Assignment 4 - Mechanics 4 Solution

Document Sample

```					                 HKIPhO Mechanics Assignment IV Solution

1. Two identical small beads with mass m are attached on a smooth circular ring that
stands vertically on the ground. Initially
the beads are resting on the top of the
ring. They then start to slide down.
Derive the relationship between the
mass of the ring and the mass of the
beads such that the ring can jump up
from the ground. Also, find the location
of the beads when the ring jumps up.

Sol:

Let the mass of the ring be M and radius be R. At time t, the angle m made with the
vertical is .

Since there is no friction between beads and ring, therefore the force acting on the beads
are always pointing along N, and Newton’s Law for circular motion implies

mv 2
mg cos  N       .
R

Conservation of energy implies

1                        mv 2
mg (2R)  mv 2  mgR(1  cos )        2mg (1  cos ) .
2                         R

Putting back in the first equation we obtain

N  3mg cos  2mg .

Notice that N becomes negative when  is large enough. In this case the reaction force
N’ of the beads on the ring will try to lift the ring up. This can happen when

2( N ' cos )  Mg .

Since N’ = - N, we have the condition for ring to jump up:

2(3mg cos  2mg ) cos  Mg , or

1        3M 
cos  1  1      . For this to have a solution, we must have
3        2m 

1
3M        M 2
1       0 or   .
2m        m 3

2. This problem models the launching of a detector from the surface of the earth to Mars.
We assume that the earth and Mars are moving around the sun in circles resting on the
same plane. The radius of the orbit of Mars is Rm, which is 1.5 times larger than that of
the earth. An economical and simple way to lounge the detector consists of 2 steps. First,
a rocket is used to accelerate the detector on the earth’s surface such that it acquires
enough kinetic energy to overcome the gravitational force of the earth and becomes a
satellite moving around the earth. Second, at a suitable time, an engine that connects to
the detector ignites for a short instant and accelerates the detector along its direction of
motion. After a short (negligible) time, the speed of the detector increases to a suitable
value such that the detector moves along an elliptical orbit that connects the earth and
Mars, with the two planets located at the end points of the ellipse (see Diagram A)

(1) In Step I, what is the minimum speed needed for the detector to become an artificial
satellite that moves along the earth’s orbit?

(2) After the detector becomes a satellite moving around the earth, on 1 March 00.00am
of a certain year, the angular distance between the detector and Mars is measured to
be 60o (diagram B). What is the date that the engine of the detector should be fired so
that the detector can fall on the surface of Mars (Correct to day)? Given: radius of the
earth: 6.4x106 m, acceleration due to gravity = 9.8m/s2 .

v 2 GMm    GM
(1) For circular orbit, m       2 v     gRe  1.12  10 4 m / s .
r   r      r

(2) Again we use a  c  Ro , and a  c  Rm , where Rm  1.5 Ro . Therefore
a  1.25 Ro for the detector orbit.

2
T2         (1 year) 2
Next we use Kepler’s third law which implies                               . Therefore the time
a3            3
Ro
T               3
taken for the detector to go to Mars is  0.5  (1.25) 2 year. During this period, the
2
v T                    Ro        Ro 2Ro
angular distance Mars travel is  m  m , where vm            ve                   .
2Rm                   Rm         Rm (1 year)
1.25 3
Therefore  m  0.5  (        ) (2 )  137 o . Therefore the angular distance between
1.5
the detector and Mars should be 180-137=43o at the moment when the detector fires. The
3
v                            v     1       2
angular velocity of earth and Mars are  e  e                  and  m  m      e ,
Ro                 1.5 
Rm
2t
respectively. The angle that earth and Mars travel in time t (in days) is  e       and
365
3
 1  2 2t
m            , respectively. Therefore, the detector should be fired t days after
 1.5  365
March 1st, when e   m  60  43  17 o . Solving the equation we obtain         t ~ 38 days,
corresponding to April 7th of the same year.

3. The perigee and apogee of the first artificial satellite of our mother country is 8754 km
and 6809 km respectively. Find the speed of the satellite when it passes through the
perigee and apogee. What is the period of the satellite?

GM a  c          GM a  c
Sol: Use v2              , v1 
2                   2
, where a + c = 8734km and a – c = 6809km,
a ac             a ac
T 2 4 2
and Kepler’s third law           .
a 3 GM

We obtain v1  6.38  10 3 m/s , v2  8.2  10 3 m/s and T = 114 mins.

4. Consider the attractive electrostatic force between an electron and a proton inside a
k
nucleus V (r )  , where k < 0 is a constant.
r

a) Assuming that the 2 particles circle around the center of mass in a circular orbit,
determine the relation between the radius of the orbit and velocities of the 2
moving particles.
b) Discuss how the Kepler’s Three Laws will be modified in this situation?
