Physics 101 Fall 2007: Pledged Problems 10 – Rotational Dynamics
Time allowed: 2 hours in one sitting
Due:
Wednesday, November 28 at 5PM in the box marked Phys101/102 in the Physics Lounge
You may use your own textbook, your notes and a non-programmed calculator. For the purposes of this problem set, you may also
use the online solutions to the corresponding suggested problems. You should consult no other help. Please follow the standard
format:
• Write legibly on one side of 8-1/2” white or lightly tinted paper.
• Staple all sheets (including this one) together in the upper left corner
• Make one vertical fold.
• On the outside (staple side up) on successive lines
– PRINT your last name in CAPITAL letters.
– PRINT your first name.
– Print the phrase ‘Pledged Problems 10’ and the due date.
– Print the times at which you started and finished the problems.
– Write and sign the Pledge, with the understanding you may consult the resources described above.
1. A ship’s anchor weighs 5000 N. Its cable passes over a roller of negligible mass and is wound around a hollow
cylindrical drum of mass 380 kg and radius 1.1 m, as shown in the figure below. The drum is mounted on
a frictionless axle. The anchor is released from rest and drops 16 m to the water.
16 m
(a) Calculate the constant angular acceleration of the cylindrical drum. (Hint: You may want to use
torque considerations to help you with this portion of the problem.)
Solution
I will use the hint applied to the cylindrical drum (an A subscript refers to Anchor, and a D subscript
refers to the drum):
~ =Iα
~τ = ~r × F
~
(1)
I = mD R 2
(2)
|τ | = R T
(T represents the tension in the cable.)
(3)
T − mA g = −mA ay
(4)
T = m A g − m A ay
(5)
Since the cable is not slipping over the cylindrical drum, we can connect the translational speed of the
cable to the rotational speed of the cylindrical drum.
v = ωR
(6)
a = αR
(7)
⇒T = mA g − mA α R = mA (g − α R)
(8)
R mA (g − α R) = I α
α=
(9)
mA g
5000N
5000N
=
=
= 5.11rad/s2
R (mD + mA )
1.1 (380kg + 509.7kg)
978.7kg m
(10)
(b) Compute the time it takes for the anchor to hit the water once it is released from rest.
Solution
Use kinematic relationships (constant-acceleration equations) to determine the time it takes the anchor
to hit the water. The linear acceleration of the anchor is related to the angular acceleration of the
drum through Eq. 7.
aA = αD R = 5.62m/s2
yf − y0 = v0,y ∆t −
r
(11)
1
aA (∆t)2
2
(12)
2h
aA
(13)
∆t = 2.39s
(14)
∆t =
(c) What is the angular velocity of the cylindrical drum at the instant the anchor hits the water?
Solution
Since we know the angular acceleration of the drum, use the constant-acceleration equations to determine the final angular velocity of the drum after ∆t seconds.
ωf = ω0 + α ∆t
ωf = α ∆t =
ωf =
s
mA g
R (mD + mA )
(15)
r
2h
aA
2 h mA g
= 12.19rad/s
R2 (mD + mA )
(16)
(17)
(d) Use energy considerations to calculate the angular velocity of the cylindrical drum at the instant the
anchor hits the water, and compare your answer to that obtained in part ( 1c).
Solution
Since there is no friction, total mechanical energy of the system (cylindrical drum, cable, and anchor)
is conserved.
∆KT + ∆UT = 0
(18)
1
1
mA vf2 + I ωf2 = mA g h
2
2
(19)
Using I = mD R2 and Eq. 7, one obtains the following equation for the final angular velocity of the
cylindrical drum.
ωf =
s
R2
2 h mA g
= 12.19rad/s
(mD + mA )
(20)
Which is identical to the previous expression obtained for the angular velocity of the cylindrical drum
at the instant the anchor has fallen 16 m from rest.