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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W11D2-1 Table Problem Cylinder on Inclined Plane Torque About Center of Mass
Solution
A uniform cylinder of outer radius R and mass M with moment of inertia about the center of
mass I cm  (1/ 2) M R 2 starts from rest and moves down an incline tilted at an angle  from the
horizontal. The center of mass of the cylinder has dropped a vertical distance h when it reaches
the bottom of the incline. Let g denote the gravitational constant. The coefficient of static
friction between the cylinder and the surface is s . The cylinder rolls without slipping down the
incline. Using the torque method about the center of mass, calculate the velocity of the center of
mass of the cylinder when it reaches the bottom of the incline.
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Solution: We shall apply the torque equation about the center of mass and the force equation for
the center of mass motion We will first find the acceleration and hence the speed at the bottom of
the incline using kinematics. The figure below shows the forces acting on the cylinder and our
choice of coordinate system.
Choose x  0 as the point where the cylinder just starts to roll. With the unit vectors shown in
the figure above, Newton’s Second Law, applied in the x - and y -directions in turn, yields
Mg sin   f s  Max ,
(1)
 N  Mg cos   0 .
(2)
Choose the center of the cylinder to compute the torque about (see figure below).
Then, the only force exerting a torque about the center of mass is the friction force, and so we
have
f s R  I cm z .
(3)
Use I cm  (1/ 2) M R 2 and the kinematic constraint for the no-slipping condition  z  ax / R in
Eq. (3) to solve for the magnitude of the static friction force yielding
f s  (1/ 2) Max .
(4)
Mg sin   (1 / 2) Max  Max
(5)
Substituting Eq. (3) into Eq. (1)
which we can solve for the acceleration
ax 
2
g sin  .
3
(6)
The displacement of the cylinder is x f  h / sin  in the time it takes to reach the bottom, t f , and
the x-component of the velocity vx at the bottom is vx , f  ax t f . The displacement in the time
interval t f satisfies x f  (1/ 2)axt f 2 . Then after eliminating t f , we have x f  vx, f 2 / 2ax , so the
magnitude of the velocity when the cylinder reaches the bottom of the inclined plane is
vx, f  2ax x f  2  (2 / 3) g sin   h / sin    (4 / 3) gh .
(7)
Although b=note part of the problem, note that if we substitute Eq. (6) into Eq. (4) the magnitude
of the friction force is
f s  (1 / 3) Mg sin  .
(8)
In order for the cylinder to roll without slipping
f s  s Mg cos  .
(9)
So combining Eq. (8) and Eq. (9) we have the condition that
(1 / 3) Mg sin   s Mg cos 
(10)
Thus in order to roll without slipping, the coefficient of static friction must satisfy
1
(11)
3
Alternatively we could calculate the torque about a fixed point that lies along the line of
contact between the cylinder and the surface
s  tan  .
Choose a fixed point that lies along the line of contact between the cylinder and the surface.
Then the torque diagram is shown below.
The gravitational force Mg  Mg sin  iˆ  Mg cos  ˆj acts at the center of mass. The vector from
the point P to the center of mass is given by rP ,mg  d P iˆ  R ˆj , so the torque due to the
gravitational force about the point P is given by
 P ,Mg  rP ,Mg  Mg  (d P iˆ  R ˆj )  ( Mg sin  iˆ  Mg cos  ˆj )
 (d P Mg cos   RMg sin  )kˆ
.
(12)
The normal force acts at the point of contact between the cylinder and the surface and is given
by N   N ˆj . The vector from the point P to the point of contact between the cylinder and the
surface is rP , N  d P iˆ . So the torque due to the normal force about the point P is given by
 P, N  rP, N  N  d P iˆ   N ˆj  d P N kˆ .
(13)
Substituting Eq. (2) for the normal force into Eq. (13) yields
 P, N  d P Mg cos  kˆ .
(14)
Therefore the sum of the torques about the point P is
 P   P,Mg   P, N  (d P Mg cos   RMg sin  )kˆ  d P Mg cos  kˆ  Rmg sin  kˆ .
(15)
The angular momentum about the point P is given by
LP  Lcm  rP ,cm  Mvcm  I cmz kˆ  (d P iˆ  R ˆj )  Mvx iˆ
 ( I cmz  RMvx ) kˆ
.
(16)
The time derivative of the angular momentum about the point P is then
dLP
 ( I cm z  RMax ) kˆ .
dt
(17)
Therefore the torque equation
P 
becomes
dLP
,
dt
(18)
RMg sin  kˆ  ( Icm z  RMax )kˆ .
(19)
Using the fact that I cm  (1/ 2) MR 2 and  x  ax / R , the z-component of Eq. (19) becomes
RMg sin   (1/ 2)MRax  Rmax  (3/ 2)MRax .
(20)
We can now solve Eq. (20) for the x-component of the acceleration
ax  (2 / 3) g sin  ,
in agreement with Eq. (6).
(21)