ N·s. Then $J=\mathrm{\Delta }p=mv$ gives $v=25/2=12.5$ m/s.

   Review: 4.2 Change in Momentum and Impulse

4. Topic 4.2 · Work it outSand falls onto a horizontal conveyor belt at 2 kg/s. The belt moves at a constant 3 m/s and the sand speeds up to match it. What horizontal force must the belt supply to the sand?

   The speed is constant but the moving mass grows, so $F=\frac{dm}{dt}v=(2)(3)=6$ N. The motor must supply this force on the sand to keep the belt moving at 3 m/s.

   Review: 4.2 Change in Momentum and Impulse

5. Topic 4.3 · Work it outA 60 kg skater and a 40 kg skater are at rest on frictionless ice and push apart. The 40 kg skater moves off at 3 m/s. Find the velocity of the 60 kg skater and of the system's center of mass.

   No net external force acts, so total momentum stays zero: $60v=40(3)$ gives $v=2$ m/s in the opposite direction. The center-of-mass velocity stays zero, since the internal forces cannot change it.

   Review: 4.3 Conservation of Linear Momentum

6. Topic 4.3 · Which one fitsA ball falls toward the ground. Is momentum conserved for the ball alone? For the ball and Earth together?

   Not for the ball alone: gravity is an external force that delivers an impulse. For the ball and Earth together the gravitational forces are internal and equal and opposite, so the total momentum is constant. Earth's share of the change is tiny.

   Review: 4.3 Conservation of Linear Momentum

7. Topic 4.4 · Work it outA 2 kg cart at 3 m/s hits a stationary 1 kg cart and they stick together. Find the final speed and the fraction of the initial kinetic energy lost.

   Momentum: $2(3)=3v$, so $v=2$ m/s. Kinetic energy goes from $\frac{1}{2}(2)(9)=9$ J to $\frac{1}{2}(3)(4)=6$ J. One third of the kinetic energy is lost.

   Review: 4.4 Elastic and Inelastic Collisions

8. Topic 4.4 · Spot the mistakeA student says that in an elastic collision each object keeps its own kinetic energy. What is wrong?

   Only the system's total kinetic energy stays the same. Energy can pass from one object to another. When a moving ball hits an equal-mass ball at rest, head on and elastically, the first stops and the second leaves with all of it.

   Review: 4.4 Elastic and Inelastic Collisions

9. Topic 4.4 · Work it outA 1 kg ball at 4 m/s hits a stationary 3 kg ball in a head-on elastic collision. Find the velocity of each ball afterward.

   Conserve momentum and kinetic energy: $v_1=\frac{1-3}{1+3}(4)=-2$ m/s and $v_2=\frac{2(1)}{1+3}(4)=2$ m/s. Check: momentum $4=-2+6$, and kinetic energy $8=2+6$.

   Review: 4.4 Elastic and Inelastic Collisions

Unit 5: Torque and Rotational Dynamics

1. Topic 5.1 · Work it outA wheel spins at 10 rad/s and slows uniformly at 2 rad/s². How long does it take to stop, and through what angle does it turn?

   $\omega ={\omega }_0-\alpha t$ gives $t=10/2=5$ s. Then ${\omega }^2={\omega }_0^2-2\alpha \mathrm{\Delta }\theta$ gives $\mathrm{\Delta }\theta =100/4=25$ rad.

   Review: 5.1 Rotational Kinematics

2. Topic 5.1 · Work it outA rotating rod has angular position $\theta (t)=2t^3-3t$ (radians, $t$ in seconds). Find its angular velocity and angular acceleration at $t=1$ s.

   $\omega =d\theta /dt=6t^2-3=3$ rad/s and $\alpha =d\omega /dt=12t=12$ rad/s². The calculus is the same as in one dimension, with angle in radians.

   Review: 5.1 Rotational Kinematics

3. Topic 5.2 · Work it outA point on the rim of a wheel of radius 0.30 m moves while the wheel has $\omega =20$ rad/s and $\alpha =5$ rad/s². Find its speed, its tangential acceleration and its centripetal acceleration.