3
a) Using Newton’s second law,

ve2     v2      k
me        mp p 
re      rp (re  rp ) 2

Considering the distances from the center of mass,

 m 
me re  m p rp  r  re  rp  1  e re
 m 
    p 

Hence by eliminating re and rp,

ve2        
1  me   k2
mpk
me                            ve                      .
r  mp  r
                           me (me  m p )r

Similarly,

me k
vp                      .
m p (me  m p )r

b) Kepler’s three laws are modified as follows:
First law: The orbits of both the electron and proton are elliptical with the center of mass
being one focus.
Second law: The vector from the center of mass to the electron sweeps out area at a
constant rate. The same for the proton.
Third law: The square of the period is proportional to the cube of the semimajor axis of
the elliptical orbits. The proportionality constant in the center-of-mass frame is modified:

4 2 re2    4 2 me m p 3
T 
2
               r .
ve2      k (me  m p )

Compared with the case that the proton is assumed stationary, the mass of the electron is
replaced by the reduced mass memp/(me + mp).

5. A solid ball with uniform density and radius R (originally resting on a smooth desk) is
hit by a horizontal force F at height h (<2R). The force lasts for a small time t.
Afterward, the ball is found to roll without sliding. What is h?

The horizontal speed of the ball after hit by the force is giving by Newton’s Law

mv   Fdt .

4
d
We also have Newton’s Law for rotation Fh  I A         .
dt

d
Therefore we have mvh  h  Fdt  I A           dt  I A , where  is the angular velocity
dt
2                7
of rotation of the ball after being hit, I A  mR 2  mR 2  mR 2 .
5                5

Since the ball did not slide, we have v  R . Therefore,

7            7
m(R)h        mR 2  h  R .
5            5

6. A small ring with radius r =10 cm falls on a desk from a height h = 20 cm. Initially,
the ring is rotating upon its axis with angular frequency o = 21 rads-1 (see figure).
Assume that the collision between the ring and desk is inelastic such that the ring stays on
the desk after collision. How long will it takes for the ring to stop rotating if the
coefficient of friction between the ring and desk is  = 0.3.

Time take for the ring to fall to the ground is

2h
to         0.2s with speed v  gt o  2m / s. During this period, the angle the ring has
g
turned is    o t o  4.2(rad).

What happens after the ring collide with the desk can be separated into two stages:

(a) What happens during the collision and (b) What happens after the collision.

First we consider (a). We assume that the collision happens in a very short interval t .
During this short period, the ring’s velocity changes from v = 2m/s to v = 0m/s (complete
stop). The resulting force between ring and desk is

mv                                        mv
F        . The corresponding friction is f   ( ) and the corresponding torque is
t                                        t

5
mv
  (      )r . The moment of inertia of the ring is I  mr 2 .
t

Therefore, from Newton’s Law for rotation, the change in angular velocity during this
period is

 v                                               v
    and the decrease in angular velocity is       6 s 1 . Notice that during
t rt                                              r

1 v
this period, the ring has rotated an angle    o t     (    )(t ) 2 which is negligible
2 rt
when t  0 .

After the collision, the angular velocity of the ring is 1   o    15s 1 . From this
point on, the force that slows the ring is the friction force with torque   mgr . The
      mgr g
rate of change of angular velocity is                30 s  2 . Therefore the
t       I    r

ring stops after time t1  1  0.5s .


1
The total angle rotated is   1t1       (t1 ) 2  3.75(rad ) .
2

The total angle of rotation is 4.2+3.75=7.95rad. Therefore the total turns of rotation is
7.95
~ 1.265 .
2

7. A circular disc with radius r and uniform density  is pinned by a wire at a distance
2r
d      from the center of the disc. The wire and the disc is set into rotation with angular
3
velocity . What is the angle  between the disc and the wire? (This problem requires
integration.)



6
Consider a horizontal strip of the disc at a distance x from the pivot; x is positive in the
downward direction.
2m
Mass of the strip: dm  2 r 2  ( x  d ) 2 dx
r
2m sin 
Momentum of the strip: dp  x sin  dm                 x r 2  ( x  d ) 2 dx
r 2
Angular momentum of the center of mass of the strip about the pivot:
2m sin  2 2
dL  xdp              x r  ( x  d ) 2 dx
r 2
The direction of this contribution to the angular momentum is normal to the disc. It has a
component in the horizontal direction that changes with time, and this component has to
be considered in the equation of motion.
