lecture12-2 2017 new - manninggroup · pdf file–ch9–ch12 –lecturesthrough...

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Physics 215 – Spring 2017 Lecture 12-2 1

Welcome back to Physics 215

Today’s agenda:

•Angular momentum•Rolling without slipping•Midterm Review


Midterm 3 – Thursday April 13th• Material covered:– Ch 9 – Ch 12 – Lectures through this Thursday– Also draws on material from Ch 1-8– HW, lectures, demos, recitations

• No office hours after exam on Thursday April 13thor Friday April 14th.

• Extra office hours will be posted online

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Women in Physics organizational meeting

• Friday (tomorrow) 3-4 pm in 233• snacks and coffee• goal is to provide mentorship and support to women undergrads and grads who are interested in physics

• all are welcome!


Angular Momentum 1.

qp

r

OPoint particle:

|L| = |r||p|sin(q) = m|r||v| sin(q)

vector form à L = r x p– direction of L given by right hand rule

(into paper here)

L = mvr if v is at 900 to r for single particle

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Angular Momentum 2.

o

w

rigid body:* |L| = Iw (fixed axis of rotation/fixed axle)* direction – along axis – into paper here


Rotational Dynamics t = I a

DL/Dt = t§ These are equivalent statements

§ If no net external torque: t = 0 à * L is constant in time * Conservation of Angular Momentum * Internal forces/torques do not contribute to external torque.

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Demo - wheels


Linear and rotational motion

• Force• Acceleration

• Momentum

• Kinetic energy

• Torque• Angular acceleration

• Angular momentum**

• Kinetic energy

Fnet =

F∑ =ma

τ net =

τ∑ = I α

L = I ωp =mv

K = 12 Iω 2

€

K = 12mv

2

** about a fixed axle or axis of symmetry

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SG A hammer is held horizontally and then released. Which way will it fall?


SG Two buckets spin around in a horizontal circle on frictionless bearings. Suddenly, it starts to rain. As a result,

A. The buckets speed up because the potential energy of the rain is transformed into kinetic energy.

B. The buckets continue to rotate at constant angular velocity because the rain is falling vertically while the buckets move in a horizontal plane.

C. The buckets slow down because the angular momentum of the bucket + rain system is conserved.

D. The buckets continue to rotate at constant angular velocity because the total mechanical energy of the bucket + rain system is conserved.

E. None of the above.

QuickCheck 12.11

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General motion of extended objects

• Net force è acceleration of CM

• Net torque about CM è angular acceleration (rotation) about CM

• Resultant motion is superposition of these two motions

• Total kinetic energy K = KCM + Krot


1. A.2. B.3. C.4. A and C.

SG Three identical rectangular blocks are at rest on a flat, frictionless table. The same force is exerted on each of the three blocks for a very short time interval. The force is exerted at a different point on each block, as shown.

After the force has stopped acting on each block, which block will spin the fastest?

Top-view diagram

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1. A.2. B.3. C.4. A, B, and C have the same C.O.M. speed.

SG Three identical rectangular blocks are at rest on a flat, frictionless table. The same force is exerted on each of the three blocks for a very short time interval. The force is exerted at a different point on each block, as shown.

After each force has stopped acting, which block’s center of mass will have the greatest speed?

Top-view diagram


EXAMPLE 12.21 Two Interacting DisksA 20-cm diameter, 2.0 kg solid disk is rotating at 200 rpm. A 20-cm-diameter, 1.0 kg circular loop is dropped straight down onto the rotating disk. Friction causes the loop to accelerate until it is riding on the disk. What is the final angular velocity of the combined system?

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Rolling without slipping

Dxcm =

Dxcm =

vcm =

vcm =


Rolling without slippingtranslation rotation

vcm =

acm =

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Rolling without slippingN

W

F

q

SF = maCM

St = Ia

Now aCM = Ra if no slipping

So, maCMand F =


Kinetic energy of rolling

• The total kinetic energy of a rolling object is the sum of its rotational and translational kinetic energies:

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5. Two cylinders with the same radius and same total mass roll down a ramp. In cylinder A, a set of 8 point masses are equally spaced in a circle with radius r1 around the cylinder’s axis of rotation, while in cylinder B, the 8 point masses are a distance r2 > r1 from the center. Which cylinder reaches the bottom of the ramp first?

