Transcript Example: Problem 8.40 - UGA Physics and Astronomy

Example Problem
The parallel axis theorem provides a useful way to
calculate I about an arbitrary axis. The theorem
states that I = Icm + Mh2, where Icm is the moment
of inertia of an object (of mass M) with an axis that
passes through the center of mass and is parallel
to the axis of interest. h is the perpendicular
distance between the two axes. Now, determine I
of a solid cylinder of radius R for an axis that lies
on the surface of the cylinder and perpendicular to
the circular ends.
Solution: The center of mass of the cylinder is on a
line defining the axis of the cylinder
R
From Table 10.2:
cm
I cm  MR
1
2
2
From the parallel axis theorem with h=R:
I  I cm  Mh  MR  MR  MR
2
Apply to thin rod:
L/2
2
1
2
1
I cm  12
ML2
I  I cm  Mh 
2
1
12
2
3
2
2
L
2
 L 1
2
ML  M    3 ML
 2
2
Rotational Work
 For translational motion, we defined the work as
W  Fs s
 For rotational motion

Fs

s
r

r
s = arc length
s
FT
W  Fs s  FT r  ( FT r ) = r
WR    Units of N m or J when  is in radians
Rotational Kinetic Energy
 For translational motion, the KE was defined
KE  mv , since vT  r
2
2 2
1
KE  m(r )  2 m r 
2
1
2
1
2
 For a point particle I = mr2, therefore
KER  I
1
2
2
 Or for a rigid body
KER 
1
2
m r 
2
2
i i
 I  KER
1
2
2
 For a rigid body that has both translational
and rotational motion, its total kinetic energy is
KEtotal  KE  KER  mv  I
1
2
2
1
2
 The total mechanical energy is then
Etotal  mv  I  mgh
1
2
2
1
2
2
2
Example
A car is moving with a speed of 27.0 m/s. Each
wheel has a radius of 0.300 m and a moment of
inertia of 0.850 kg m2. The car has a total mass
(including the wheels) of 1.20x103 kg. Find (a) the
translational K of the entire car, (b) the total KR of
the four wheels, and (c) the total K of the car.
Solution:
Given: vcar = 27.0 m/s, mcar = 1.20x103 kg,
rw = 0.300 m, Iw = 0.850 kg m2
a)
b)
c)
K  mcar v  (1.2x10 )(27.0)
K  4.37x10 J
2
2




v
v
2
K R  12 I  12 I w  t   12 I w  car 
 rw 
 rw 
2
1
K R  2 (0.850)(27.0 / 0.300)
3
 3.44x10 J
4
K wheels  4 K R  1.38x10 J
K total  K  K wheels
5
4
5
 4.37x10  1.38x10  4.51x10 J
1
2
2
car
5
1
2
3
2
Example Problem
A tennis ball, starting from rest, rolls down a hill
into a valley. At the top of the valley, the ball
becomes airborne, leaving at an angle of 35° with
respect to the horizontal. Treat the ball as a thinwalled spherical shell and determine the horizontal
distance the ball travels after becoming airborne.
0
3
1.8 m
1
2

4
x
Solution:
Given: v0 = 0, 0 = 0, y0 = 1.8 m = h, y1 = y2 =
y4 = 0, 2 = 35°, x2 = 0, I=(2/3)MR2
Find: x4 ?
Method: As there is no friction or air resistance in
the problem, therefore no non-conservative
forces, we can use conservation of mechanical
2
2
energy
1
1
E  mv  I  mgh
total
2
2
E0  m gh
2
2
1
1
E1  2 mv1  2 I1  m g(0)
2
2
1
1
E2  2 mv 2  2 I 2  m g(0)  E1
 v 1  v 2 , 1   2
S in ceE0  E1  E2
2
2
1
1
m gh  2 mv 2  2 I 2 , bu t v t  r
v t  v 2 Velocity of ball equals tangential
velocity at edge of ball
m gh  mv  ( m r )(v 2 / r )
2
1 2
gh  2 v 2  v  v 2
 v 2  6 gh / 5
 6(9.80)(1.
8)/5  4.6 m /s
1
2
2
2
1
3
1
2
2
2
2
3
5
6
2
2
Now, use 2D kinematic equations for projectile
motion
v 2 x  v 2 cos 2  v 4 x
v2
v 2 y  v 2 sin  2   v 4 y
v4
v 4 y  v 2 y  g (t 4  t 2 ), t 2  0
 v 2 y  v 2 y  gt4
 2v 2 y   gt4  t 4  2v2 y / g
x4  x2  12 ( v 2 x  v 4 x )t 4  v 2 x t 4
x4  v 2 x 2v 2 y / g  2v2 cos 2 v 2 sin  2 / g
2
2v2 cos 2 sin  2 6 gh 2 cos 2 sin  2
x4 

g
5
g
12h cos 2 sin  2 12(1.8) cos35sin 35
x4 

5
5
x4  2.0 m