   $v=r\omega =0.30\times 20=6$ m/s. $a_T=r\alpha =0.30\times 5=1.5$ m/s². $a_c={\omega }^{2}r=400\times 0.30=120$ m/s², which equals $v^2/r$.

   Review: 5.2 Connecting Linear and Rotational Motion

4. Topic 5.2 · Say it in contextTwo points sit on a spinning rigid disk, one near the center and one near the rim. Compare their angular speeds, linear speeds and centripetal accelerations.

   Every point of a rigid system has the same angular speed. The rim point has the larger linear speed ($v=r\omega$) and the larger centripetal acceleration ($a_c={\omega }^{2}r$), both proportional to the distance from the axis.

   Review: 5.2 Connecting Linear and Rotational Motion

5. Topic 5.3 · Work it outA 40 N force acts 0.25 m from a bolt, at 30 degrees to the line from the bolt to the point of application. Find the torque. What perpendicular force at that same point gives the same torque?

   $\tau =rF\sin \theta =0.25\times 40\times 0.5=5$ N·m. A force perpendicular to the position vector has $\sin {90}^{\circ }=1$, so $F=5/0.25=20$ N gives the same torque.

   Review: 5.3 Torque

6. Topic 5.3 · Spot the mistakeA large force pulls on a door handle straight toward the hinge, along the line through the axis. A student says a large force must give a large torque. What is wrong?

   Torque comes only from the force component perpendicular to the position vector. A force along the line through the axis has $\theta =0$, so $\tau =rF\sin \theta =0$ and the lever arm is zero.

   Review: 5.3 Torque

7. Topic 5.4 · Work it outA light rod carries a 1 kg point mass 0.5 m from the rotation axis and a 3 kg point mass 2 m from it, on the same line. Find the rotational inertia about the axis.

   $I=\mathrm{\Sigma }{m}_i{r}_i^2=1(0.5{)}^2+3(2{)}^2=0.25+12=12.25$ kg·m². The far mass dominates, since distance is squared.

   Review: 5.4 Rotational Inertia

8. Topic 5.4 · Work it outA uniform thin rod of mass $M$ and length $L$ has $I_{cm}=\frac{1}{12}M{L}^2$ about its center. Find its rotational inertia about one end with the parallel axis theorem.

   The end is $d=L/2$ from the center, so $I=I_{cm}+M{d}^2=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\frac{1}{3}M{L}^2$. The center-of-mass axis always gives the smallest value.

   Review: 5.4 Rotational Inertia

9. Topic 5.5 · Work it outA uniform 2 m plank of mass 10 kg balances on a pivot at its center. A 5 kg block sits 0.6 m from the pivot on the left. Where must a 4 kg block go on the right for rotational equilibrium?

   The plank's weight acts at the pivot and has no torque. Net torque zero: $5(0.6)=4x$, so $x=0.75$ m to the right of the pivot. The pivot force must also balance the weights for translational equilibrium.

   Review: 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form

10. Topic 5.5 · Say it in contextGive one example of a system in rotational equilibrium but not translational equilibrium, and one that is in translational equilibrium but not rotational.

    A ball in free fall that is not spinning has a net force but no net torque about its center. A free wheel pushed by two equal and opposite forces tangent to its rim at opposite ends of a diameter (a couple) has zero net force but a net torque, so its angular velocity changes.

    Review: 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form

11. Topic 5.6 · Work it outA 2 kg uniform disk of radius 0.5 m ($I=\frac{1}{2}M{R}^2$) turns freely on its axle. A 10 N force acts tangentially at the rim. Find the angular acceleration and the angular speed after 3 s from rest.

    $\tau =FR=5$ N·m and $I=\frac{1}{2}(2)(0.25)=0.25$ kg·m². Then $\alpha =\tau /I=20$ rad/s², and $\omega =\alpha t=60$ rad/s.

