Lecture 23 1/24

Physics 111Lecture 23 (Walker: 10.6, 11.1)Conservation of Energy in Rotation

TorqueMarch 30, 2009

Lecture 23 2/24

Kinetic Energy of Rolling ObjectTotal kinetic energy of a rolling object is the sum of its linear and rotational kinetic energies:

The second equation makes it clear that the kinetic energy of a rolling object is a multiple of the kinetic energy of translation.

Lecture 23 3/24

Example: Rolling Disk

1.20 kg disk with radius of 10.0 cm rolls without slipping. The linear speed of disk is v = 1.41 m/s.(a) Find the translational kinetic energy.(b) Find the rotational kinetic energy.(c) Find the total kinetic energy.

1 12 22 2 (1.20 kg)(1.41 m/s) 1.19 JtK mv= = =

1 1 1 12 2 2 22 2 2 4( )( / ) (1.20 kg)(1.41 m/s) 0.595 JrK I mr v rω= = = =

(1.19 J) (0.595 J) 1.79 Jt rK K KΣ = + = + =

Lecture 23 4/24

Question

A solid sphere and a hollow sphere of the same mass and radius roll forward without slipping at the same speed.

How do their kinetic energies compare?

(a) Ksolid > Khollow(b) Ksolid = Khollow(c) Ksolid < Khollow(d) Not enough information to tell

Lecture 23 5/24

Rolling Down an Inclinei i f fK U K U+ = +

1 2 22 (1 / )K mv I mr= +

U mgh=

1 2 22 (1 / )mgh mv I mr= +

22 / (1 / )v gh I mr= +

i

f

Lecture 23 6/24

2Hollow Cylinder : ; I mr v gh= =1 422 3Solid Cylinder: ; I mr v gh= =2 623 5Hollow Sphere: ; I mr v gh= =

2 1025 7Solid Sphere: ; I mr v gh= =

Lecture 23 7/24

Question

Which of these two objects, of the same mass and radius, if released simultaneously, will reach the bottom first? Or, is it a tie? Assume rolling without slipping.

(a) Hoop; (b) Disk; (c) Tie; (d) Need to know massand radius.

Lecture 23 8/24

The Great Downhill RaceA sphere, a cylinder, and a hoop, all of mass M and radius R, are released from rest and roll down a ramp of height hand slope θ. They are joined by a particle of mass Mthat slides down the ramp without friction.

Who wins the race? Who is the big loser?

Lecture 23 9/24

The Winners

Lecture 23 10/24

Example: Spinning WheelBlock of mass m attached to string wrapped around circumference of a wheel of radius R and moment of inertia I, initially rotating with angular velocity ω. The block rises with speed v = rω . The wheel rotates freely about its axis and the string does not slip. To what height h does the block rise?

i fE E=1 1 1 1 12 2 2 2 2 22 2 2 2 2( / ) (1 / )iE mv I mv I v R mv I mRω= + = + = +

fE mgh=

2

212v Ihg mR

⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Lecture 23 11/24

Example: Flywheel-Powered CarYour vehicle’s braking mechanism transforms translational kinetic energy into rotational kinetic energy of a massive flywheel. Flywheel has I = 11.1 kg m2 and a maximum angular speed of 30,000 rpm. At the minimum highway speed of 40 mi/h, air drag and rolling friction dissipate energy at 10.0 kW. If you run out of gas 15 miles from home, with flywheel spinning at maximum speed, can you make it back?

1 12 2 22 2 (11.1 kg m )(3,142 rad/s) 54.9 MJK Iω= = =

( )( )30,000 rev/min 2 rad/rev / (60 s/min)=3,142 rad/sω π=

/ (15 mi) / (40 mi/h) 0.375 h 1350 st x v∆ = ∆ = = =

Energy needed =Wnc=(10 kW)(1350 s)=13.5 MJYou make it.

Lecture 23 12/24

Torque (τ)To make an object start rotating, a force is needed; the position and direction of the force matter as well. The perpendicular distance r⊥from the axis of rotation to the line along which the force acts is called the lever arm.

Lecture 23 13/24

TorqueFrom experience, we know that the same force will be much more effective at rotating an object such as a nut or a door if our hand is not too close to the axis.

This is why we have long-handled wrenches, and why doorknobs are not next to hinges.

Lecture 23 14/24

TorqueWe define a quantity called torque which is a measure of “twisting effort”. For force F applied perpendicular distance r⊥ from the rotation axis:

The torque increases as the force increases and also as the perpendicular distance r⊥ increases.

Lecture 23 15/24

Other Torque Units• Common US torque unit: foot-pound (ft-lb)• Seen in wheel lug nut torques for vehicles:Ford Aspire 1994-97 85 ft-lbsFord Contour 1999-00 95 ft-lbsFord Crown Victoria 2000-05 100 ft-lbs

“Torque Amplifier”

Lecture 23 16/24

A longer lever arm is very helpful in rotating objects.

Lecture 23 17/24

Finding the Lever Arm r⊥Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.

Lecture 23 18/24

Torque in Terms of Tangential Force

The torque is defined as:

Could also write this as τ = r F⊥ where r is the distance from the axis to where the force is applied and F⊥ is the tangential force component.

Lecture 23 19/24

TorqueThis leads to a different definition of torque (as given in the textbook):

where F is the force vector, r is the vector from the axis to the point where force is applied, and θ is the angle between r and F.

(Angle measured CCW from r direction to F direction.)

Lecture 23 20/24

Sign of Torque

If the torque causes a counterclockwise angular acceleration, it is positive; if it causes a clockwise angular acceleration, it is negative.

CCW: 0CW: 0

ττ

><

Lecture 23 21/24

Two Interpretations of Torque T o rq ue can be co nsid ered as th e

fo rce actin g a t a p erpen d icu lard is tance s in (called th e

) fro m th e p ivo t p o in t

to th e th ro u gh . T h us, ( s in ) .

Fd r

m om en t a rm

line o f action FF r F d

φ

τ φ

=

= =

r

r

Lecture 23 22/24

Two Interpretations of Torque Torque can be considered as thedistance from the pivot to the point

of action of the force F times thetangential force component sin . Thus, ( sin ) .

t

t

r

F Fr F rF

φτ φ

=

= =

r

Lecture 23 23/24

QuestionFive different forces are applied to the

same rod, which pivots around the black dot. Which force produces the smallest torque about the pivot?

Lecture 23 24/24

End of Lecture 23For Wednesday, read Walker, 11.2-4

Homework Assignment 10b is due at 11:00 PM on Wednesday, April 1.