Homework 5 Due: 7:30pm on Tuesday, June 2, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy
Exercise 9.16 At a grind grinding ing whee wheell has an angular velocit velocity y of 30.0 . It has a constant angular angular acceleration of 30.0 until a circuit bre breaker aker trips at time = 2.50 2.50 . From From then on, it turns turns through through an ang angle le 433 as it coasts coasts to a stop at constant angular angu lar acceleration.
Part A Through what tot total al angle did the wheel turn between
and the ti time me it st stopped? opped?
ANSWER: =
602 60 2
Correct
Part B At what time did it stop? ANSWER: =
10.7 10 .7
Correct
Part C What was its acceleration as it slowed down? ANSWER: =
-12.7
Correct
Problem 9.74 A sphere consists c onsists of a solid woo wooden den ball of uniform density 800 of le lead ad fo foil il with are rea a de dens nsity ity 20 .
Part A
and radius 0.30
and is covered with a thin coating
Calculate the moment of inertia of this sphere about an axis passing through its center. ANSWER: =
4.61
Correct
Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is through point p, and
, where is the perpendicular distance from the center of mass to the axis that passes is the mass of the object.
Part A Suppose a uniform slender rod has length
and mass
. The moment of inertia of the rod about about an axis that is
perpendicular to the rod and that passes through its center of mass is given by
. Find
, the
moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express
in terms of
and
. Use fractions rather than decimal numbers in your answer.
Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod. ANSWER: =
Correct
ANSWER: =
Correct
Part B Now consider a cube of mass
with edges of length
. The moment of inertia
its center of mass and perpendicular to one of its faces is given by about an axis p through one of the edges of the cube
of the cube about an axis through . Find
, the moment of inertia
Express in terms of and . Use fractions rather than decimal numbers in your answer.
Hint 1. Find the distance from the Find the perpendicular distance
axis to the
axis
from the center of mass axis to the new edge axis (axis labeled p in the figure).
ANSWER: =
ANSWER: =
Correct
Kinetic Energy of a Dumbbell This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy of a dumbbell of mass when it is rotating with angular speed and its center of mass is moving translationally with speed . Denote the dumbbell's moment of inertia about its center of mass by . Note that if you approximate the spheres as point masses of mass each located a distance from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by , but this fact will not be necessary for this problem.
Part A Find the total kinetic energy
of the dumbbell.
Express your answer in terms of
,
,
, and
.
Hint 1. How to approach the problem Compute separately the rotational and translational kinetic energies of the dumbbell. Then add the two to find the total kinetic energy.
Hint 2. Find the rotational kinetic energy What is the dumbbell's rotational kinetic energy
?
Give your answer in terms of some or all of the variables
,
, and
.
Hint 1. Formula for rotational kinetic energy The formula for the rotational kinetic energy rotating with angular velocity about it, is
of a body with moment of inertia
.
ANSWER: =
Hint 3. Find the translational kinetic energy What is the dumbbell's translational kinetic energy
?
Give your answer in terms of some or all of the variables ANSWER:
,
, and
.
about some axis,
=
ANSWER: =
Correct
Part B The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic energy term is called the kinetic energy of the center of mass. You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell. It is important to understand the applicability of the formula necessary for the formula to be valid?
. Which of the following conditions are
Check all that apply. ANSWER: The velocity vector
must be perpendicular to the axis of rotation.
The velocity vector
must be perpendicular or parallel to the axis of rotation.
The moment of inertia must be taken about an axis through the center of mass.
Correct
Exercise 10.3 A square metal plate 0.180 plate .
on each side is pivoted about an axis through point
at its center and perpendicular to the
Part A Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are = 24.0 , = 14.6 , and = 16.7 . The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise. ANSWER: =
1.28
Correct
Exercise 10.13 A textbook of mass 1.95 rests on a frict ionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.160 , to a hanging book with mass 3.07 . The syst em is released from rest, and the books are observed to move a distance 1.15 over a time interval of 0.750 .
Part A What is the tension in the part of the cord attached to the textbook? ANSWER: 7.97 N
Correct
Part B What is the tension in the part of the cord attached to the book? Take the free fall acceleration to be
= 9.80
.
ANSWER: 17.5 N
Correct
Part C What is the moment of inertia of the pulley about its rotation axis? Take the free fall acceleration to be
= 9.80
.
ANSWER: 1.50×10−2
Correct
Exercise 10.45 A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 and a total mass of 140 . The turntable is initially rotating at 4.00 about a vertical axis through its center. Suddenly, a 75.0parachutist makes a soft landing on the turntable at a point near the outer edge.