    Review: 5.6 Newton’s Second Law in Rotational Form

12. Topic 5.6 · Work it outA 2 kg mass hangs from a string wrapped around a uniform disk pulley of mass 4 kg ($I=\frac{1}{2}M{R}^2$). Use $g=10$ m/s². Find the acceleration of the mass and the string tension.

    Mass: $mg-T=ma$. Pulley: $TR=I\alpha =\frac{1}{2}M{R}^2(a/R)$, so $T=\frac{1}{2}Ma$. Then $20=(2+2)a$ gives $a=5$ m/s² and $T=10$ N, less than the weight because the mass accelerates.

    Review: 5.6 Newton’s Second Law in Rotational Form

Unit 6: Energy and Momentum of Rotating Systems

1. Topic 6.1 · Work it outA uniform disk of mass 10 kg and radius 0.4 m ($I=\frac{1}{2}M{R}^2$) spins at 30 rad/s about its stationary center. Find its rotational kinetic energy. Does it have kinetic energy even though its center is at rest?

   $I=\frac{1}{2}(10)(0.16)=0.8$ kg·m², so $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.8)(900)=360$ J. Yes. Every point off the axis is moving, so the disk has kinetic energy.

   Review: 6.1 Rotational Kinetic Energy

2. Topic 6.1 · Say it in contextA thin hoop and a uniform disk have the same mass and radius and spin at the same angular speed. Which has more rotational kinetic energy, and by what factor?

   The hoop. Its mass is all at radius $R$, so $I=M{R}^2$, while the disk has $I=\frac{1}{2}M{R}^2$. Since $K=\frac{1}{2}I{\omega }^2$, the hoop has twice the rotational kinetic energy.

   Review: 6.1 Rotational Kinetic Energy

3. Topic 6.2 · Work it outA constant 3 N·m torque turns a wheel through 4 rad. Then a different torque $\tau =2\theta$ N·m (θ in radians) acts from $\theta =0$ to $\theta =3$ rad. Find the work each does.

   Constant torque: $W=\tau \mathrm{\Delta }\theta =3\times 4=12$ J. Variable torque: $W={\int }_0^{3}2\theta d\theta ={\theta }^2{|}_0^3=9$ J, the area under the torque-angle graph.

   Review: 6.2 Torque and Work

4. Topic 6.3 · Work it outA 2 kg particle moves in a straight line at 3 m/s along the line $y=4$ m, parallel to the $x$ axis. Find its angular momentum about the origin. Does it change as the particle moves?

   $L=mvr\sin \theta =mv{r}_{⟂}=2\times 3\times 4=24$ kg·m²/s. It does not change: no torque acts, and $r_{⟂}=4$ m for every point on the line.

   Review: 6.3 Angular Momentum and Angular Impulse

5. Topic 6.3 · Work it outA torque $\tau =6t$ N·m acts on a wheel with $I=3$ kg·m² that starts at rest. Find the angular impulse over the first 2 s and the final angular speed.

   Angular impulse is ${\int }_0^{2}6tdt=3t^2{|}_0^2=12$ N·m·s. It equals the change in angular momentum, $I\omega =12$, so $\omega =12/3=4$ rad/s.

   Review: 6.3 Angular Momentum and Angular Impulse

6. Topic 6.4 · Work it outA skater spins at 2 rad/s with $I=5$ kg·m². She pulls in her arms, so $I=2$ kg·m². Find her new angular speed and the change in her kinetic energy. Where does the energy come from?

   $L=I\omega$ is constant: $5(2)=2\omega$, so $\omega =5$ rad/s. $K$ rises from $\frac{1}{2}(5)(4)=10$ J to $\frac{1}{2}(2)(25)=25$ J, a gain of 15 J. Her muscles do work pulling the arms inward.

   Review: 6.4 Conservation of Angular Momentum

7. Topic 6.4 · Check the conditionsA horizontal disk ($I=0.5$ kg·m²) turns at 6 rad/s on a fixed vertical axle. A 1 kg lump of clay dropped straight down sticks to the rim, 0.5 m from the axle. Why can you conserve angular momentum, and what is the new angular speed?

