Torque and Rotation

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Review: Inertia – Measure of how difficult it is to an object's

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Moment of Inertia

Both barbells have the same mass, why do they have a DIFFERENT moment of inertia?

Barbell Comparison

The inertia of rotating objects depends on the object's mass AND the way that the mass is around the

Mass Distribution

Which would have a greater moment of inertia with a vertical axis of rotation?

Mass Comparison
Teacher Workspace: Moment of Inertia Concepts
Same mass, but in one barbell the mass sits farther from the axis of rotation.
$I = \sum mr^{2}$: mass far from the axis counts with a bigger $r^{2}$.
Larger moment of inertia: the one with the masses farthest from the vertical axis (the masses at the ends).

You won't have to memorize the moment of inertia equations for any shape or arrangement. It will be given.

Torque and Moment of Inertia

Torque = $\Sigma \tau = I\alpha$    [2nd Law]

$\alpha$       $I$  

$\tau = F\sin\theta[r]$

Ex 1) A string is wrapped around a 2.0 kg cylinder with a .10 m radius. A force of 20. N is applied down on the string. Find the angular acceleration of the cylinder.

The moment of inertia of a cylinder is $\frac{1}{2}mR^{2}$

Teacher Workspace: Ex 1 – String on a Cylinder
$I = \frac{1}{2}mR^{2} = \frac{1}{2}(2.0\text{ kg})(0.10\text{ m})^{2} = 0.010\text{ kg·m}^{2}$
$\tau = F\sin\theta\, r = (20\text{ N})(\sin 90^{\circ})(0.10\text{ m}) = 2.0\text{ N·m}$
$\Sigma\tau = I\alpha \Rightarrow \alpha = \frac{2.0}{0.010}$
$\alpha = 200\text{ rad/s}^{2}$

Ex 2) If a 2.0 kg is hung from the same cylinder, what would its acceleration be?

Do the FBD

2.0 kg block for the FBD
Teacher Workspace: Ex 2 – Hanging Mass
Block: $mg - T = ma$
Cylinder: $TR = I\alpha = \frac{1}{2}MR^{2}\left(\frac{a}{R}\right) \Rightarrow T = \frac{1}{2}Ma$
$mg = \left(m + \frac{1}{2}M\right)a \Rightarrow (2.0)(10) = (2.0 + 1.0)a$
$a = 6.7\text{ m/s}^{2}$   (and $T = \frac{1}{2}Ma = 6.7\text{ N}$)
Module 1: Teacher CER Key
EX 1 - AP DERIVATION: 1. $I = \frac{1}{2}MR^{2} = \frac{1}{2}(2.0\text{ kg})(0.10\text{ m})^{2} = 0.01\text{ kg}\cdot\text{m}^{2}$.
2. $\tau = Fr\sin(90^{\circ}) = (20\text{ N})(0.10\text{ m}) = 2.0\text{ N}\cdot\text{m}$.
3. $\alpha = \tau/I = 2.0/0.01 = \mathbf{200\text{ rad/s}^{2}}$.
EX 2 - AP DERIVATION: 1. Cylinder: $TR = I\alpha = (\frac{1}{2}MR^{2})(a/R) \implies T = \frac{1}{2}Ma$.
2. Hanging Mass: $mg - T = ma \implies mg - \frac{1}{2}Ma = ma$.
3. Solve for $a$: $mg = a(m + M/2) \implies a = \frac{(2)(10)}{2 + 1} = \mathbf{6.67\text{ m/s}^{2}}$.

Quick Check

3 questions on Moment of Inertia — pick an answer to check yourself.

1. Two barbells have the same total mass. In barbell A the weights sit near the center; in barbell B they sit at the ends. Spun about the center, which has the larger moment of inertia?

2. A string wrapped around a 2.0 kg solid cylinder (radius 0.10 m, I = ½mR²) is pulled with 20 N. What is the cylinder's angular acceleration?

3. A 2.0 kg mass hangs from a string wrapped around a 2.0 kg solid cylinder (I = ½MR²) on a fixed axle. What is the hanging mass's acceleration? (g = 10 m/s²)

Rotational Inertia 2

I. Rotational Inertia AKA Moment of Inertia

Inertia A Inertia B

Symbol for the moment of inertia

Common Moments of Inertia (You will be given)

Equation: $I = \sum mr^{2}$    m - r -

Ex 1)

a) Find the moment of inertia of two 5.0 kg bowling balls joined by a meter-long rod of negligible mass when rotated about the center of the rod.

Teacher Workspace: Ex 1 a) – About the Center
Each ball is $r = 0.50\text{ m}$ from the axis
$I = \sum mr^{2} = 2(5.0)(0.50)^{2}$
$I = 2.5\text{ kg·m}^{2}$

b) Find the moment of inertia of the object when it is rotated about one of the masses.