There is also the contribution to the angular momentum due to the rotation of the strip
about the center of mass of the strip itself. This component is directed vertically and does
not change with time during the rotation.
Total angular momentum:

d r
2m sin  
L              x r  ( x  d ) dx
2 2        2

r 2
d r

Let x  d  r sin  . Then

 /2
2m sin  
L            (d  r sin  ) cos d
2   2

       / 2
 /2
2m sin   d 2 d2       2dr         r2  r2      
              sin 2 
 2                cos3     sin 4 

           4         3          8   32        / 2
 2 r2 
 m sin   d  

    4

Using Newton’s law for rotational motion,

dL

dt
mgd sin   L cos
gd         24 g
 cos                  
2 2   r  25 r
2        2
 d  

     4 

8. Consider a long, solid, rigid, square prism. The mass of the prism is M and it is
uniformly distributed. The length of each side of the cross-sectional square is a. The
moment of inertia of the square prism about its central axis is

7
1
I     Ma2 .
6

Answer parts a) to e) of the 1998 Physics Olympiad question titled “Rolling of a
hexagonal prism”, replacing the hexagonal prism with the square prism.

Before impact                         After impact

P                                      P
θ                                         θ

a) Consider the angular momentum about the corner of impact P.
Before the impact:
1
Angular momentum due to the rotation about the centre of mass =       Ma2i .
6
Angular momentum due to the translational motion of the centre of mass = 0.
1
Total initial angular momentum = Ma2i .
6
After the impact:
1               2

Angular momentum due to rotation about P =       Ma 2  M  a   f  2 Ma 2 f .
    
6          2         3

Since the impulse passes through P, the angular impulse about P is 0. Hence angular
momentum is conserved.
1            2
Ma2i  Ma2 f
6            3
1
 f  s i where s  .            (answer)
4
b) Since the kinetic energy is proportional to 2,
1
K f  rK i where r        . (answer)                                45o–
16                                 θ
c) The kinetic energy Kf after the impact
must be sufficient to lift the centre of mass
to its highest position. This energy is
a
E0  Mg       (1  cos(450   )) .                           P
2                                           θ
Hence we get the condition
a
K f  rK i  E0  Mg       (1  cos(450   ))
2

8
Thus Ki,min = Mga, where
1
            (1  cos(45 0   )) (answer)
r 2
d) Between successive impacts, the height of the centre of mass of the prism decrease by
a sin  and its kinetic energy increases by Mga sin . We therefore have
K i ,n 1  rK i ,n  Mga sin  .
In the limit of very large n, the kinetic energy approaches the limit Ki0, where
K i , 0  rK i , 0  Mga sin 
Mga sin                           sin 
K i ,0             Mga where                . (answer)
1 r                             1 r
d) For indefinite continuation, the limit value of Ki in part d) must be larger than the
minimum value for continuation found in part c):
Mga sin           a
 Mg         (1  cos(45 0   )) .
1 r           r 2
r 2       2
Let A               . Then we have
1  r 15
A sin   1  cos(45 0   )  1  cos 45 0 cos  sin 45 0 sin 
       1             1
A        sin         cos  1
        2             2
Multiplying both sides by 2 ,
17
sin   cos  2
15
17 / 15                       1                       2
sin                     cos 
(17 / 15) 2  1           (17 / 15) 2  1        (17 / 15) 2  1
               
Hence define u  arcsin                    41.42 0
1
 (17 / 15)  1 
2
               
2
sin(  u )                   0.9357
(17 / 15) 2  1
   0 where  0  arcsin 0.9357  41 .42 0  69 .34 0  41 .42 0  27 .92 0 .   (answer)

9.   A bowler throws a bowling ball of radius R down a lane. The ball slides on the lane,
with initial speed v0 and initial angular speed 0. The coefficient of kinetic friction
between the ball and the lane is k. When the speed of the ball has deceased enough
and the angular speed has increased enough, the ball stops sliding and then rolls
smoothly.
(a) How far does the ball slide?
(b) What is the speed of the ball when smooth rolling begins?

9
f
(a) Let a = linear acceleration of the ball,  = angular acceleration of the ball.
When the ball slides on the lane, the frictional force is f   k Mg
Newton’s law for linear motion:  k Mg  Ma            a   k g
(Here we use the convention that the right direction is positive.)
k MgR
Newton’s law for rotation about the centre of mass:   k MgR  I          
I
(Here we use the convention that the anticlockwise direction is positive.)
