Download Exam II Spring 2011 with solutions

 Physics 218 Exam 2
Spring 2011, Sections 513-­‐515, 526, 528 Do not fill out the information below until instructed to do so! Name :____Solutions in RED_____ Signature :______________________________ Student ID :______________________________ E-­‐mail :______________________________ Section # :______________________________ Rules of the exam: 1.
2.
3.
4.
5.
You have the full class period to complete the exam. When calculating numerical values, be sure to keep track of units. You may use this exam or come up front for scratch paper. Be sure to put a box around your final answers and clearly indicate your work to your grader. Clearly erase any unwanted marks. No credit will be given if we can’t figure out which answer you are choosing, or which answer you want us to consider. 6. Partial credit can be given only if your work is clearly explained and labeled. 7. All work must be shown to get credit for the answer marked. If the answer marked does not obviously follow from the shown work, even if the answer is correct, you will not get credit for the answer. Put your initials here after reading the above instructions:__________ 1 Table to be filled by the graders Part 1: (25p) Forces and Dynamics Part Score Problem 1.1: A triangular shape object of mass m1 and aperture angle α is Part 1
(
25) between a wall and square block of mass m2 as shown in the figure below. Part 2 (25) There is friction between the wall and the triangular shape object with Part 3 (25) coefficient uK, and there is no friction anywhere else. As the triangular Part 3 (25) shape object wedge its way down the Bonus (10) block gets displaced in the horizontal Exam Total µk Y position. Question 1.1.1: (5p) As a first step, draw a coordinate system and based on the geometry derive an expression g m1 between the vertical acceleration of the triangular object and the horizontal acceleration of the square block. α
m2 It is clear that !x=-!y tan(! ), " ax = #ay tan(! ) X
Question 1.1.2: (5p) Draw the free body diagram of both the triangular shape object and the square block, properly indicating relevant angles. Fk N2 α
m1 α
w1 N3 Nw
w
m2 α
N2 w2 Question 1.1.3: (15p) Using Newton’s laws find the acceleration of the square block in the horizontal direction.
From the free body diagram's above we obtain:
x̂ : N w ! N 2 cos(! ) = 0
" N w = N 2 cos(! )
ŷ :
Fk ! m1g + N 2 sin(! ) = m1ay
x̂ :
N 2 cos(! ) = m2 ax
ŷ :
N 3 ! N 2 sin(! ) ! m2 g = 0
" µ k N 2 cos(! ) ! m1g + N 2 sin(! ) = m1ay = !m1
ax
tan(! )
Where we've used ax = !ay tan(! ) in the 2nd equation (right). We are left with a system of 4 equations
and 4 unknown variables (N 2 , N 3, N w , ax ). Replacing N w and N 2 from the first and third equations into
the second equation we get:
µ k m2 ax ! m1g + m2 ax tan(! ) = !m1
ax
ax
" µ k m2 ax + m2 ax tan(! ) + m1
= m1g
tan(! )
tan(! )
#
m1 &
m1g
ax % µ k m2 + m2 tan(! ) +
( = m1g " ax =
tan(! ) '
$
m2 (µ k + tan(! )) +
m1
tan(! )
=
m1g tan(! )
m2 tan(! )(µ k + tan(! )) + m1
2 Part 2: (25p) Equilibrium Problem 2.1: A block of mass m is hanging from two ropes that are connected to a single rope, which in turn is attached to the wall as shown in the figure below. The angles of the ropes α, β, 90° are shown in the picture and are known. Question 2.1.1: (5p) very clearly draw the relevant g free body diagrams including relevant angles and force Y names. β
3-­‐rope point: block: T3 α
m β
α
X
α
T1 T1 T2 w T2 Nw Question 2.1.2: (15p) Find the tension in the rope at an angle α as a function of known parameters. T1 + T2 sin(! ) ! mg = 0
