Download Example 5-8 A Rock on a String

Example 5-8 A Rock on a String
You tie one end of a lightweight string of length L around a rock of mass m. You
hold the other end of the string in your hand and make the rock swing in a horizontal
circle at constant speed. As you swing the rock, the string makes an angle u with the
vertical (Figure 5-18). Derive an expression for the speed at which the rock moves
around the circle. Ignore the effect of drag forces.
String, length L
Figure 5-18 ​​Moving in a circle How fast must the rock move to make the string hang at an
Speed v = ?
angle u from the vertical?
Set Up
Because the rock moves at constant speed
in a horizontal circle, its acceleration vector
points horizontally toward the center of the
circle. We choose the positive x axis to point in
this direction, and the positive y axis to point
upward.
Only two forces act on the rock: the
s exerted by the string and the
tension force T
s rock. The
downward gravitational force w
tension force is directed along the string and so
points at an angle u to the vertical. The vertical
s balances the downward force
component of T
of gravity, while the horizontal component of
s provides the centripetal ­acceleration. Our
T
goal is to solve for the rock’s speed v.
Solve
Newton’s second law equation for the
rock:
v2
r
O
y
arock
x
Centripetal acceleration:
toward center
of the circle
(5-10)
Write Newton’s second law for the rock in
component form. Note that the quantity v,
which is what we’re trying to find, doesn’t
­appear in these equations. We’ll introduce it
in the next step through the expression for
the rock’s centripetal acceleration.
Newton’s second law in
component form applied to
the rock:
Use Equation 5-10 to express the acceleration
of the rock. Note that because the string is at
an angle, the radius of the rock’s circular path
is not equal to the length of the string.
Radius of the rock’s circular path:
wrock
Tx = T sin O
y
x: T sin u = mrock arock,cent
y: T cos u + (2w)
= T cos u 2 mrock g = 0
T
O
Ty = T cos O
x
wrock,y = –wrock
r = L sin u
O
L
so the rock’s centripetal
acceleration is
arock,cent =
Substitute the expression for centripetal
acceleration into the Newton’s second law
equation for the x direction and solve for v.
T
s
s
s rock = mrock s
arock
a Fext on rock = T + w
arock,cent =
Rock,
mass m
q
v2
v2
=
r
L sin u
radius = L sin O
Newton’s second law equations for the rock become
x: T sin u = mrock arock,cent =
mrockv 2
L sin u
y: T cos u = mrock g
Solve the y component equation for T:
T =
mrockg
cos u
Substitute this expression into the x component equation and solve
for v:
mrock g
mrock v 2
sin u =
cos u
L sin u
g sin u
v2
=
cos u
L sin u
v2 =
v =
Reflect
gL sin2 u
cos u
gL sin2 u
C cos u
You can try this motion for yourself by tying a piece of string around an eraser or other small object and whirling it in a
horizontal circle. You’ll find that the faster the object moves, the greater the angle u. Let’s check that our answer agrees
with this conclusion.
If u is a small angle close to zero so that the string hangs down almost vertically, sin u is small and cos u is nearly
equal to 1. (Remember that sin 0 = 0 and cos 0 = 1.) So the ratio 1sin2 u2 > 1cos u2 is a small number, and the speed v
will be small. If u is a large angle close to 90° so that the string is nearly horizontal, sin u is a little less than 1 and cos u is
small (recall sin 90° = 1 and cos 90° = 0). In this case, the ratio 1sin2 u2 > 1cos u2 is a large number and the speed v will
be large. So our formula agrees with the experiment!