Physics 2210 Fall 2015

smartPhysics 16 Rotational Dynamics

11/13/2015

Example 15.2 (1/2) A rotor consists of a thin rod of length ℓ=60 cm, mass 𝑚=10.0 kg, with two spheres attached to the ends. Each sphere has radius 𝑅=10 cm, and mass 𝑀=5.0 kg. A force of magnitude 𝐹=48 N is applied at a distance of ℓ/4 to the right of the axle in the direction shown (a) Calculate the moment-of-inertia of the rotor. (b) Find the angular acceleration of the rotor. Solution (a) The moment of inertia of the rotor (about the axle shown) is the sum of the 3 parts

𝐼 = 𝐼𝑟𝑟𝑟 + 2𝐼𝑠𝑠𝑠𝑠𝑟𝑠 where

𝐼𝑟𝑟𝑟 = 𝑚ℓ2 12⁄ since the axel goes through its CM perpendicularly, and each sphere has (by parallel axis theorem)

𝐼𝑠𝑠𝑠𝑠𝑟𝑠 = 𝐼𝐶𝐶 + 𝑀𝐷2 = 2𝑀𝑅2 5⁄ + 𝑀 𝑅 + ℓ 2⁄ 2 And to total is therefore

𝐼 =1

12𝑚ℓ2 +

45𝑀𝑅

2 + 2𝑀 𝑅 +ℓ2

2

=10 kg 0.60 m 2

12 +4 5.0 kg 0.10 m 2

5 + 2 5.0 kg 0.10 m + 0.30 m 2

= 0.30 kg ∙ m2 +0.04 kg ∙ m2 + 1.60 kg ∙ m2 = 1.94 kg ∙ m2

�⃗�

ℓ

ℓ 4⁄

𝑅

120°

A rotor consists of a thin rod of length ℓ=60 cm, mass 𝑚=10.0 kg, with two spheres attached to the ends. Each sphere has radius 𝑅=10 cm, and mass 5.0 kg. A force of magnitude 𝐹=48 N is applied at a distance of ℓ/4 to the right of the axle in the direction shown (a) Calculate the moment-of-inertia of the rotor. (b) Find the angular acceleration of the rotor.

And so the torque is

𝜏 =14 ℓ𝐹 sin𝜑 =

0.60 m 48 N sin 120°4 = 6.24 N ∙ m

The angular acceleration is given by

𝛼 =𝜏𝐼 =

6.24 N ∙ m1.94 kg ∙ m2 = 3.21 rad/s2

*** Units: N ∙ m

kg ∙ m2 =kg ∙ m 𝑠2⁄ ∙ m

kg ∙ m2 =kg ∙ m

s2 ∙m

kg ∙ m2 =1s2 =

rads2

�⃗�

𝑟 = ℓ 4⁄

ℓ

ℓ 4⁄

𝑅

120°

�⃗�

𝜑 = 120°

120°

Solution (b): The torque on the rotor is given by 𝜏 = 𝑟𝐹 sin𝜑

From the zoom of the diagram we have 𝑟 = ℓ 4⁄ and 𝜑 = 120° (not immediately obvious)

Example 15.2 (2/2)

Rotational Dynamics

Unfortunately as is usually the case with smartPhysics, they give you a nice convenient formula and almost invites you to memorize it… BUT IT IS MISLEADING and not entirely correct

Work done by a torque (often not associated with a conservative force):

𝑾 = � 𝝉𝝉𝝉𝝉𝟐

𝝉𝟏, ∆𝑲 = 𝑾

Remembering, however, that 𝑲 = 𝑲𝑪𝑪 + 𝑲∗ … for rigid bodies 𝑲 = 𝟏

𝟐𝑪𝑽𝑪𝑪𝟐 + 𝟏

𝟐𝑰𝑪𝑪𝝎𝟐

Often more convenient to use (𝑾𝑵𝑪=work done by none-conservative forces) ∆𝑬 = ∆𝑲 + ∆𝑼 = 𝑾𝑵𝑪

