No Title
PHYSICS II
Solution to Exam 1
                                                                                                           
Monday 12/14/15 Name
O233: periods 1,2
JFW CM:
Section:














                    
                                         (1)
(25)
(2)
(25)
(3)
(25)
(4)
(25)
TOTAL
(100)
  1. A pulley (modeled as a solid disk of uniform mass distribution) in mounted vertically with its rotational axis fixed in space and passing through the center of mass.
    mpulley=28.0 kg
    Rpulley=12.0 cm

      (a) (5 pts) Calculate the rotational inertia, Ip, of the pulley.

      I
      =
      1

      2
      mR2
      =
      1

      2
      (28.0 kg)(0.120 m)2
      =
      0.202 kg·m2

      (b) (5 pts) At t=2.00 s, the tangential acceleration at a point on the rim of the wheel is atan=0.240 m/s2. Calculate the angular acceleration, a, of the pulley.

      a
      =
      atan

      R
      =
      0.240 m/s2

      0.120 m
      =
      2.00 rad/s2

      (c) (5 pts) Calculate the torque, t, exerted on the pulley.

      t
      =
      Ia
      =
      (0.202 kg·m2)(2.00 rad/s2)
      =
      4.03 N·m

      (d) (5 pts) At t=2.00 s, calculate the angular speed, w, of the pulley. Assume that the pulley starts from rest.

      w
      =
      vtan

      R
      =
      vi+atant

      R
      =
      0+(0.240 m/s2)(2.00 s)

      0.120 m
      =
      4.00 rad/s

      (e) (5 pts) Calculate the angular displacement of the pulley in the first two seconds?

      Dq
      =
      1

      2
      at2
      =
      1

      2
      (2.00 rad/s2)(2.00 s)2
      =
      4.00 rad

    • (a) (10 pts) A hula hoop of mass m=4.00 kg and radius R=40.0 cm rolls along a horizontal floor in a vertical plane without slipping. The center of mass speed is a constant 3.00 m/s. Calculate the total kinetic energy of the hoop.

      Ktot
      =
      1

      2
      mv2cm+ 1

      2
      Iw2
      =
      1

      2
      mv2cm+ 1

      2
      (mR2)(vcm/R)2
      =
      mv2cm
      =
      (4.00 kg)(3.00 m/s)2
      =
      36.0 J

      (b) (10 pts) The hoop is now positioned in a vertical plane on a ramp inclined at q = 30.0o and released from rest. When it reaches the bottom of the ramp, the hoop has the same total kinetic energy as in part (a). Calculate the distance the center of mass traveled down the ramp.

      mgh
      =
      Ktot
      mgs sinq
      =
      Ktot
      (4.00 kg)(9.80 N/kg)s sin30.0o
      =
      36.0 J
      s
      =
      1.84 m

      (c) (5 pts) At the bottom of the ramp, is the center of mass speed the same as or different from the center of mass speed stated in part (a)?
      ANS: The center of mass speed is the same.
      mgh
      =
      mv2cm
      Solving for vcm gives
      vcm = 3.00 m/s
  2. A constant force Fy=9.00 N acts on a particle of mass m=3.00 kg. The motion is along the x=4.00 m direction. When the particle is at the location (x=4.00 m,y=0 m) where vy=2.00 m/s,

      (a) (7 pts) what torque does an observer located at the origin measure? Include the direction.

      t
      =
      r×F
      =
      xFy 
      ^
      k
       
      =
      (4.00 m)(9.00 N) 
      ^
      k
       
      =
      +36.0 N·m 
      ^
      k
       

      (b) (7 pts) what angular momentum does the observer measure? Include the direction.

      L
      =
      r×p
      =
      mxv^ 
      ^
      k
       
      =
      (3.00 kg)(4.00 m)(2.00 m/s) 
      ^
      k
       
      =
      24.0 kg·m2/s 
      ^
      k
       

      (c) (11 pts) The angular momentum of the particle at any position along the direction of travel is
      L = mx(viy+ayt) 
      ^
      k
       
      Apply Newton's second law for rotational motion (t = dL/dt) to show that one obtains the same result as found in part (a).

      t
      =
      dL

      dt
      =
      mxay 
      ^
      k
       
      =
      xFy 
      ^
      k
       
      =
      36.0 N·m 
      ^
      k
       
  3. A light string connects a hanging mass, m, to a pulley of rotational inertia, Ip. The hanging mass is released from rest.
    Ip=0.160 kg·m2
    Mp=8.00 kg
    Rp=20.0 cm
    m=4.00 kg

      (a) (15 pts) Set up the two equations needed for determing the acceleration of the hanging mass and solve them to find the numerical value.

      mg-T = may

      TR
      =
      Ia
      =
      1

      2
      MpR2 (ay/R)
      T
      =
      1

      2
      Mpay
      Elimimnate the tension between the first and third equations and solve for ay. The result is
      ay
      =
      2mg

      2m+Mp
      =
      2(4.00 kg)(9.80 m/s2)

      2(4.00 kg)+8.00 kg
      =
      4.90 m/s2

      (b) (4 pts) Calculate the speed of the hanging mass after 1.80 s has elapsed.

      vf
      =
      vi +ayt
      =
      0+(4.90 m/s2)(1.80 s)
      =
      8.82 m/s

      (c) (6 pts) Calculate the angular momentum of the pulley after 1.80 s has elapsed.

      L
      =
      Iw
      =
      (MR2/2)vf

      R
      =
      1

      2
      MRvf
      =
      1

      2
      (8.00 kg)(0.200 m)(8.82 m/s)
      =
      7.06 kg·m2/s
      The direction of the vector is into the paper.
FORMULAS
s=rq
v=rw
atan=ra
arad=w2r
w = dq/dt
a = dw/dt
wavg=Dq/ Dt
wavg=(wf+wi)/2
aavg=Dw/Dt
atan=dvtan/dt
Dq = wit+at2/2
wf=wi+at
wf2=wi2+2aDq
Ihoop=mR2
Krot=Iw2/s
Ktrans=mv2/2
tnet = Ia
t=r×F
Fnet=ma
vf2=vi2+2aDx
vf=vi+at
U=mgh
Ktot=Icmw2/2+mvcm2/2
Icyl=mR2/2
L=r×p



File translated from TEX by TTH, version 3.85.
On 18 Dec 2015, 16:22.