Bowling Rod
Teacher Workspace: Ex 1 b) – About One Ball
That ball has $r = 0$; the other ball is $r = 1.0\text{ m}$ away
$I = (5.0)(1.0)^{2} = 5.0\text{ kg·m}^{2}$

Ex 2) Two masses of 3.1 kg and 4.6 kg are attached to either end of a thin, light rod (assume massless) of length 1.8 m. Compute the moment of inertia for:

  1. The rod is rotated about its midpoint.
    2. Teacher Workspace: Ex 2 (1) – Midpoint
    Both masses at $r = 0.90\text{ m}$: $I = (3.1 + 4.6)(0.90)^{2}$
    $I = 6.2\text{ kg·m}^{2}$
  2. The rod is rotated at a point 0.30 m from the 3.1 kg mass.
    3. Teacher Workspace: Ex 2 (2) – 0.30 m from the 3.1 kg Mass
    $r_{3.1} = 0.30\text{ m}$, $r_{4.6} = 1.5\text{ m}$
    $I = (3.1)(0.30)^{2} + (4.6)(1.5)^{2}$
    $I = 0.28 + 10.4 = 10.6\text{ kg·m}^{2}$
  3. The rod is rotated about the 4.6 kg mass.
    4. Teacher Workspace: Ex 2 (3) – About the 4.6 kg Mass
    Only the 3.1 kg mass is away from the axis: $r = 1.8\text{ m}$
    $I = (3.1)(1.8)^{2} = 10.\text{ kg·m}^{2}$

II. Newton's 2nd Law for Rotation

$\tau_{\text{net}}$   $I$   $\alpha$

Moment of Inertia Race: every shape has the same mass and radius, and every string pulls on the same hub. Use the I equations to predict which weight lands 1st, 2nd, 3rd… then release them.