   The axle force acts through the axis and gravity is parallel to it, so the net external torque about the axle is zero, and the falling clay starts with no angular momentum about it. Then $0.5(6)=(0.5+1\times {0.5}^2)\omega$, giving $\omega =4$ rad/s. Mechanical energy is not conserved.

   Review: 6.4 Conservation of Angular Momentum

8. Topic 6.5 · Work it outA solid sphere ($I=\frac{2}{5}M{R}^2$) rolls from rest without slipping down a ramp 1.75 m high. Use $g=10$ m/s². Find its speed at the bottom and compare it with a block sliding down a frictionless ramp of the same height.

   $Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{7}{10}M{v}^2$, so $v=\sqrt{2(10)(1.75)/1.4}=5$ m/s. The block has all its energy as translation: $\sqrt{35}\approx 5.9$ m/s. The sphere is slower.

   Review: 6.5 Rolling

9. Topic 6.5 · Say it in contextA bowling ball is released so that it slides along the lane before it begins to roll. Does friction dissipate mechanical energy while it slips? Does it once the ball rolls without slipping?

   While it slips, the contact point moves relative to the lane, so kinetic friction dissipates energy. Once it rolls without slipping, the contact point is momentarily at rest, so friction dissipates none in the ideal case and $v_{cm}=r\omega$ holds.

   Review: 6.5 Rolling

10. Topic 6.6 · Say it in contextA satellite of mass $m$ circles a planet of mass $M$ at radius $r$. Write its kinetic energy, the system's potential energy and the total energy, and say what the sign of the total tells you.

    $K=\frac{GMm}{2r}$, $U=-\frac{GMm}{r}$ (zero at infinite separation) and $E=K+U=-\frac{GMm}{2r}$. The total is negative, so the satellite is bound and cannot escape unless it gains energy.

    Review: 6.6 Motion of Orbiting Satellites

11. Topic 6.6 · Work it outA planet's surface gravitational field is 10 N/kg and its radius is $6.4\times {10}^6$ m. Find the escape speed from its surface. Does it depend on the mass of the escaping object?

    Escape means total energy zero: $\frac{1}{2}m{v}^2=GMm/R$, so $v=\sqrt{2GM/R}=\sqrt{2gR}$ with $GM=g{R}^2$. This gives $\sqrt{2\times 10\times 6.4\times {10}^6}\approx 1.1\times {10}^4$ m/s. The object's mass cancels.

    Review: 6.6 Motion of Orbiting Satellites

12. Topic 6.6 · Which one fitsA satellite moves on an ellipse around a planet. Which of these stay constant: its speed, its angular momentum about the planet, the total mechanical energy of the system?

    The angular momentum and the total mechanical energy are constant, because gravity exerts no torque about the planet and no energy leaves the system. The speed and kinetic energy change, and the satellite is fastest at its closest point.

    Review: 6.6 Motion of Orbiting Satellites

Unit 7: Oscillations

1. Topic 7.1 · Which one fitsFor each net force along $x$, decide whether the motion is SHM and where it oscillates about: $F=-3x$, $F=-3x^3$, and $F=-3x+5$.

   $F=-3x$ is SHM about $x=0$. $F=-3x^3$ makes the object oscillate about $x=0$, but the force is not proportional to displacement, so it is not SHM. $F=-3x+5=-3(x-5/3)$ is SHM about $x=5/3$, where the net force is zero.

   Review: 7.1 Defining Simple Harmonic Motion (SHM)

2. Topic 7.2 · Work it outA 0.5 kg block on a spring with $k=50$ N/m oscillates. Find its period and frequency. Does either change if the amplitude is doubled, or if the system is moved to the Moon?

   $T=2\pi \sqrt{m/k}=2\pi \sqrt{0.5/50}\approx 0.63$ s, so $f\approx 1.6$ Hz. Neither depends on amplitude, and neither depends on $g$, so the Moon changes nothing for a horizontal or vertical spring oscillator.

