Physics 110 Homework Solutions Week #8 – Tuesday
10/29 and Wednesday10/30
Tuesday October 29, 2013
Chapter 7
Questions
- none
Multiple-Choice
- none
Problems
7.3 The kinetic energy is
.
7.27 A physics demo
a. The angular velocity is
.
b. We first need to determine the angular acceleration and then calculate the torque
from Newton’s 2nd law in rotational form. The angular acceleration is
. Therefore the torque is
.
c. If the angular velocity remains constant then the time it takes is
.
d. The total kinetic energy of the rolling wheel is the sum of the translational and
rotational kinetic energies. Thus we have
.
7.35
To find the distance where the board tips, measure the distance the cat walks from
the edge of the table toward the left. We take the sum of the torques (equal to
zero just before the board tips) and we have
. Thus
the cat can walk 25cm from the edge of the table out on the plank before it tips.
7.39 A ball at the end of a massless rod
a. If the rod is assumed massless then the initial torque is given by
.
b. The initial angular acceleration is given from Newton’s 2nd law in rotational form.
We have
.
c. Applying conservation of energy we have (taking the zero of the gravitational
potential energy to be when the ball is at its lowest point and starting the ball from
rest)
. Since
we can solve for the
angular velocity of the ball at its lowest point and we find
.
d. The angular momentum at the lowest point is
7.43 The angular momentum is the product of the angular velocity and the moment of
inertia of the system. The moment of inertia is the sum of the two point mass caps
and the rod spun about its center. We have for the moment of inertia of the system
.
Therefore the angular momentum is
.
7.51 A ladder against a wall
a. Answers will vary.
b. The free-body diagram is shown below.
c. Using Newton’s 2nd law and the free-body diagram we have
d. Applying Newton’s 2nd law in rotational form we have
FN,wallg
O
person
mp g
F
g
floor N,floor
θ
mL g
Ffr
wall
Wednesday, October 30, 2013
Chapter 7
Questions
- none
Multiple-Choice
7.21 A
7.22 A
Problems
7.18 A space station
a. The centripetal acceleration is
toward the center of the space
station.
b. The linear velocity is tangent to the outer rim and has a magnitude of
.
c. The rotational kinetic energy of the space station is
.
d. Here we have an inelastic collision and momentum is conserved. Applying
conservation of momentum we have
.
e. Applying conservation of angular momentum we can calculate the final rotational
speed of the combined system. We have
7.19
The world’s largest yo-yo
a. The number of radians is given by
.
b. After the yo-yo has fallen 10m its angular velocity is given by conservation of
energy. Thus we have
.
c. The translational velocity of the yo-yo is given by
vertically down.
7.20 Two cylinders
a. By applying conservation of angular momentum during the collision we have
b. The rotational kinetic energies before and after the collision are given by
and
. Taking the ratio of the final to initial
kinetic energy we see that
, which gives KEf as a
fraction of KEi, or the kinetic energy decreases.
c. The kinetic energy decreases due to friction between the two surfaces brining the
two cylinders to the same rotational speed.
7.21
In order to determine the new angle that the pumpkin leaves we apply
conservation of energy between the top of the dome and when the pumpkin loses
contact with the surface. Here, in contrast to problem 5.28, we need to include
the rotational kinetic energy of the pumpkin. From the free body diagram we
have
, where we
have set FN = 0 when the pumpkin leaves the surface. In order to
find the angle we need to use conservation of energy:
. The initial energy is mgR
(taking the dome level as the zero of the potential energy) and the
final energy when the pumpkin leaves the surface is
+
2
½ mv + mgh, where h = Rcosθ is the height of the pumpkin
above the base of the dome. Therefore we have
R
and
FN
θ
FW
cancelling the common factors of the mass and radii of the pumpkin and the dome
and the pumpkin leaves at a larger
angle 54o compared to when it simply slides and loses contact at 48.2o.
7.22 The gravitational collapse of a star
a. The initial magnitude of the Sun’s angular velocity is
.
b. The angular velocity of the pulsar is obtained from
c. Assuming that the Sun does not lose any mass during the collapse, we find the
radius of the neutron star formed by applying conservation of angular momentum.
We have
d. We calculate the change in the rotational kinetic energy of the Sun to be
So there is a change in the rotational kinetic energy of the Sun and this is due to
gravity collapsing the Sun to form a neutron star.
7.23 A rotating space ship
a. The linear speed of a person is found from the centripetal acceleration. Thus we
have
.
b. Since we know the linear velocity the angular velocity can be determined from
.
c. The angular momentum of the ship is
d. Applying conservation of angular momentum we can solve for the final angular
speed. We have
.
e. The new radial acceleration is
.
7.25 A student on a stool
a. Applying conservation of angular momentum we have
b. The kinetic energy before is
.
while after is