  1. What torque needs to be applied to an antique sewing machine spinning wheel of radius 0.28 m and mass 3.1 kg (model it as a hoop, with $I = MR^{2}$) to give it an angular acceleration of 4.8 rad/s$^{2}$?
    2. Teacher Workspace: Sewing Machine Wheel
    $I = MR^{2} = (3.1)(0.28)^{2} = 0.243\text{ kg·m}^{2}$
    $\tau = I\alpha = (0.243)(4.8)$
    $\tau = 1.2\text{ N·m}$
  2. What is the angular acceleration of a 75 g lug nut when a lug wrench applies a 135 N-m torque to it? Model the lug nut as a hollow cylinder of inner radius 0.85 cm and outer radius 1.0 cm ($I = \frac{1}{2}M(r_{1}^{2} + r_{2}^{2})$). What is the tangential acceleration at the outer surface? Why are these numbers so high – what factor was not considered?
    3. Teacher Workspace: Lug Nut
    $I = \frac{1}{2}M(r_{1}^{2} + r_{2}^{2}) = \frac{1}{2}(0.075)(0.0085^{2} + 0.010^{2}) = 6.5\times10^{-6}\text{ kg·m}^{2}$
    $\alpha = \frac{\tau}{I} = \frac{135}{6.5\times10^{-6}}$
    $\alpha = 2.1\times10^{7}\text{ rad/s}^{2}$
    $a_{t} = \alpha r = (2.1\times10^{7})(0.010)$
    $a_{t} = 2.1\times10^{5}\text{ m/s}^{2}$
    So high because the friction from the threads of the bolt (opposing torque) was ignored.
  3. A baseball player swings a bat, accelerating it uniformly from rest to 4.2 revolutions/second in 0.25 s. Assume the bat is modeled as a uniform rod ($I = 1/3 ML^{2}$), and has m = 0.91 kg and is 0.86 m long. Find the torque applied by the player to the bat.
    4. Teacher Workspace: Baseball Bat
    $\omega = 4.2\text{ rev/s} \times 2\pi = 26.4\text{ rad/s}$,   $\alpha = \frac{\Delta\omega}{t} = \frac{26.4}{0.25} = 106\text{ rad/s}^{2}$
    $I = \frac{1}{3}ML^{2} = \frac{1}{3}(0.91)(0.86)^{2} = 0.224\text{ kg·m}^{2}$
    $\tau = I\alpha = (0.224)(106)$
    $\tau = 24\text{ N·m}$
  4. A large pulley of mass 5.21 kg (its mass cannot be neglected) is rotated by a constant tension force of 19.6 N in the counterclockwise direction. The rotation is resisted by the frictional torque of the axle on the pulley. The frictional torque is a constant 1.86 N-m in the clockwise direction. The pulley accelerates from 0 to 27.2 rad/s in 4.11 s. Find the moment of inertia of the pulley.
    5. Teacher Workspace: Pulley (not assigned)
  5. A baton twirler has a baton of length 0.42 m with masses of 0.53 kg at each end. Assume the rod itself is massless. The rod is first rotated about its midpoint. It is then rotated about one of its ends, and in both cases uniformly accelerates from 0 rad/s to 1.8 rad/s in 3.0 s.
    1. Find the torque exerted by the twirler on the baton when it is rotated about its middle.
      2. Teacher Workspace: Baton Twirler (a) – Middle
      $\alpha = \frac{\Delta\omega}{t} = \frac{1.8}{3.0} = 0.60\text{ rad/s}^{2}$
      $I = 2(0.53)(0.21)^{2} = 0.0467\text{ kg·m}^{2}$
      $\tau = I\alpha = (0.0467)(0.60) = 0.028\text{ N·m}$
    2. Find the torque exerted by the twirler on the baton when it is rotated about its end.
      3. Teacher Workspace: Baton Twirler (b) – End
      Same $\alpha = 0.60\text{ rad/s}^{2}$
      One mass at $r = 0.42\text{ m}$: $I = (0.53)(0.42)^{2} = 0.0935\text{ kg·m}^{2}$
      $\tau = I\alpha = (0.0935)(0.60) = 0.056\text{ N·m}$
  6. What torque needs to be applied to a hula hoop of radius 0.58 m and mass 2.5 kg (model it as a hoop, with $I = MR^{2}$) to give it an angular acceleration of 6.8 rad/s$^{2}$?
    7. Teacher Workspace: Hula Hoop
    $I = MR^{2} = (2.5)(0.58)^{2} = 0.841\text{ kg·m}^{2}$
    $\tau = I\alpha = (0.841)(6.8)$
    $\tau = 5.7\text{ N·m}$
