Physics I: 33–131
Fall 2012
Wednesday October 31, 2012
Name (Printed)
Section
Signature:
A=8:30am B=9:30am C=10:30am D=11:30am
(a) A spherical object of mass m and radius R is initially at rest on a surface. The object
has a string wrapped around the outside of it, and the end of the string is pulled some
force. You observe that at all times, the angular velocity, ω, of the sphere is related
to the center-of-mass speed, vcm , of the sphere through the relation ω R = vcm . After
some time, you observe that the object has translational kinetic energy of Ktran and
rotational kinetic energy of Krot . If you assume that all the work done on the sphere
goes into translational and rotational kinetic energy, what is the moment of inertia,
I, of the spherical object in terms of m, R, Ktran and Krot ?
We start by noting that the translational kinetic energy of the sphere will be given
by
Ktran =
1
2
m vcm
,
2
and that the rotational kinetic energy will be given by
Krot =
1
I ω2 ,
2
but we know that ω = vcm /R, so we have that
Krot =
1 I 2
v .
2 R2 cm
From this, we get that
2Ktran
m
=
2R2 Krot
,
I
or solving for I, we get that
I = mR2
Krot
.
Ktran
(b) You are told that the force used to pull the string is a constant, F , and in getting
to the final state above, that the string is pulled through a distance 5R/3. During
this, you note that a distance 2R/3 of this has unwound from the object. What is
the moment of inertia of the object, I, in erms of m, R, and F ?
1
If the sphere is chosen as the system, we note that the total work done on the sphere
is given as
5
RF .
3
Wext =
we also note that the center-of-mass of the object has moved a distance of R, so we
have that
Wcm = RF .
The latter gives us that
Ktran = RF .
If we assume, as suggested, that the only energy in the system is kinetic, then the
former expresion tells us that
Ktotal =
5
RF ,
3
and we know that Ktotal = Ktran + Krot , so we have that
Krot =
2
RF .
3
From part (a), we then have that
2
mR2
3
I =
2