Download Physics 1307 Practice Quiz 7 Chapter 10 3.6cm2 1/2

Physics 1307 Practice Quiz 7
Chapter 10
Problem I : Bone Fracture
A compressive force of 3.6×10 4 N is exerted on the end of a 20­cm­long bone of cross sectional area of 3.6 cm2 , (a) will the bone break, and (b) if not, by how much does it shorten?
6
( The ultimate compressive strength of bone is about 170×10
9
order of 15×10
N
m
2
N
m
2
and the Young modules of bone is of the )
Problem II: Jane to the rescue
Tarzan stands on a branch as a leopard threatens. Fortunately, Jane is on a nearby branch of the same hight holding a 25­m­long vine attached directly above the point midway between her and Tarzan. How soon does she reach Tarzan.
Problem III: Piston motion
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression

x=5.00cm cos2t  where x is in centimeters and t is in seconds. At t=0, (a) find the position of the 6
piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion. SOLUTIONS
Problem I
To know if the bone is going to break we need to find if the stress is bigger than the ultimate compressive strength of bone. This is easily done:
F 3.6×10 4 N
N
comp = =
=108 2
−4 2
A 3.6 x 10 m
m
6
which is less than 170×10
N
m
2
. Thus the bone will not break. To find how much shortens we need to use the definition of the Young's modulus: Y =
F/A
 l/l o
. Solving for  l we find:
1/2
 l=
F l0
AY
=
3.6×10 4 N 0.20 m
9
15×10
N
m2
3.6×10 m
−4
=0.0013 m=1.3 mm
2
Problem II
This problem is about a pendulum, which we identify as consisting of Jane and the vine. The period of the pendulum is the time for a full swing back and forth, so the answer we are after is half the period. Then:
1
25 m
L
T= 2
=
=5.0 s 2
2
2
g
9.8 m/ s
1
 
Problem III
Note that to obtain the correct values of the trigonometric functions you need to work with the angles given in radians. 
(a) The position at any time is x=5.00 cmcos 2t 6  . At t=0, the position of the piston is given by 
x t=0 =5.00 cmcos =4.33cm , i.e we just evaluated the function in the particular time required by the 6
problem.
(b) The velocity is given by the derivative of the position with respect to time:

vt =−5.00 cm2Hzsin2t 6  .
Thus the velocity at time t=0, is equal to:

v t=0 =−5.00 cm2Hzsin  6 =−8.66 cm/ s .

2
(c) Similarly for acceleration we have a t=− 5.00 cm2Hz cos 2t 6  and therefore 
a t=0 =−5.00 cm2Hz2 cos  =−17.32 cm/ s2 .
6
(d) The amplitude is given by the coefficient of the trigonometric function in x(t). Then A=5.00 cm. The period is 2 2
=s . (Remember that the angular frequency  is given by the factor that multiplies the just T =  =
2 Hz
time t in the argument of the trigonometric function)
2/2