  7. A cricket batsman swings his bat, accelerating it uniformly from rest to 17.3 rad/s in 0.21 s. Assume the bat is modeled as a flat plate ($I = 1/3 Mh^{2} + 1/2 Mw^{2}$), and has m = 1.36 kg, h=0.97 m, and w = 0.11 m. Find the torque applied by the batsman to the bat.
    8. Teacher Workspace: Cricket Bat (not assigned)
    $\alpha = \frac{17.3}{0.21} = 82.4\text{ rad/s}^{2}$
    $I = \frac{1}{3}(1.36)(0.97)^{2} + \frac{1}{2}(1.36)(0.11)^{2} = 0.435\text{ kg·m}^{2}$
    $\tau = I\alpha = 36\text{ N·m}$
  8. Two masses of 4.2 kg and 5.8 kg are attached to either end of a thin, light rod (assume massless) of length 2.4 m. Compute the moment of inertia for:
    1. The rod is rotated about its midpoint.
      2. Teacher Workspace: Two Masses (a) – Midpoint
      Both masses at $r = 1.2\text{ m}$: $I = (4.2 + 5.8)(1.2)^{2}$
      $I = 14\text{ kg·m}^{2}$
    2. The rod is rotated at a point 0.50 m from the 4.2 kg mass.
      3. Teacher Workspace: Two Masses (b) – 0.50 m from the 4.2 kg Mass
      $r_{4.2} = 0.50\text{ m}$, $r_{5.8} = 1.9\text{ m}$
      $I = (4.2)(0.50)^{2} + (5.8)(1.9)^{2}$
      $I = 1.05 + 20.9 = 22\text{ kg·m}^{2}$
    3. The rod is rotated about the end where the 5.8 kg mass is located.
      4. Teacher Workspace: Two Masses (c) – About the 5.8 kg End
      Only the 4.2 kg mass is away from the axis: $r = 2.4\text{ m}$
      $I = (4.2)(2.4)^{2} = 24\text{ kg·m}^{2}$
  9. A large pulley of mass 6.91 kg (its mass cannot be neglected) is rotated by a constant torque of 22.3 Nm in the counterclockwise direction. The rotation is resisted by the frictional torque of the axle on the pulley. The frictional torque is a constant 2.12 N-m in the clockwise direction. The pulley accelerates from 0 to 31.2 rad/s in 5.63 s. Find the moment of inertia of the pulley.
    10. Teacher Workspace: Pulley with Friction
    $\alpha = \frac{\Delta\omega}{t} = \frac{31.2}{5.63} = 5.54\text{ rad/s}^{2}$
    $\tau_{\text{net}} = 22.3 - 2.12 = 20.2\text{ N·m}$
    $I = \frac{\tau_{\text{net}}}{\alpha} = \frac{20.2}{5.54}$
    $I = 3.6\text{ kg·m}^{2}$
  10. A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is first rotated about its midpoint. It is then rotated about one of its ends, and in both cases uniformly accelerates from 0 rad/s to 2.4 rad/s in 3.6 s.
    1. Find the torque exerted by the twirler on the baton when it is rotated about its middle.
      2. Teacher Workspace: Baton Twirler (a) – Middle
      $\alpha = \frac{2.4}{3.6} = 0.667\text{ rad/s}^{2}$
      $I = 2(0.48)(0.18)^{2} = 0.0311\text{ kg·m}^{2}$
      $\tau = I\alpha = (0.0311)(0.667) = 0.021\text{ N·m}$
    2. Find the torque exerted by the twirler on the baton when it is rotated about its end.
      3. Teacher Workspace: Baton Twirler (b) – End
      Same $\alpha = 0.667\text{ rad/s}^{2}$
      One mass at $r = 0.36\text{ m}$: $I = (0.48)(0.36)^{2} = 0.0622\text{ kg·m}^{2}$
      $\tau = I\alpha = (0.0622)(0.667) = 0.041\text{ N·m}$
Module 2: Teacher CER Key
LUG NUT (Q2) ANALYSIS: 1. $I = \frac{1}{2}(0.075\text{ kg})((0.0085\text{ m})^{2} + (0.01\text{ m})^{2}) = 6.46 \times 10^{-6}\text{ kg}\cdot\text{m}^{2}$.
2. $\alpha = \tau/I = 135\text{ Nm} / 6.46 \times 10^{-6} = \mathbf{2.09 \times 10^{7}\text{ rad/s}^{2}}$.
3. $a_{t} = \alpha R = (2.09 \times 10^{7})(0.01) = \mathbf{2.09 \times 10^{5}\text{ m/s}^{2}}$.
Reasoning: These values are high because the model ignores the massive wrench's inertia and thread friction.

Quick Check

3 questions on Rotational Inertia II — pick an answer to check yourself.

