Physics Lecture 26

Lecture 26More on energy in rotational motion

March 26, 2015

Pulley systems revisited

previously, we ignored the massof the pulley

in reality, the pulley would havea moment of inertia

this would decrease theacceleration of the system

the pulley contributes arotational kinetic energy in theenergy conservation expression

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Example

A block of mass m is attached to one end of a string. The other end iswrapped around a frictionless cylindrical pulley. If the block is releasedfrom rest, what is its speed as it hits the ground?

Source: Young and Freedman, University Physics, 13th ed.

Lecture 26 March 26, 2015 3 / 18


Solution:Set up conservation of energy. The initial state is when the block is at restat a height h. The final state is when the block is about to hit the ground.Let the ground be the y = 0 height.

Ui +Ki = Uf +Kf

mgh+ 0 = 0 +1

2mv2 +

1

2I2f

Use the fact that I = 12MR2 and that the angular speed of the cylinder

and the linear speed of the block is related by v = R.

mgh =1

2mv2f +

1

2

(1

2MR2

)(vfR

)2

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Solve for vf :

vf =

2gh

1 +M/2m.

Check for the limiting cases:What happens when M is very large? vf 0What happens when m is very large? vf

2gh

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Example

Two metal disks, one with radius R1 andmass M1 and the other with radius R2 andmass M2, are welded together and mountedon a frictionless axis through their commoncenter. A block of mass m is wound on thesmaller disk. If the block is released fromrest at a distance h above the ground, whatis its speed just before it hits the ground?

Source: Young and Freedman, University Physics

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Solution:Set up conservation of energy. The initial state is when the block is at restat a height h. The final state is when the block is about to hit the ground.Let the ground be the y = 0 height.

Ui +Ki = Uf +Kf

mgh+ 0 = 0 +1

2mv2f +

1

2I2f

Use the fact that I = 12M1R21 +

12M2R

22 and that the angular speed of the

cylinder and the linear speed of the block is related by v = R1.

mgh =1

2mv2f +

1

2

(1

2M1R

21 +

1

2M2R

22

)(vfR1

)2

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Solve for vf :

vf =

2gh1 + M12m

(1 +

M2R22M1R21

) .

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Quiz!

A uniform thin rod of length L and mass M , pivoted at one end, is heldhorizontal and then released from rest. Assuming that effects due tofriction and air resistance are negligible, find the angular speed of the rodas it sweeps through the vertical position. The moment of inertia of a rodpivoted at one end is I = 13ML

2.

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Quiz!

Solution:Set up conservation of energy. The initial state is when the rod ishorizontal. The final state is when the rod is vertical. Let the initial heightbe the y = 0 height.

Ui +Ki = Uf +Kf

0 + 0 = Mg(L/2) + 12I2f

MgL = I2f

Subst. the correct expression for the moment of inertia, then solve for f .

MgL =

(1

3ML2

)2f

f =

3g

L.

(What is vf for the center of mass of the rod?)Lecture 26 March 26, 2015 10 / 18

Combined translation and rotation


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The kinetic energy of a body with both rotational and translational motionis just the sum of...

(a) the translational kinetic energy 12Mv2cmassociated with the motion of

the center of mass, and

(b) the rotational kinetic energy 12Icm2 about the center of mass.

K =1

2Mv2cm +

1

2Icm

2

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Rolling without slipping

Source: Halliday, Resnick, Fundamentals of Physics, 9th ed.

[distance travelled by center of mass] = R

Taking the time rate of change of the above equation, we get

vcm = R

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Caution!

There is a difference between

vcm = R

and

v = r

The latter pertains to the linear speed of a point on an object a distance rfrom its axis. Its valid only when the axis is fixed.

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Rolling race

Example

Various objects are about to race along an inclined plane. Which shapewill win the race?


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Rolling race

Solution:We dont need to know the time at which each shape finish the race. Theshape with the fastest speed at the bottom must have won. Set upconservation of energy. The initial state is when the object is at a heighth. The final state is when the object is at the floor. Let the floor be they = 0 height.

Ui +Ki = Uf +Kf

Mgh+ 0 = 0 +1

2Mv2cm,f +

1

2Icm

2f

Let Icm = cMR2, where c is a constant (e.g. c = 1/2 for a solid cylinder).

Since the object is rolling without slipping, vcm = R.

Mgh =1

2Mv2cm,f +

1

2(cMR2)

(vcm,fR

)2Lecture 26 March 26, 2015 17 / 18

Rolling race

Solve for vcm,f :

vcm,f =

2gh

1 + c

Lecture 26 March 26, 2015 18 / 18

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Physics Lecture 26