2
Here the moment of inertia of the ball about the centre of mass  I  MR2
5
Hence at time t, the linear and angular velocity of the ball is:
v  v0  at
  0   t
When the ball stops sliding, v  R . Hence
v0  at   R(0  t )
v0  R0        v0  R0     2  v  R0 
t                               0        .
 a  R k g  k gMR / I 7  k g 
2
         
Distance traveled by the ball
2
1         2  v  R0  1        2 v  R0      2
 v0t  at 2  v0  0  g  2   k g  0
 7  g   49  g (v0  R0 )( 6v0  R0 ) .

2         7       k                  k          k

(b) When the ball stops sliding, the velocity of the ball
 2 v  R0  5       2
 7  g   7 v0  7 R0 . (answer)
 v0  at  v0  k g  0         
      k    

10. A sphere of radius R is placed at rest on a table. After a slight push, the table rolls off
the table edge.
(a) What is the angle  when the sphere loses contact with the table?
(b) What is the velocity of the sphere when it loses contact with the table?



(a) Let M = mass of the sphere,

10
 = angular velocity of the sphere when it loses contact with the table.
Conservation of energy:
1
MgR  MgR cos  I 2 .
2
Here the moment of inertia of the sphere about the point of contact
2                7
 I  MR2  MR2  MR2
5                5
2MgR(1  cos )
2 
I
Let N = normal reaction.
Centripetal acceleration =R2.
Newton’s law of motion in the radial direction: Mg cos   N  MR 2 .
When the sphere loses contact with the sphere, N = 0. Therefore
2MR 2 (1  cos  )
Mg cos   MR  Mg
2
.
I
10
cos  (1  cos ) .
7
10
cos             540 .    (answer)
17
2MgR(1  cos ) g 10                   10 g
(b)  2                           (1  cos ) 
I          R 7                  17 R
10
Velocity of the sphere  R          gR .      (answer)
17

11. On an inclined plane of 30o, a wooden block of mass m2 = 4 kg is connected with a
solid cylinder of mass m1 = 8 kg and radius r = 5 cm. The coefficient of kinetic
friction between the wooden block and the inclined plane is  = 0.2. When the
system is released, the wooden block slides and the cylinder rolls down the inclined
plane. What is the acceleration of the system?

m2
m1

30o

Let a = acceleration of the system,
T = tension in the connection between the cylinder and the wooden block,
f = frictional force between the cylinder and the inclined plane.
For the wooden block, normal reaction  m2 g cos
frictional force between the wooden block and the inclined plane  m2 g cos

11
Newton’s law of linear motion: m2 g sin   m2 g cos  T  m2 a                          (1)
For the cylinder,
Newton’s law of linear motion: m1 g sin   f  T  m1a                                    (2)
a
Newton’s law of rotation: fR  I                                                           (3)
R
1
Here, the moment of inertia of the cylinder about its centre is I              m1 R 2 .
2
I
Combining (2) and (3): m1 g sin         a  T  m1 a                                     (4)
R2
I
(1) + (4): (m1  m2 ) g sin   m2 g cos  2 a  (m1  m2 )a
R
(m1  m 2 ) sin   m2 cos      (m1  m 2 ) sin   m 2 cos
ag                                  g
m1  m2  I / R 2                   3m1 / 2  m 2
(8  4) sin 30 0  0.2  4 cos 30 0
 9.8                                        3.25 m/s 2 .    (answer)
38 / 2  4

12. A uniform rod of mass M and length L is placed like a ladder against a frictionless
wall and frictionless horizontal floor. It is released from rest, making an angle  with
the vertical. Show that the initial reaction of the wall and the floor are

3                             3          
RW      Mg cos sin  ,  R F  Mg 1  sin 2  ,
4                             4          
2      
and the angle at which the rod will leave the wall is cos 1  cos  . [Hint: Note that
3      
the distance between the centre of mass and the point O is a constant.]
Forces                 Linear and angular accelerations
Y                                   Y
RW                                 

                                             
Mg                                      L 2 / 2

RF                                   
L / 2
O                         X         O                            X
(a) Let  be the angle between the rod and the wall. The distance between the centre of
mass of the rod and O is equal to L/2 for all inclinations of the rod. Hence the motion of
the rod can be considered as the superposition of a uniform circular motion of the centre
of mass, and the rotation of the rod about the centre of mass.

The centre of mass has a centripetal acceleration of L 2 / 2 and a tangential acceleration
of L / 2 . The angular acceleration is  .
                                   
L          
Newton’s law of horizontal motion: RW  M (cos   2 sin  ) .           (1)
2

12
L           
Newton’s law of vertical motion: Mg  RF  M      ( sin    2 cos ) .      (2)
2
Newton’s law of rotation about the centre of mass:
L           L
RF sin   RW cos  I .          (3)
2           2
1
Here, the moment of inertia of the ladder about the centre of mass  I        ML2 .