1. Two 5.0 kg balls are joined by a 1.0 m massless rod. What is the moment of inertia about the rod's center?

2. Two equal masses on a light rod are moved so each is twice as far from the rotation axis. The moment of inertia becomes…

3. What torque gives a hoop (I = MR²) of mass 2.0 kg and radius 0.50 m an angular acceleration of 4.0 rad/s²?

Advanced Rotation Problems

  1. A very light cotton tape is wrapped around the outside surface of a uniform cylinder of mass M and radius R. The free end of the tape is attached to the ceiling. The cylinder is released from rest and as it descends it unravels from the tape without slipping. The moment of inertia of the cylinder about its center is $I = \frac{1}{2}MR^{2}$.
    Cylinder of mass M and radius R hanging from a tape Circle for the free body diagram
    1. On the circle above, shows all the forces applied on the cylinder.
      2. Teacher Workspace: Tape (1) – Forces on the Cylinder
      Weight $Mg$ down at the center; tension $T$ up at the left edge (a distance $R$ from the center).
    2. Find the acceleration of the center of the cylinder when it moves down.
      3. Teacher Workspace: Tape (2) – Acceleration
      Translation: $Mg - T = Ma$
      Rotation: $TR = \frac{1}{2}MR^{2}\left(\frac{a}{R}\right) \Rightarrow T = \frac{1}{2}Ma$
      $Mg - \frac{1}{2}Ma = Ma$
      $a = \frac{2}{3}g$
    3. Find the tension force in the tape.
      4. Teacher Workspace: Tape (3) – Tension
      $T = \frac{1}{2}Ma$ with $a = \frac{2}{3}g$
      $T = \frac{1}{3}Mg$
  2. A uniform cylinder of mass M and radius R is fixed on a frictionless axle at point C. A block of mass m is suspended from a light cord wrapped around the cylinder and released from rest at time t = 0. The moment of inertia of the cylinder is $I = \frac{1}{2}MR^{2}$.
    Video for Questions 2 - 3
    Cylinder of mass M and radius R on a fixed axle at C with a hanging block of mass m Circle and square for the free body diagrams
    1. On the circle and the square above, show all the applied forces on the cylinder and the block.
      2. Teacher Workspace: Axle and Block (1) – Forces
      Block: $T$ up, $mg$ down.   Cylinder: $Mg$ down, $T$ down at the right edge, axle force up at C.
    2. Find the acceleration of the block as it moves down.
      3. Teacher Workspace: Axle and Block (2) – Acceleration
      Block: $mg - T = ma$
      Cylinder: $TR = \frac{1}{2}MR^{2}\left(\frac{a}{R}\right) \Rightarrow T = \frac{1}{2}Ma$
      $a = \frac{mg}{m + \frac{1}{2}M} = \frac{2mg}{2m + M}$
    3. Find the tension in the lower cord.
      4. Teacher Workspace: Axle and Block (3) – Lower Cord Tension
      $T = \frac{1}{2}Ma$
      $T = \frac{mMg}{2m + M}$
    4. Express the angular momentum of the cylinder as a function of time t.
  3. A block, A of mass M, is suspended from a light string that passes over a pulley and is connected to block B of mass 2M. Block B sits on the surface of a smooth table. Block C, of mass 3M, sits on the top of block B. The surface between block C and block B is not frictionless. When the system of three blocks is released from rest, block A accelerates downward with a constant acceleration, a, and the two blocks on the table move relative to each other. The moment of inertia of the pulley is $I = 1.5 MR^{2}$. Present all results in terms of M, g, and a.
    Multi-Block
    1. Find the tension force in the vertical section of the string.
      2. Teacher Workspace: Three Blocks (1) – Vertical String
      Block A: $Mg - T_{1} = Ma$
      $T_{1} = M(g - a)$
    2. Find the tension force in the horizontal section of the string.
      3. Teacher Workspace: Three Blocks (2) – Horizontal String
      Pulley: $(T_{1} - T_{2})R = I\alpha = 1.5MR^{2}\left(\frac{a}{R}\right)$
      $T_{1} - T_{2} = 1.5Ma$
      $T_{2} = M(g - a) - 1.5Ma = M(g - 2.5a)$
  4. A pulley of radius, R, and moment of inertia, $I = 2 MR^{2}$ is mounted on an axle with negligible friction. Block A, with a mass M, and Block B, with a mass of 3M, are attached to a light string that passes over the pulley. Assuming that the string doesn't slip on the pulley, answer the following questions in terms of M, R, and fundamental constants.
    Atwood Pulley
    1. What is the acceleration of the two blocks?
      2. Teacher Workspace: Atwood (1) – Acceleration
      A (left, $M$) moves up; B (right, $3M$) moves down.
      $3Mg - Mg = \left(M + 3M + \frac{I}{R^{2}}\right)a = 6Ma$
      $a = \frac{g}{3}$
    2. What is the tension force in the left section of the string?
      3. Teacher Workspace: Atwood (2) – Left Tension
      Block A: $T_{A} - Mg = Ma$
      $T_{A} = M(g + a) = \frac{4}{3}Mg$
    3. What is the tension force in the right section of the string?
      4. Teacher Workspace: Atwood (3) – Right Tension
      Block B: $3Mg - T_{B} = 3Ma$
      $T_{B} = 3M(g - a) = 2Mg$
    4. What is the angular acceleration of the pulley?
      5. Teacher Workspace: Atwood (4) – Angular Acceleration
      No slipping: $a = \alpha R$
      $\alpha = \frac{a}{R} = \frac{g}{3R}$
Module 3: Advanced CER Key
UNRAVELING CYLINDER: 1. Net Force: $Mg - T = Ma$.
2. Net Torque: $TR = I\alpha = (\frac{1}{2}MR^{2})(a/R) \implies T = \frac{1}{2}Ma$.
3. Substitute $T$: $Mg - \frac{1}{2}Ma = Ma \implies Mg = \frac{3}{2}Ma \implies \mathbf{a = \frac{2}{3}g}$.
4. Tension: $T = \frac{1}{2}M(\frac{2}{3}g) = \mathbf{\frac{1}{3}Mg}$.