Rotational Dynamics

Problem: It is not easy to go from the 2nd box to the 3rd

NOT clear which equation they are talking about

https://www.youtube.com/watch?v=Mc7Nq63wGTQ 3:53—4:03

In general, the linear motion of the center-of-mass and the rotational motion about the center-of-mass are independent of one another

Rolling (without slipping)

• In rolling, rotational and translational are locked together (no slipping!!!): • Displacement of CM in ONE rotation (𝟐𝝅 rad) = ∆𝑋𝐶𝐶 = Circumference = 2𝜋𝑅 CM velocity and angular velocity about the CM are related by: 𝑉𝐶𝐶 = 𝑅𝜔 • Static friction force 𝑓𝑆 is perpendicular to the direction of motion at all times: 𝒇𝑺 does NO work

𝑓𝑆

See https://www.youtube.com/watch?v=ohP0clzK5rc

Gravitational Potential Energy of an Extended Body

For a collection of particles with mass 𝑚1,𝑚2,⋯𝑚𝑁 at positions 𝑥1,𝑦1 , 𝑥2,𝑦2 ,⋯ 𝑥𝑁,𝑦𝑁 , where the +𝒚 direction is assumed to be UP, and

the reference ZERO for potential energy is taken at 𝑦 = 0, The gravitational potential energy of the body is the SUM of those of the individual particles:

𝑈𝑔 = � 𝑈𝑔𝑔𝑁

𝑔=1= � 𝑚𝑔𝑔𝑦𝑔

𝑁

𝑔=1= 𝑔� 𝑚𝑔𝑦𝑔

𝑁

𝑔=1

= 𝑔1𝑀∙ 𝑀� 𝑚𝑔𝑦𝑔

𝑁

𝑔=1= 𝑀𝑔

1𝑀� 𝑚𝑔𝑦𝑔

𝑁

𝑔=1

𝑈𝑔 = 𝑀𝑔𝑌𝐶𝐶

i.e. The gravitational potential energy of a body = gravitational potential energy of the center-of-mass (treated like a point particle)

Poll 11-13-01

A block and a ball have the same mass M and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping.

Which one makes it furthest up the ramp?

A. Block B. Ball C. Both reach the same height.

M

M

Poll 11-13-02

A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping.

Which one reaches the bottom first?

A. Cylinder B. Hoop C. Both reach the bottom at the same time.

Poll 11-13-03

A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping.

Which one reaches the bottom first?

A. Small cylinder B. Large cylinder C. Both reach the bottom at the same time.

Example 16.1 (1/4) A round “wheel” of mass 𝑀, radius 𝑅 and 𝑰𝑪𝑪 = 𝜸𝑪𝑹𝟐 (0 < 𝛾 < 1) rolls without slipping down an incline at angle 𝜙 from the horizontal. Find the linear acceleration of the wheel down the incline.

Solution: Solution: the kinetic energy of the “wheel” is given by

𝐾 =12𝑀𝑉𝐶𝐶2 +

12𝐼𝐶𝐶𝜔2

For rolling without slipping: 𝑉𝐶𝐶 = 𝑅𝜔, and we have 𝐼𝐶𝐶 = 𝛾𝑀𝑅2:

𝐾 =12𝑀𝑉𝐶𝐶

2 +12 𝛾𝑀𝑅

2𝜔2 =12𝑀𝑉𝐶𝐶

2 +12 𝛾𝑀(𝑅2𝜔2)

=12𝑀𝑉𝐶𝐶

2 +12 𝛾𝑀𝑉𝐶𝐶

2 =12 1 + 𝛾 𝑀𝑉𝐶𝐶2

If we start from rest, then by conservation of energy: ∆𝐸 = ∆𝐾 + ∆𝑈 = 0

∆𝐾 =12 1 + 𝛾 𝑀𝑉𝐶𝐶2 = −∆𝑈 = −𝑀𝑔∆𝐻𝐶𝐶

Taking down the ramp to be positive 𝑥 direction then −∆𝐻𝐶𝐶 = ∆𝑋𝐶𝐶 sin𝜙. But: ∆𝑋𝐶𝐶= ∆𝑥 and 𝑉𝐶𝐶 ≡ 𝑣𝑥

