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
Lecture 26 March 26, 2015 2 / 18
Pulley systems revisited
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
Pulley systems revisited
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
Lecture 26 March 26, 2015 4 / 18
Pulley systems revisited
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
Lecture 26 March 26, 2015 5 / 18
Pulley systems revisited
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
Lecture 26 March 26, 2015 6 / 18
Pulley systems revisited
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
Lecture 26 March 26, 2015 7 / 18
Pulley systems revisited
Solve for vf :
vf =
2gh1 + M12m
(1 +
M2R22M1R21
) .
Lecture 26 March 26, 2015 8 / 18
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.
Lecture 26 March 26, 2015 9 / 18
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
Source: Young and Freedman, University Physics, 13th ed.
Lecture 26 March 26, 2015 11 / 18
Combined translation and rotation
Source: Young and Freedman, University Physics, 13th ed.
Lecture 26 March 26, 2015 12 / 18
Combined translation and rotation
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
Lecture 26 March 26, 2015 13 / 18
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
Lecture 26 March 26, 2015 14 / 18
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.
Lecture 26 March 26, 2015 15 / 18
Rolling race
Example
Various objects are about to race along an inclined plane. Which shapewill win the race?
Source: Young and Freedman, University Physics, 13th ed.
Lecture 26 March 26, 2015 16 / 18
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