Quick Check

3 questions on Advanced Rotation — pick an answer to check yourself.

1. A uniform cylinder (I = ½MR²) unwinds from a tape tied to the ceiling as it falls, without slipping. Its downward acceleration is…

2. For that same unwinding cylinder (I = ½MR²), the tape tension is Mg/3. What is the tension if M = 3.0 kg? (g = 10 m/s²)

3. A block hangs from a cord wrapped around a cylinder on a frictionless fixed axle. As the block accelerates downward, the cord tension is…

Ramp Problems

Cylinder of radius R at the top of a ramp of height h and angle theta
  1. A solid uniform sphere of mass, M, and radius, R is placed on an inclined plane at a distance, h from the base of the incline. The inclined plane makes an angle, $\theta$ with the horizontal. The sphere is released from rest and rolls down the incline without slipping. The moment of inertia of the sphere is $I = \frac{2}{5} MR^{2}$.
    1. Determine the translational kinetic energy of the sphere when it reaches the bottom of the inclined plane.
      2. Teacher Workspace: Sphere (1) – Translational KE
      $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}\left(\frac{2}{5}MR^{2}\right)\left(\frac{v}{R}\right)^{2} = \frac{7}{10}Mv^{2}$
      $KE_{\text{trans}} = \frac{1}{2}Mv^{2} = \frac{5}{7}Mgh$
    2. Determine the rotational kinetic energy of the sphere when it reaches the bottom of the inclined plane.
      3. Teacher Workspace: Sphere (2) – Rotational KE
      $KE_{\text{rot}} = \frac{1}{5}Mv^{2}$   (or $Mgh - \frac{5}{7}Mgh$)
      $KE_{\text{rot}} = \frac{2}{7}Mgh$
  2. A solid uniform cylinder of mass M and radius R is placed on an inclined plane at a distance h from the base of the incline. The inclined plane makes an angle $\theta$ with the horizontal. The cylinder is released from rest and rolls down the incline without slipping. The moment of inertia of the cylinder is $I = \frac{1}{2} MR^{2}$.
    Cylinder of radius R at the top of a ramp of height h and angle theta
    1. Determine the translational kinetic energy of the cylinder when it reaches the bottom of the inclined plane.
      2. Teacher Workspace: Cylinder (1) – Translational KE
      $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}\left(\frac{1}{2}MR^{2}\right)\left(\frac{v}{R}\right)^{2} = \frac{3}{4}Mv^{2}$
      $KE_{\text{trans}} = \frac{1}{2}Mv^{2} = \frac{2}{3}Mgh$
    2. Determine the rotational kinetic energy of the cylinder when it reaches the bottom of the inclined plane.
      3. Teacher Workspace: Cylinder (2) – Rotational KE
      $KE_{\text{rot}} = \frac{1}{4}Mv^{2}$   (or $Mgh - \frac{2}{3}Mgh$)
      $KE_{\text{rot}} = \frac{1}{3}Mgh$
  3. In a physics experiment, students made a lab cart of a wooden block of mass 5m and four wheels each of mass, m, and radius r. The moment of inertia of each wheel is $I = \frac{1}{2} mr^{2}$. The cart is released from rest and rolls without slipping from the top of an inclined plane of height h. After the cart reaches the bottom of the inclined plane, it collides with an elastic spring with negligible mass and a spring constant k.
    Lab Cart
    1. Determine the moment of inertia of all four wheels.
      2. Teacher Workspace: Lab Cart (1) – Moment of Inertia
      Each wheel: $\frac{1}{2}mr^{2}$; four wheels: $4 \times \frac{1}{2}mr^{2}$
      $I_{\text{wheels}} = 2mr^{2}$
    2. Determine the speed of the cart at the bottom of the inclined plane.
      3. Teacher Workspace: Lab Cart (2) – Speed at the Bottom
      Total mass $= 5m + 4m = 9m$
      $9mgh = \frac{1}{2}(9m)v^{2} + \frac{1}{2}(2mr^{2})\left(\frac{v}{r}\right)^{2} = 5.5mv^{2}$
      $v = \sqrt{\frac{18gh}{11}}$
Module 4: Ramp CER Key
SPHERE ROLLING: Energy Conservation: $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}$.
For Sphere ($I = \frac{2}{5}MR^{2}$): $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{5}Mv^{2} = \frac{7}{10}Mv^{2}$.
1. Translational KE: $\frac{1}{2}Mv^{2} = \frac{5}{10}Mv^{2} = \mathbf{\frac{5}{7}Mgh}$.
2. Rotational KE: $\frac{1}{5}Mv^{2} = \frac{2}{10}Mv^{2} = \mathbf{\frac{2}{7}Mgh}$.

Quick Check

3 questions on Ramp Problems — pick an answer to check yourself.

1. A solid sphere (I = ⅖MR²) rolls without slipping from rest down a height h. At the bottom, its translational kinetic energy is…

2. A solid cylinder (I = ½MR²) rolls without slipping from rest down a ramp of height 2.7 m. Its speed at the bottom? (g = 10 m/s²)

3. A solid sphere and a solid cylinder of different masses and radii roll without slipping from rest down the same ramp. Which reaches the bottom first?