12 1 + 𝛾 𝑀𝑣𝑥2 = 𝑀𝑔∆𝑥 sin𝜙

𝜙

𝑀𝑔 𝑅 𝐼𝐶𝐶 = 𝛾𝑀𝑅2

𝜙

∆𝑥 ∆𝐻

Example 16.1 (2/4) A round “wheel” of mass 𝑀, radius 𝑅 and 𝑰𝑪𝑪 = 𝜸𝑪𝑹𝟐 (0 < 𝛾 < 1) rolls without slipping down an incline at angle 𝜙 from the horizontal. Find the linear acceleration of the wheel down the incline.

Solution (continued): Differentiating both sides with respect to time: 12 1 + 𝛾 𝑀 ∙ 2𝑣𝑥 ∙

𝑑𝑣𝑥𝑑𝑑 = 𝑀𝑔

𝑑𝑥𝑑𝑑

sin𝜙 → 1 + 𝛾 𝑀𝑣𝑥𝑎𝑥 = 𝑀𝑔𝑣𝑥 sin𝜙

→ 𝑎𝑥 =𝑀𝑔𝑣𝑥 sin𝜙1 + 𝛾 𝑀𝑣𝑥

→ 𝑎𝑥 =𝑔 sin𝜙 1 + 𝛾

Example given in Main Point: Solid Ball/Sphere

𝐼 =25𝑀𝑅

2 → 𝛾 =25 → 1 + 𝛾 =

75 →

11 + 𝛾 =

75

𝑎𝑥 =75𝑔 sin𝜙

Note that we used 𝜙 here for the angle of the inclined ramp instead of 𝜃 to avoid confusion between the rotation angle of the ball and the angle of the incline

for a solid ball

Example 16.1 (3/4) A round “wheel” of mass 𝑀, radius 𝑅 and 𝑰𝑪𝑪 = 𝜸𝑪𝑹𝟐 (0 < 𝛾 < 1) rolls without slipping down an incline at angle 𝜙 from the horizontal. Find the linear acceleration of the wheel down the incline.

Alternate Solution: Let +x be down slope and +y point perpendicularly out of the incline. We will chose CCW As the positive rotation sense. For rolling without slipping then we have:

𝑣𝑥 = −𝑅𝜔, 𝑎𝑥 = −𝑅𝛼 Where𝑣𝑥 and 𝑎𝑥 are the velocity and acceleration components down slope. Applying Newton’s Second Law on the CM, first in the y-direction

𝑀𝑎𝑦 = 𝑁 −𝑀𝑔 cos𝜙 = 0 → 𝑁 = 𝑀𝑔 cos𝜙 Which is not really used in this problem except possibly to verify that 𝑓𝑠 < 𝜇𝑠𝑁 In the x-direction :

𝑀𝑎𝑥 = 𝑀𝑔 sin𝜙 − 𝑓𝑠 … 1 Next we look at rotation about the CM:

𝐼𝐶𝐶𝛼 = 𝑅𝑓𝑠 sin(−90°) = −𝑅𝑓𝑠 *** We have used the fact that, for the purpose of calculating torque, the force of gravity acts at the center-of-mass (CM). This follows from 𝐼𝐶𝐶𝛼 = 𝑅𝑓𝑠 sin(−90°) = −𝑅𝑓𝑠

𝜙

𝑀𝑔

𝑁 𝑓𝑠

𝑅 𝐼𝐶𝐶 = 𝛾𝑀𝑅2

𝑥

𝑦

Example 16.1 (4/4) A round “wheel” of mass 𝑀, radius 𝑅 and 𝑰𝑪𝑪 = 𝜸𝑪𝑹𝟐 (0 < 𝛾 < 1) rolls without slipping down an incline at angle 𝜙 from the horizontal. Find the linear acceleration of the wheel down the incline.

