# 0.50 rad/s**2. How many rotations does the disk make before

```Review Problems
From
Chapter 10&amp;11
1) At t=0, a disk has an angular velocity of 360 rev/min, and constant
angular acceleration of -0.50 rad/s**2. How many rotations does the disk
make before coming to rest?
A1
A2
A3
A4
A5
226
180
360
90
113
360  2
0 
 120 rad / s,   0.50 rad / s 2 ,   0
60
 0
   0  t  t 
 find t

  t  12 t 2   12 t 2  find  in rad / s
divide by 2 to get  in revolutions.
B
B  2A 
2
A
a1   A R
2
a2   B R
2
2
a2  B R  B 
 2

 4
a1  A R  A 
2
Let's call the 2kg body 1, the 4 kg body 2 and the pulley p;
v p  v 1  v 2  2 m / s , v p is the velocity of any point at the rim of the pulley.
vp
2
R 0.03
1
1
K1  m1v 12   2  22  4 J
2
2
1
1
K 2  m 2v 2 2   4  22  8 J
2
2
1
1
K p  I o p 2   0.0045  66.72  10 J
2
2
K  K1  K 2  K p  4  8  10  22 J
p 

4) A uniform rod (M = 2.0 kg, L = 2.0 m) is held vertical about a pivot at point P, a
distance L/4 from one end (as in the Figure). The rotational inertia of the rod
about P is 1.17 kg*m**2. If it starts rotating from rest, what is the linear speed of
the lowest point of the rod as it passes again through the vertical position (v)?
A1
A2
A3
A4
A5
8.7 m/s
4.8 m/s
17 m/s
2.4 m/s
zero
L
K i  0.0, U i  Mg ( ) (taking our U  0.0 at the point P)
4
L
 Ei  K i  U i  Mg ( )
4
L
L
K f  12 I 2 , U f   Mg ( )  E f  K f  U f  12 I 2  Mg ( )
4
4
L
L
but Ei  E f  Mg ( )  12 I 2  Mg ( )
4
4
1
1
MgL
v
Mg L   I 2   

 find v
3L / 4
2
2
I
5) A uniform solid sphere of radius 0.10 m rolls smoothly across a
horizontal table at a speed 0.50 m/s with total kinetic energy 0.70 J. Find the
mass of the sphere.
A1
A2
A3
A4
A5
4.0
8.0
2.0
1.0
5.0
kg
kg
kg
kg
kg
R  0.10 m, K  0.70 J
1
1
2
Mv com
 I 2 (1)
2
2
vcom
2
2
sub. in (1) for I  MR ( sphere ) and for  
(rolling )
5
R
to get M .
K
acom

R
  I
1
2   acom 
f s R   MR   

2
  R 
1
1
f s  Macom   3.0  2.0  3.0 N
2
2
F
fs
7) A 2.0 kg particle is moving such that its position vector (r) relative to the
origin is r =(-2.0*t**2 i + 3.0 j) m. What is the torque (about the origin) acting
on the particle at t=2.0 s?
A1 24 k N.m
A2 -36 k N.m
A3 -24 k N.m
A4 -48 k N.m
A5 0
dr
v
 4t i
dt
dv
a
 4.0 i
dt
F  ma  2.0(4.0)i  8.0 i
r (2.0)  8.0 i  3.0 j
 ( 2. 0)  r  F
 (8.0 i  3.0 j )   8.0 i 
 (3.0 j )   8.0 i 
 24 k
8) In the figure, m1 = 0.50 kg, m2 = 0.40 kg and the pulley has a disk shape of
radius 0.05 m and mass M = 1.5 kg. What is the linear acceleration of the
block of mass m2?
T1
T2
A1 0.59 m/s**2
A2 0.42 m/s**2
m1
m2
a
a
A3 1.46 m/s**2
A4 0.21 m/s**2
A5 0.0
m2 g
m1g
m1 g - T1  m1a
T1  m1 g - m1a (1)
T1
T2
a
R
 net  I  I  
I
1
MR 2
2
a
T1 R  T2 R  I   (3)
R
from (1) and (2) we get :
T1 - T2  (m1 - m2 )g - (m1  m2 )a
put this in (3) to get the value of a
T2 - m2 g  m2 a
T2  m2 a  m2 g (2)
9) Consider two thin rods each of length (L = 1.5 m) and mass 30 g, arranged
on a frictionless table as shown in the figure. The system rotates about a
vertical axis through point O with constant angular speed of 4.0 rad/s. What
is the angular momentum of the system about O?
A1
A2
A3
A4
A5
0.18 kg*m**2/s
0.54 kg*m**2/s
1.5 kg*m**2/s
0.27 kg*m**2/s
0.0
I  2 I rod around O
1
L 2
1
2
2
 2( ML  M ( ) )  2( ML )
12
2
3
L( around O )  I
1
 I 11  I 
2
 I 22  4I 
Li 
1

2
 5I 
I tot  I 1  I 2  3I
Lf  I tot f  3I f
Li  Lf
5I   3I f
5
f  
3
```