Rotational KE - Energy of Rotation

KE = $\frac{1}{2} I \omega^{2}$
YouTube Video 1 - 4

$I$ = $\omega$ =

If an object has translational and rotational KE, Its total KE is ….
KE = $\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}$

  1. A solid cylinder ($I = \frac{1}{2}MR^{2}$) of mass 0.56 kg and radius 0.042 m rolls, without slipping, down an incline of height 0.67 m. What is the speed of the cylinder at the bottom of the incline? Does its speed depend on the mass and radius of the cylinder?
    2. Teacher Workspace: Solid Cylinder on an Incline
    $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{4}Mv^{2} = \frac{3}{4}Mv^{2} \Rightarrow v = \sqrt{\frac{4gh}{3}}$
    $v = \sqrt{\frac{4(10)(0.67)}{3}}$
    $v = 3.0\text{ m/s}$
    No – $M$ and $R$ cancel, so the speed does not depend on mass or radius.
  2. Two uniform spheres ($I = \frac{2}{5}MR^{2}$) roll, without slipping, down an incline of height 0.72 m. Sphere 1 has a mass of 1.1 kg and a radius of 0.18 m and Sphere 2 has a mass of 1.8 kg and a radius of 0.14 m. Which sphere gets to the bottom of the incline quicker? What is the velocity of each sphere?
    3. Teacher Workspace: Two Spheres on an Incline
    Any solid sphere: $Mgh = \frac{7}{10}Mv^{2} \Rightarrow v = \sqrt{\frac{10gh}{7}}$ (no $M$ or $R$)
    $v = \sqrt{\frac{10(10)(0.72)}{7}}$
    Both: $v = 3.2\text{ m/s}$ – they reach the bottom at the same time (a tie).
  3. What is the rotational kinetic energy of a 0.82 kg sphere ($I = \frac{2}{5} MR^{2}$), with a radius of 0.058 m, rolling with an angular velocity of 5.2 rad/s?
    4. Teacher Workspace: Rotational KE of a Sphere
    $I = \frac{2}{5}MR^{2} = \frac{2}{5}(0.82)(0.058)^{2} = 1.10\times10^{-3}\text{ kg·m}^{2}$
    $KE_{\text{rot}} = \frac{1}{2}I\omega^{2} = \frac{1}{2}(1.10\times10^{-3})(5.2)^{2}$
    $KE_{\text{rot}} = 0.015\text{ J}$
  4. A 0.36 kg, 0.11 m radius, thin hoop ($I = MR^{2}$) is rotating, without slipping, while moving linearly with an angular velocity of 4.8 rad/s along a path. What is its total kinetic energy (translational plus rotational)?
    5. Teacher Workspace: Rolling Hoop – Total KE
    $v = \omega R = (4.8)(0.11) = 0.528\text{ m/s}$
    $KE = \frac{1}{2}Mv^{2} + \frac{1}{2}(MR^{2})\omega^{2} = Mv^{2} = (0.36)(0.528)^{2}$
    $KE = 0.10\text{ J}$
  5. A solid cylinder rolls down a hill without slipping. How much work does the frictional force between the hill and the cylinder do on the cylinder as it is rolling? Why?
    6. Teacher Workspace: Friction on a Rolling Cylinder (not assigned)
    Zero work: rolling without slipping means static friction acts at the contact point, which is momentarily at rest, so friction does no work.
  6. How much work is required to uniformly slow a merry-go-round of mass 1850 kg and a radius of 8.30 m from a rotational rate of 1 revolution per 7.40 s to a stop? Model the merry-go-round as a solid cylinder ($I = \frac{1}{2} MR^{2}$). If the merry-go-round is stopped in 7.40 s, what power is exerted?
    7. Teacher Workspace: Merry-Go-Round
    $\omega = \frac{2\pi}{7.40\text{ s}} = 0.849\text{ rad/s}$
    $I = \frac{1}{2}MR^{2} = \frac{1}{2}(1850)(8.30)^{2} = 6.37\times10^{4}\text{ kg·m}^{2}$
    $W = \Delta KE = \frac{1}{2}I\omega^{2} = \frac{1}{2}(6.37\times10^{4})(0.849)^{2}$
    $W = 2.3\times10^{4}\text{ J}$ (removed from the merry-go-round)
    $P = \frac{W}{t} = \frac{2.3\times10^{4}}{7.40} = 3.1\times10^{3}\text{ W}$
  7. A thin hoop ($I = MR^{2}$) of mass 0.56 kg and radius 0.042 m rolls, without slipping, down an incline of height 0.67 m. What is the speed of the hoop at the bottom of the incline? Does its speed depend on the mass and radius of the hoop?
    8. Teacher Workspace: Hoop on an Incline
    $Mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}(MR^{2})\left(\frac{v}{R}\right)^{2} = Mv^{2} \Rightarrow v = \sqrt{gh}$
    $v = \sqrt{(10)(0.67)}$
    $v = 2.6\text{ m/s}$
    No – $M$ and $R$ cancel.
  8. Two solid cylinders ($I = \frac{1}{2}MR^{2}$) roll, without slipping, down an incline of height 0.85 m. Sphere 1 has a mass of 1.1 kg and a radius of 0.11 m, and Sphere 2 has a mass of 2.1 kg and a radius of 0.14 m. Which cylinder gets to the bottom of the incline quicker? What is the velocity of each cylinder?
    9. Teacher Workspace: Two Cylinders on an Incline
    Any solid cylinder: $v = \sqrt{\frac{4gh}{3}}$ (no $M$ or $R$)
    $v = \sqrt{\frac{4(10)(0.85)}{3}}$
    Both: $v = 3.4\text{ m/s}$ – they tie.
  9. What is the rotational kinetic energy of a 1.3 kg solid cylinder ($I = \frac{1}{2} MR^{2}$), with a radius of 0.043 m, rolling with an angular velocity of 4.9 rad/s?
    10. Teacher Workspace: Rotational KE of a Cylinder
    $I = \frac{1}{2}MR^{2} = \frac{1}{2}(1.3)(0.043)^{2} = 1.20\times10^{-3}\text{ kg·m}^{2}$
    $KE_{\text{rot}} = \frac{1}{2}I\omega^{2} = \frac{1}{2}(1.20\times10^{-3})(4.9)^{2}$
    $KE_{\text{rot}} = 0.014\text{ J}$
  10. A 0.42 kg, 0.09 m radius sphere ($I = \frac{2}{5} MR^{2}$) is rotating, without slipping, while moving linearly with an angular velocity of 5.2 rad/s along a path. What is its total kinetic energy (translational plus rotational)?
    11. Teacher Workspace: Rolling Sphere – Total KE
    $v = \omega R = (5.2)(0.09) = 0.468\text{ m/s}$
    $KE = \frac{1}{2}Mv^{2} + \frac{1}{2}\left(\frac{2}{5}MR^{2}\right)\omega^{2} = \frac{7}{10}Mv^{2} = 0.7(0.42)(0.468)^{2}$
    $KE = 0.064\text{ J}$
Instructional Video 6 - 8
Module 5: Energy CER Key
CYLINDER SPEED: Total $KE = \frac{3}{4}Mv^{2}$. Energy Eq: $Mgh = \frac{3}{4}Mv^{2} \implies v = \sqrt{\frac{4}{3}gh}$.
Calculated $v = \sqrt{1.333 \times 10 \times 0.67} = \mathbf{2.99\text{ m/s}}$.
Reasoning: Mass and radius cancel out in the energy conservation equation, so final speed depends only on height and shape.