Solution (continued): From last page 𝑀𝑎𝑥 = 𝑀𝑔 sin𝜙 − 𝑓𝑠 … 1 𝐼𝐶𝐶𝛼 = 𝑅𝑓𝑠 sin(−90°) = −𝑅𝑓𝑠

----------- But 𝑎𝑥 = −𝑅𝛼,→ 𝛼 = −𝑎𝑥 𝑅⁄ , and so we have

𝛾𝑀𝑅2 ∙−𝑎𝑥𝑅 = 𝑅𝑓𝑠, → −𝛾𝑀𝑎𝑥 = 𝑓𝑠 … (2)

substituting (2) back into (1) for 𝑓𝑠: 𝑀𝑎𝑥 = 𝑀𝑔 sin𝜙 − 𝛾𝑀𝑎𝑥, → 1 + 𝛾 𝑀𝑎𝑥 = 𝑀𝑔 sin𝜙

And so we have:

𝑎𝑥 =𝑔 sin𝜙 1 + 𝛾

Example 16.2 (1/6) A 1152 kg car is being unloaded by a winch. At the moment shown below, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 344 kg·m2 and that of the pulley is 3 kg·m2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

Solution: Looking for the speed of something as it drops through a certain distance. Does it remind you of a problem to solve using conservation of energy? Except in this case, as the car acquires speed, both the pulley and the drum acquire angular speed (no slipping!!!): Drum: 𝑅𝐷𝜔𝐷 = 𝑣𝐶 ... (1), Pulley: 𝑅𝑃𝜔𝑃 = 𝑣𝐶 … (2). The total kinetic energy of the system is the sum of those of the three objects:

𝐾 = 𝐾𝐷 + 𝐾𝑃 + 𝐾𝐶 =12 𝐼𝐷𝜔𝐷

2 +12 𝐼𝑃𝜔𝑃

2 +12𝑀𝐶𝑣𝐶2 … (3)

Substituting (1) and (2) into (3) gives us

𝐾 =12 𝐼𝐷

𝑣𝐶𝑅𝐷

2

+12 𝐼𝑃

𝑣𝐶𝑅𝑃

2

+12𝑀𝐶𝑣𝐶2 =

12

𝐼𝐷𝑅𝐷2

+𝐼𝑃𝑅𝑃2

+ 𝑀𝐶 𝑣𝐶2

Example 16.2 (2/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water.

𝐾 =12

𝐼𝐷𝑅𝐷2

+𝐼𝑃𝑅𝑃2

+ 𝑀𝐶 𝑣𝐶2

No friction: total energy is conserved. The only potential energy is the gravitational potential energy of the car: so we have ∆𝐸 = ∆𝐾 + ∆𝑈 = 0. We started from rest: 𝐾𝑔 = 0 So the final kinetic energy (after 5.0m drop) is then

𝐾𝑓 = 𝐾𝑔 + ∆𝐾 = −∆𝑈 = − 𝑀𝐶𝑔∆ℎ , ∆ℎ = −5.0m And so

12

𝐼𝐷𝑅𝐷2

+𝐼𝑃𝑅𝑃2

+ 𝑀𝐶 𝑣𝐶𝑓2 = − 𝑀𝐶𝑔∆ℎ

𝑣𝐶𝑓2 =−2𝑀𝐶𝑔∆ℎ

𝐼𝐷𝑅𝐷2

+ 𝐼𝑃𝑅𝑃2

+ 𝑀𝐶

=−2 ∙ 1152kg ∙ 9.8 m s2⁄ ∙ (−5.0m)344 kg·m2

0.80 m 2 + 3 kg·m2

0.30 m 2 + 1152kg=

112896 kg ∙ m2 s2⁄1723.8 kg

= 65.53m2

s2

𝑣𝐶𝑓 = 8.10 m s⁄

Example 16.2 (1/6) A 1152 kg car is being unloaded by a winch. At the moment shown below, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 344 kg·m2 and that of the pulley is 3 kg·m2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