12
Substituting RW from (1) and RF from (2) into (3):
L           L2                             L2                           1
                                                
Mg sin   M (sin 2    2 sin  cos )  M (cos2    2 sin  cos )  ML2
2            4                              4                            12
L        1        3g sin  .
Mg sin   ML2 ,                        (4)
2        3              2L

At  = ,   0 .
L            3
(1): RW  M cos  Mg sin  cos . (answer)
2            2
L              3
(2): RF  Mg  M  sin   Mg(1  sin 2  ) .       (answer)
2                4

(b) The expression of  can be obtained from the conservation of energy:
2
L              L         1  L  1  1           
Mg cos  Mg cos  M      ML2  2 .
2              2         2  2  2  12           
    3g
 2  (cos  cos ) .          (5)
L
Substituting (4) and (5) into (1):
L  3g                                                3Mg
sin  3 cos  2 cos  
3g                3g
RW  M          sin  cos      cos  sin      cos sin   
2  2L               L                 L               4
When the rod leaves the wall, RW = 0. Hence 3cos  2 cos  0 ,
2        
  arccos cos   .           (answer)
3        

13. A solid sphere of radius r is placed at the bottom of a spherical bowl of radius R.
When it is pushed gently, it rolls about the bottom. The rolling motion is described
by the equation

   2  0,
where θ is the angle between the vertical and the line joining the centres of the
sphere and the bowl.
(a) What is the value of 2?

13
(b) (For students who have learned simple harmonic motion:) What is the period of the
rolling motion?
(c) What is the frictional force at the angle θ?
[Hint: Let  be the angular displacement of the sphere with respect to the vertical line.
First find the relation between  and θ.]

f
r             θ
R

N       θ       mg

(a) Let m be the mass of the sphere.
Let  be the angular displacement of the sphere with respect to the vertical line. It is
related to  by R = r( + ), or (R – r) = r.
Let N be the normal reaction between the bowl and the sphere.
Let f be the frictional force between the bowl and the sphere.
Newton’s law of tangential motion: f  mg sin   m( R  r ) .          (1)
Newton’s law of radial motion: N  mg cos  m( R  r )      2 .         (2)
Newton’s law of rotational motion about the centre of the sphere:
I ( R  r ) 

fr   I              .            (3)
r
2
Here the moment of inertia of the sphere about its centre  I  mr 2 .
5
I ( R  r ) 
Eliminating f from (1) and (3):          2

  mg sin   m( R  r )
r
7m( R  r ) 
When  is small, sin   . The equation becomes                       mg  0 .
5
5g

Hence    2  0 where  2                 . (answer)
7( R  r )
2            7( R  r )
(b) The period of the rolling motion                2                . (answer)
                5g
I ( R  r )  2               2
(c) f           2
  m( R  r ) 2  mg .               (answer)
r          5               7

14. A uniform rod of mass m and length l is free to rotate in a vertical plane about the
pivot through its centre. It is initially oriented in a horizontal direction. A spider, also
of mass m, falls down with a vertical velocity v0 and lands on the midpoint between
the pivot and one end of the rod. After landing, the spider immediately climbs along
the rod in such a way that the angular speed of the rod-spider system remains a
constant.

14
(a) Show that the distance of the spider from the pivot takes the form x(t) = A sin Bt + C.
What are the values of A, B and C?
(b) If the spider can reach the end of the rod before the rod reaches the vertical position,
what is the largest possible value of v0?

l



x
mg

(a) When the spider lands on the rod, the initial angular velocity 0 is given by the
conservation of angular momentum:
l 1                l 
2

mv 0   ml  m  0
2

4  12
              4   
12v0                          12v0 t
0          . Hence    0 t             .
7l                            7l
When the spider climbs along the rod to a distance x from the centre, the angular
1                 
momentum becomes   ml 2  mx 2 0 .
 12               
Hence the rate of change of angular momentum  2mxv 0 , where v is the velocity of the
spider along the rod.
Newton’s law of rotation about the centre:
mgx cos  2mxv 0 .
g                    dx
cos  0 t  v       .
2 0                  dt
g
Integrating, x            sin  0 t  C .
2 0
2

Initial condition: x = l/4 at t = 0, therefore C = l/4.
g     49 gl 2            12v0        l
Hence x = A sin Bt + C, where A                            , B  0       and C  .
20 288v0
2         2
7l         4
(b) The spider can reach the end of the rod if A + C  l/2.
49 gl 2 l l                           7 gl
2
  , yielding v0                    .
288v0 4 2                             6 2

15

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 5 posted: 1/1/2013 language: Unknown pages: 15