A. Cylinder AB. Cylinder BC. They both reach the bottom at the same timeD. Not enough information to tell


Example problem: Calculate the acceleration of the center of mass of a cylindrical object that rolls without slipping down a ramp of height h and angle q. The object has mass M and radius R, with a moment of inertia I = cMR2about its center of mass/axis of rotation.

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midterm review


A toy race car of mass 20 g is traveling on a frictionless loop-the-loop track. The diagram shows the track from the side. The radius of the loop track is R=0.1 m. The initial speed of the car at the bottom of the loop is 2.5 m/s. What is the speed of the car at the top of the loop? What is the normal force on the car at the top?

4.(16 pts total)

A toy race car of mass 20 g is traveling on a frictionless loop-the-loop track. The diagram shows the track from the side. The radius of the loop track is R=0.1 m. The initial speed of the car at the bottom of the loop is 2.5 m/s.

a) (2 pts) What is the kinetic energy of the car at the bottom of the loop

b) (2 pts) What is the change in the gravitational potential energy of the car if it makes it to the top of the loop ? (assume that g=9.8 m/s2).

c) (3 pts) Using conservation of energy calculate the speed of the car at the top of the loop

d) (5 pts) Draw a free body diagram for the car at the top of the loop. Using this diagram, calculate the normal force acting on the car from the track at that point.

c) (4 pts) If the normal force is zero at the top of the loop the car will leave the track. Calculate the smallest possible speed that the car should have at the bottom of the loop if it is to complete the loop without falling off.

4

R

Monday, November 12, 12

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A sled slides along a horizontal surface with a coefficient of kinetic friction of 0.2. Its velocity at point A is 8.0 m/s and its velocity at point B is 5.0 m/s. How long does it take the sled to travel from point A to point B? How far did it slide? Use the work-kinetic energy and the impulse-momentum theorems.


! ! ! PHY211 Exam 3 - PracticeName (please print) ___________________________ !TA/section:__________________________________ Document your work. Use the other side of the sheet if you need more space. Answers should be given to 3 significant figures unless otherwise directed and should carry appropriate units.

1. (16 pts total)

A block of mass m=2.0kg is set into motion down a rough inclined plane with speed 2.0 m/s as shown in the picture. The coefficient of kinetic friction between the block and the plane is 0.4 (take g=9.8 m/s2)

a) Draw a free body diagram showing all the forces acting on the block.

b) Calculate the work done on the block by friction after the block has traveled 1.0 m down the slope.

c) Calculate the work done on the block by gravity at this point. How is the latter related to the change in gravitational potential energy ?

d) What is the work done on the block by the normal force ?

e) Use the work-kinetic energy theorem to calculate the speed of the block after the block has moved 1.0 m down the slope.

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v=2.0m/s

300

Wednesday, November 7, 12

A block of mass m=2.0kg is set into motion down a rough inclined plane with speed 4.0 m/s as shown in the picture. The coefficient of kinetic friction between the block and the plane is 0.4 (take g=9.8 m/s2). What is the speed of the block after it has moved 1.0 m down the incline?

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A child sits on a seesaw of length L = 2m with his father and his pet dog. The mass of the seesaw is 20 kg, the mass of the father is 50 kg, the mass of the dog is 5 kg and the mass of the child is 20 kg. The father sits on the right edge of the seesaw, while the child sits on the left edge. The dog is 0.5 m from the father. If the seesaw is in equilibrium, what is the distance from the pivot to the child? Confirm that the net torque about this pivot point is zero.

lecture12-2 2017 new - manninggroup · pdf file–ch9–ch12 –lecturesthrough...

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