Solution: *** The figure has been annotated *** (%i1) /* see the free body diagram for the car

We choose down to be the positive direction for y

Newton's 2nd Law in the y direction is then */

eqn1: m*a = m*g - T1;

(%o1) a m = g m - T1

(%i2) /* solve for T1 */

soln1: solve(eqn1, T1);

(%o2) [T1 = (g - a) m]

(%i3) T1a: rhs(soln1[1]);

(%o3) (g - a) m

... continued

Example 16.2 (2/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water.

(%i4) /* next observe the free body diagram for the disk pulley. We choose Clock-wise (C) to be the positive sense for rotation. Let Rp = radius of pulley. angular acceleration of the pulley is then linked to the acceleration of the car by */ alpha_p: a/Rp; a (%o4) -- Rp (%i5) /* Rotational Newton's 2nd Law for the pulley */ eqn2: Ip*alpha_p = Rp*T1 - Rp*T2; a Ip (%o5) ---- = Rp T1 - Rp T2 Rp (%i6) /* substitute T1 from eqn1 into eqn2 */ eqn2a: eqn2, T1=T1a; a Ip (%o6) ---- = (g - a) m Rp - Rp T2 Rp ... continued

Example 16.2 (3/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water.

(%i7) /* solve fopr T2 */

soln2a: solve(eqn2a, T2);

2

(g - a) m Rp - a Ip

(%o7) [T2 = --------------------]

2

Rp

(%i8) T2a: rhs(soln2a[1]);

2

(g - a) m Rp - a Ip

(%o8) --------------------

2

Rp

... continued

Example 16.2 (4/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i9) /* Now look at the free body diagram for the drum. Again let the positive direction of rotation be clock-wise. Let Rd be the radius of the drum. Again the angular acceleration of the drum is linked to the linear acceleration of the car by the cord */ alpha_d: a/Rd; a (%o9) -- Rd (%i10) /* Newton's 2nd law for the drum */ eqn3: Id*alpha_d = Rd*T2; a Id (%o10) ---- = Rd T2 Rd (%i11) /* substittute in T2 from eqn2a */ eqn3a: eqn3, T2=T2a; 2 a Id Rd ((g - a) m Rp - a Ip) (%o11) ---- = ------------------------- Rd 2 Rp ... continued

Example 16.2 (5/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i12) /* now we can solve for "a" */

soln3a: solve(eqn3a, a);

2 2

g m Rd Rp

(%o12) [a = -------------------------]

2 2 2

(m Rd + Id) Rp + Ip Rd

(%i13) aa: rhs(soln3a[1]);

2 2

g m Rd Rp

(%o13) -------------------------

2 2 2

(m Rd + Id) Rp + Ip Rd

(%i14) /* solve for a value */

aa_val: aa, g=9.81, m=1152, Rd=0.80, Rp=0.30, Id=344, Ip=3;

(%o14) 6.559613040534004

... continued

Example 16.2 (6/6) A 1152 kg car falls: no slipping between the (massless) rope, the pulley (has mass and rotates with the string), and the winch drum. Moment of inertia of the winch drum: 344 kg·m2, of pulley: 3 kg·m2. Radius of the winch drum is 0.80 m, pulley: 0.30 m. Find the speed of the car as it hits the water. (%i15) /* For this last part we can use

vf^2 - vi^2 = 2*a*Dy */

kill(a);

(%o15) done

(%i16) /* and since vi=0 then we have */

vf: sqrt(2*a*Dy);

(%o16) sqrt(2) sqrt(a Dy)

(%i17) vf, a=aa_val, Dy=5.0, numer;

(%o17) 8.099143806930461

Answer: vf = 8.10 m/s