Derive the formula of velocity for rolling motion and also discuss for ring, solid cylinder and sphere
Answers
Answer:
To analyze rolling without slipping, we first derive the linear variables of velocity and acceleration of the center of mass of the wheel in terms of the angular variables that describe the wheel’s motion. The situation is shown in (Figure).
Figure a shows a free body diagram of a wheel, including the location where the forces act. Four forces are shown: M g is a downward force acting on the center of the wheel. N is an upward force acting on the bottom of the wheel. F is a force to the right, acting on the center of the wheel, and f sub s is a force to the left acting on the bottom of the wheel. The force f sub s is smaller or equal to mu sub s times N. Figure b is an illustration of a wheel rolling without slipping on a horizontal surface. Point P is the contact point between the bottom of the wheel and the surface. The wheel has a clockwise rotation, an acceleration to the right of a sub C M and a velocity to the right of v sub V M. The relations omega equals v sub C M over R and alpha equals a sub C M over R are given. A coordinate system with positive x to the right and positive y up is shown. Figure c shows wheel in the center of mass frame. Point P has velocity vector in the negative direction with respect to the center of mass of the wheel. That vector is shown on the diagram and labeled as minus R omega i hat. It is tangent to the wheel at the bottom, and pointing to the left. Additional vectors at various locations on the rim of the wheel are shown, all tangent to the wheel and pointing clockwise.
Figure 11.3 (a) A wheel is pulled across a horizontal surface by a force
→
F
. The force of static friction
→
f
S
,
|
→
f
S
|
≤
μ
S
N
is large enough to keep it from slipping. (b) The linear velocity and acceleration vectors of the center of mass and the relevant expressions for
ω
and
α
. Point P is at rest relative to the surface. (c) Relative to the center of mass (CM) frame, point P has linear velocity
−
R
ω
^
i
.
From (Figure)(a), we see the force vectors involved in preventing the wheel from slipping. In (b), point P that touches the surface is at rest relative to the surface. Relative to the center of mass, point P has velocity
−
R
ω
^
i
, where R is the radius of the wheel and
ω
is the wheel’s angular velocity about its axis. Since the wheel is rolling, the velocity of P with respect to the surface is its velocity with respect to the center of mass plus the velocity of the center of mass with respect to the surface:
→
v
P
=
−
R
ω
^
i
+
v
CM
^
i
.
Since the velocity of P relative to the surface is zero,
v
P
=
0
, this says that
v
CM
=
R
ω
.
Thus, the velocity of the wheel’s center of mass is its radius times the angular velocity about its axis. We show the correspondence of the linear variable on the left side of the equation with the angular variable on the right side of the equation. This is done below for the linear acceleration.
If we differentiate (Figure) on the left side of the equation, we obtain an expression for the linear acceleration of the center of mass. On the right side of the equation, R is a constant and since
α
=
d
ω
d
t
,
we have
a
CM
=
R
α
.
Furthermore, we can find the distance the wheel travels in terms of angular variables by referring to (Figure). As the wheel rolls from point A to point B, its outer surface maps onto the ground by exactly the distance travelled, which is
d
CM
.
We see from (Figure) that the length of the outer surface that maps onto the ground is the arc length
R
θ
. Equating the two distances, we obtain
d
CM
=
R
θ
.