Part A Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) ANSWER: =
1.93
Correct
Part B Compute the kinetic energy of the system before the parachutist lands. ANSWER: =
2240
Correct
Part C
Compute the kinetic energy of the system after the parachutist lands. ANSWER: =
1080
Correct
Part D Why are these kinetic energies not equal? ANSWER: 3701 Character(s) remaining Friction does negative work in order to keep the parachutist on the turntable, which slows it down.
Submitted, grade pending
Exercise 11.30 A vertical solid steel post of diameter = 25 and length = 2.20 is required t o support a load of mass You can ignore the weight of the post. Take free fall acceleration to be .
Part A What is the stress in the post? Express your answer using two significant figures. ANSWER: = 1.5×106
Correct
Part B What is the strain in the post? Express your answer using two significant figures. ANSWER: = 7.5×10−6
= 7500
.
Correct
Part C What is the change in the post's length when the load is applied? Express your answer using two significant figures. ANSWER: = 1.6×10−5
Correct
Video Tutor: Walking the plank First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.
Part A In the video, the torque due to the mass of the plank is used in the calculations. For this question, ignore the mass of the board. Rank, from largest to smallest, the mass needed to keep the board from tipping over. To rank items as equivalent, overlap them.
Hint 1. How to approach the problem In equilibrium, the sum of the clockwise torques about an axis must equal the sum of the counterclockwise torques.
ANSWER:
Correct
Precarious Lunch A uniform steel beam of length and mass is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle , as shown. Through the hinge, the wall exerts an unknown force, , on the beam. A workman of mass sits eating lunch a distance from the building.
Part A Find
, the tension in the cable. Remember to account for all the forces in the problem.
Express your answer in terms of
,
,
,
,
, and , the magnitude of the acceleration due to gravity.
Hint 1. Pick the best origin This is a statics problem so the sum of torques about any axis a will be zero. In order to solve for , you want to pick the axis such that will give a torque, but as few as possible other unknown forces will enter the equations. So where should you place the origin for the purpose of calculating torques? ANSWER: At the center of the bar At the hinge At the connection of the cable and t he bar Where the man is eating lunch
Hint 2. Calculate the sum torques Now find the sum of the torques about center of the hinge. Remember that a positive torque will tend to rotate objects counterclockwise around the origin. Answer in terms of
,
,
,
,
,
, and
.
ANSWER:
= 0 =
ANSWER:
=
Correct
Part B Find , the -component of the force exerted by the wall on the beam ( attention to the direction that the wall exerts the force. Express your answer in terms of
), using the axis shown. Remember to pay
and other given quantities.
Hint 1. Find the sign of the force The beam is not accelerating in the -direction, so the sum of the forces in the given coordinate sys tem, is going to have to be positiv e or negative?
-direction is zero. Using the
ANSWER: =
Correct
Part C Find , the y-component of force that the wall exerts on the beam ( attention to the direction that the wall exerts the force. Express your answer in terms of
,
,
,
), using the axis shown. Remember to pay
, and .
ANSWER: =
Correct If you use your result from part (A) in your expression for part (C), you'll notice that the result simplifies somewhat. The simplified result should show that the further the luncher moves out on the beam, the lower the magnitude of the upward force the wall exerts on the beam. Does this agree with your intuition?
Problem 11.76 You are trying to raise a bicyc le wheel of mass force
and radius
up over a curb of height
. To do this, you apply a horizontal
.
Part A What is the least magnitude of the force at the center of the wheel? ANSWER:
that will succeed in raising the wheel onto the curb when the force is applied
=
Correct
Part B What is the least magnitude of the force at the top of the wheel?
that will succeed in raising the wheel onto the curb when the force is applied
ANSWER:
=
Correct
Part C In which case is less force required? ANSWER: case A case B
Correct
Problem 11.83 A garage door is mounted on an overhead rail (the figure ). The wheels at A and B have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.52. The distance between the wheels is 2.00 , and each is 0.50 from the vertical sides of the door. The door is uniform and weighs 871 . It is pushed to the left at constant speed by a horizontal force .
Part A If the distance
is 1.41
, what is the vertical component of the force exerted on the wheel A by the track?
is 1.41
, what is the vertical component of the force exerted on the wheel B by the track?
ANSWER: =
116
Correct
Part B If the distance ANSWER: =
755
Correct
Part C Find the maximum value
can have without causing one wheel to leave the track.
ANSWER: =
1.92
Correct Score Summary: Your score on this assignment is 99.3%. You received 22.85 out of a possible total of 23 points.