Quick Check

3 questions on Rotational KE — pick an answer to check yourself.

1. A solid cylinder rolls without slipping down an incline from rest. Its speed at the bottom depends on…

2. A solid cylinder (I = ½MR²) rolls from rest without slipping down an incline of height 0.60 m. What is its speed at the bottom? (g = 10 m/s²)

3. A solid cylinder and a thin hoop of the same mass and radius are released together and roll down the same incline. Which reaches the bottom first?

Angular Momentum

Figure skater with arms out and arms in: pulling mass toward the rotation axis reduces moment of inertia and increases angular speed. Why do skaters spin faster? Click to enlarge the illustrated explanation
When net external torque is negligible, angular momentum stays constant. Pulling the arms inward decreases moment of inertia, so angular speed increases.
Video questions 1 - 4

Equation: L = in terms of v,   L =

Units

Rotational Momentum is during interactions

  1. You spin a ball of mass 0.18 kg that is attached to a string of length 0.98 m at $\omega$ = 5.2 rad/s in a circle. What is the ball's angular momentum? Answer 0.90 kgm$^{2}$/s
    2. Teacher Workspace: Ball on a String
    $I = mr^{2} = (0.18)(0.98)^{2} = 0.173\text{ kg·m}^{2}$
    $L = I\omega = (0.173)(5.2)$
    $L = 0.90\text{ kg·m}^{2}\text{/s}$
  2. A student is standing, with her arms outstretched, on a platform that is rotating at 1.6 rev/s. She pulls her arms in and the platform now rotates at 2.2 rev/s. What is her final moment of inertia (I) in terms of her original moment of inertia ($I_{0}$)? Answer: I = 0.73$I_{0}$
    3. Teacher Workspace: Spinning Student
    $L$ is conserved: $I_{0}\omega_{0} = I\omega$
    $I = I_{0}\frac{\omega_{0}}{\omega} = I_{0}\frac{1.6}{2.2}$
    $I = 0.73I_{0}$
  3. An LP record is spinning on an old-fashioned record player with an angular velocity of $\omega$. The record changer drops an identical record on top of the spinning record. What is the new angular velocity of both records? Answer $\omega = \omega_{0}/2$
    4. Teacher Workspace: Dropped Record
    $I\omega_{0} = (I + I)\omega$
    $\omega = \frac{\omega_{0}}{2}$
  4. Calculate the angular momentum of a ballet dancer who is spinning at 1.5 rev/sec. Model the dancer as a cylinder ($I = \frac{1}{2} MR^{2}$) with a mass of 62 kg, a height of 1.6 m, and a radius of 0.16 m. Answer 7.5 kgm$^{2}$/s
    5. Teacher Workspace: Ballet Dancer
    $I = \frac{1}{2}MR^{2} = \frac{1}{2}(62)(0.16)^{2} = 0.794\text{ kg·m}^{2}$
    $\omega = 1.5 \times 2\pi = 9.42\text{ rad/s}$
    $L = I\omega = 7.5\text{ kg·m}^{2}\text{/s}$
  5. A student of mass 42 kg is standing at the center of a merry-go-round of radius 3.4 m and a moment of inertia of 840 kg-m$^{2}$ that is rotating at $\omega$ = 1.8 rad/s. The student walks to the outer edge of the merry-go-round. What is the angular velocity of the merry-go-round when he reaches the edge? Answer: 1.1 rad/s
    6. Teacher Workspace: Student Walks to the Edge
    At the center the student adds no inertia: $L = (840)(1.8)$
    At the edge: $I = 840 + (42)(3.4)^{2} = 1326\text{ kg·m}^{2}$
    $(840)(1.8) = (1326)\omega$
    $\omega = 1.1\text{ rad/s}$
  6. A ball of mass 0.14 kg attached to a string of length 0.64 m is spun in a circle with $\omega$ = 4.9 rad/s. What is the ball's angular momentum? Answer: 0.28 kgm$^{2}$/s
    7. Teacher Workspace: Ball on a String
    $I = mr^{2} = (0.14)(0.64)^{2} = 0.0573\text{ kg·m}^{2}$
    $L = I\omega = (0.0573)(4.9) = 0.28\text{ kg·m}^{2}\text{/s}$
  7. A platform is rotating at 2.2 rev/s and a student is standing in the middle of it with his arms at his side. He extends his arms straight out and the platform now rotates at 1.4 rev/s. What is his final moment of inertia (I) in terms of his original moment of inertia ($I_{0}$)? Answer: I = 1.6$I_{0}$
    8. Teacher Workspace: Arms Out on a Platform
    $I_{0}\omega_{0} = I\omega \Rightarrow I = I_{0}\frac{2.2}{1.4}$
    $I = 1.6I_{0}$
  8. A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 kg directly onto the middle of the wheel. The clay is in the shape of a pancake and has a radius of 0.19 m. Assume both the wheel and the chunk of clay can be modeled as solid cylinders ($I = \frac{1}{2} MR^{2}$). What is the new tangential velocity of the wheel and the clay? Answer: 1.8 m/s
    9. Teacher Workspace: Potter's Wheel (not assigned)
    $I_{w} = \frac{1}{2}(4.2)(0.35)^{2} = 0.257$,   $I_{c} = \frac{1}{2}(2.9)(0.19)^{2} = 0.0523\text{ kg·m}^{2}$
    $\omega_{0} = 0.98 \times 2\pi = 6.16\text{ rad/s}$;   $I_{w}\omega_{0} = (I_{w} + I_{c})\omega \Rightarrow \omega = 5.12\text{ rad/s}$
    Rim: $v = \omega R = (5.12)(0.35) = 1.8\text{ m/s}$
  9. What is the angular momentum of a roller skater who is spinning at 1.5 rev/seconds? Model the skater as a cylinder ($I = \frac{1}{2} MR^{2}$) with a mass of 81 kg, a height of 1.8 m, and a radius of 0.18 m. Answer: 12 kgm$^{2}$/s
    10. Teacher Workspace: Roller Skater
    $I = \frac{1}{2}MR^{2} = \frac{1}{2}(81)(0.18)^{2} = 1.31\text{ kg·m}^{2}$
    $\omega = 1.5 \times 2\pi = 9.42\text{ rad/s}$
    $L = I\omega = 12\text{ kg·m}^{2}\text{/s}$
  10. A student of mass 59 kg is standing at the edge of a merry-go-round of radius 4.2 m and a moment of inertia of 990 kg-m$^{2}$ that is rotating at $\omega$ = 2.1 rad/s. The student walks to the middle of the merry-go-round. What is the angular velocity of the merry-go-round when he reaches the middle? Answer: 4.3 rad/s
    11. Teacher Workspace: Student Walks to the Middle
    Start (edge): $I = 990 + (59)(4.2)^{2} = 2031\text{ kg·m}^{2}$
    End (middle): $I = 990\text{ kg·m}^{2}$
    $(2031)(2.1) = (990)\omega$
    $\omega = 4.3\text{ rad/s}$
YouTube Video Questions 5 - 9
Module 6: Momentum CER Key
CONSERVATION (Q2): $\Sigma L_{i} = \Sigma L_{f} \implies I_{0} \omega_{i} = I_{f} \omega_{f}$.
$I_{0} (1.6) = I_{f} (2.2) \implies I_{f} = \frac{1.6}{2.2} I_{0} = \mathbf{0.73I_{0}}$.
Reasoning: Moving mass inward reduces $I$, forcing $\omega$ to increase to keep $L$ constant.

Check Your Understanding

5 quick questions — pick an answer to see if you’re right.

1. The moment of inertia of an object depends on

2. Two barbells have equal mass. Which has the larger moment of inertia about its center?

3. Two 5.0 kg balls sit at the ends of a 1.0 m massless rod. What is the moment of inertia about the rod's center?

4. A 20. N force pulls a string wrapped on a cylinder (I = 0.010 kg·m², r = 0.10 m). What is its angular acceleration?

5. A 2.0 kg mass hangs from a string wrapped around a 2.0 kg cylinder (I = ½MR²) on a fixed axle. Its acceleration is

What, If Anything, Is Wrong?No calculator
The Rolling Race

A solid cylinder and a hoop have the same mass M and the same radius R. Both are released from rest at the top of the same ramp and roll without slipping.

solid cylinder (M, R) hoop (M, R) both roll without slipping

Student: "It's a tie. They have the same mass and the same radius, and mass doesn't matter on a ramp, so they reach the bottom at the same time with the same speed."

What, if anything, is wrong? If something is wrong, explain the error and fix it. If nothing is wrong, explain why it is correct.

More Torque and Rotation Concept Challenges →

More practice: Torque & Rotation QuizMaster
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Why do skaters spin faster?

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Angular momentum and moment of inertia illustrated with a figure skater. L = I omega: if moment of inertia halves, angular speed doubles when net external torque is negligible. The skater does work to increase rotational kinetic energy.