   Review: 7.2 Frequency and Period of SHM

3. Topic 7.2 · Work it outA simple pendulum has a period of 2.0 s where $g=10$ m/s², for small angles. Find its length. What is the new period if the length is quadrupled?

   $T=2\pi \sqrt{\ell /g}$ gives $\ell =g{T}^2/(4{\pi }^2)=10(4)/39.5\approx 1.0$ m. Since $T\propto \sqrt{\ell }$, four times the length doubles the period to 4.0 s. The bob's mass is irrelevant.

   Review: 7.2 Frequency and Period of SHM

4. Topic 7.3 · Say it in contextA mass on a spring moves as $x=A\cos (\omega t)$. Where in the motion are the speed and the magnitude of the acceleration each at a maximum, and how large are they?

   Speed is maximum at $x=0$: $v_{max}=A\omega$, where the acceleration is zero. Acceleration is maximum at $x=\pm A$: $a_{max}=A{\omega }^2$, directed toward the center, where the speed is zero. They never peak together.

   Review: 7.3 Representing and Analyzing SHM

5. Topic 7.3 · Say it in contextA child on a swing is pushed by a steady rhythm. When does the amplitude grow the most, and why does pushing at a much different frequency work poorly?

   Amplitude grows most when the pushes come at the swing's natural frequency, called resonance. Pushes at other frequencies are out of step with the motion, so they sometimes oppose it and transfer much less energy.

   Review: 7.3 Representing and Analyzing SHM

6. Topic 7.3 · Work it outA mass moves as $x(t)=0.2\cos (5\pi t)$ meters. Find the amplitude, period, maximum speed and maximum acceleration.

   $A=0.2$ m and $\omega =5\pi$ rad/s, so $T=2\pi /\omega =0.4$ s. Then $v_{max}=A\omega =\pi \approx 3.1$ m/s and $a_{max}=A{\omega }^2=5{\pi }^2\approx 49$ m/s².

   Review: 7.3 Representing and Analyzing SHM

7. Topic 7.4 · Work it outA 0.4 kg mass on a frictionless spring ($k=100$ N/m) is released from rest at $x=0.1$ m. Find the total energy, the maximum speed, and the speed at $x=0.05$ m.

   $E=\frac{1}{2}k{A}^2=0.5$ J. At $x=0$ it is all kinetic: $v_{max}=\sqrt{2(0.5)/0.4}\approx 1.6$ m/s. At $x=0.05$ m, $U=0.125$ J, so $K=0.375$ J and $v=\sqrt{2(0.375)/0.4}\approx 1.4$ m/s.

   Review: 7.4 Energy of Simple Harmonic Oscillators

8. Topic 7.4 · Say it in contextThe amplitude of an oscillating mass on a spring is doubled. What happens to its period, its maximum speed and its total energy?

   The period does not change. The maximum speed doubles, since $v_{max}=A\omega$. The total energy, $\frac{1}{2}k{A}^2$, becomes four times as large.

   Review: 7.4 Energy of Simple Harmonic Oscillators

9. Topic 7.5 · Work it outA uniform thin rod 1.2 m long is pivoted at one end and swings with a small amplitude. Use $g=10$ m/s² and $I=\frac{1}{3}M{L}^2$ about the end. Find its period.

   The center of mass is $d=L/2$ from the pivot, so $T=2\pi \sqrt{\frac{I}{Mgd}}=2\pi \sqrt{\frac{2L}{3g}}=2\pi \sqrt{\frac{2(1.2)}{30}}\approx 1.8$ s.

   Review: 7.5 Simple and Physical Pendulums

10. Topic 7.5 · Which one fitsName the restoring torque in a simple pendulum, a physical pendulum and a torsion pendulum, and say which of them need the small-angle approximation to be SHM.

    Simple and physical: gravity on the center of mass gives $\tau =-mgd\sin \theta$, which is $\approx -mgd\theta$ only for small angles. Torsion: the twisted wire gives $\tau =-k\mathrm{\Delta }\theta$ already proportional to the angle, so no approximation is needed.

    Review: 7.5 Simple and Physical Pendulums