NW ! T2 cos(! ) = 0 " NW = T2 cos(! )
T2 cos(! ) ! T3 sin(" ) = 0 " T3 = T2
T3 cos(" ) ! T1 ! T2 sin(! ) = 0
cos(! )
sin(" )
" T2
cos(! )
cos(" ) ! T1 ! T2 sin(! ) = 0
sin(" )
# cos(! )
&
" T2 %
! sin(! )( = T1
$ tan(" )
'
Plugging the last equation into the first one we get:
# cos(! )
&
# cos(! )
&
mg tan(" )
T2 %
! sin(! )( + T2 sin(! ) ! mg = 0 " T2 %
! sin(! ) + sin(! )( = mg " T2 =
cos(! )
$ tan(" )
'
$ tan(" )
'
Question 2.1.3: (5p) Find the normal force that the wall needs to withstand so that the block does not sink into it. From the previous point we obtained NW = T2 cos(! ), therefore
Nw =
mg tan(" )
cos(! ) ! N w = mg tan(" )
cos(! )
3 Part 3: (25p) Work and Energy Problem 3.1: A block of mass m is at the bottom of an inclined plane of length d and angle α . The block is being pulled up by a special winch located at the top of the plane and at a distance L from the top of the plane as shown in the diagram below. There is friction with coefficient of kinetic friction µ k only in the inclined portion of the plane, and no friction at the top of the plane. The winch provides a constant force T to the rope and will work until the block reaches the winch. Question 3.1.1: (5p) Find the work of the L g tension force on the block done in bringing it all the way up to the winch. d Since the rope is always in the direction of the displacement we get
rom the previous point we obtained NW = T2 cos(! ), therefore
Y WT = Td + TL = T (d + L)
m
µk
α
X Question 3.1.2: (5p) Find the work of the friction force on the block done in bringing it all the way up to the winch. Since the friction force is always against direction of the displacement and
we only have friction force at the inclined portion of the plane we get
Wfriction = !Fd = !µ k Nd.
The normal force is obtained from a free body diagram a N=mgcos(! ) so we get
Wfriction = !Fd = !µ k mg cos(! )d.
Question 3.1.3: (15p) Using the work energy theorem find the velocity of the block at the moment it hits the winch. Using a coordinate system in which the Y axis goes up from zero at the ground we get:
mv 2f
mv 2
= T (d + L) ! µ k mg cos(! )d = mgd sin(! ) + f "
2
2
2
mv f
2T (d + L)
= T (d + L) ! µ k mg cos(! )d ! mgd sin(! ) " v f =
! µ k g cos(! )d ! gd sin(! )
2
m
WT + W friction = mgh +
" vf =
2T (d + L)
! gd (µ k cos(! ) ! sin(! ))
m
4 Part 4: (25p) Down the ramp Problem 4.1: A skier is at the top of a circular hill of radius R with horizontal velocity v0. Question 4.1.1: (5p) Knowing that there is no friction use the work energy theorem to express the velocity of the skier as a g function of R, v0 and the angle of the skier θ . v0 Using a coordinate system with vertical axis going up and zero at the center of the circle
r mv02
mv 2
= mgR cos(! ) +
!
2
2
mv 2
mv 2
mv 2
= mgR+ 0 " mgR cos(! ) = 0 + mgR(1" cos(! )) ! v 2 = v02 + 2gR(1" cos(! ))
2
2
2
mgR+
R θ θ
Skier: Question 4.1.2: (3p) Write a useful coordinate system and the N free-­‐body diagram of the skier with relevant angles. r θ
θ
W Question 4.1.3: (2p) In term of forces what happens when the skier looses contact with the hill? The normal force is zero. Question 4.1.4: (10p) Find the angle θ ’ at which the skier looses contact with the hill as a function of g, R and v0. Using the coordinate system depicted above we get in the r̂ direction:
v2
R
At the angle ! ' in which the skier lose contact with the ground we have N=0, and therefore
N ! mg cos(! ) = !m
2
2
Rg cos(! ') = v and using the v obtained from conservation of energy we get
Rg cos(! ') = v02 + 2Rg(1! cos(! ')) " Rg cos(! ') + 2Rg cos(! ') = v02 + 2Rg " 3Rg cos(! ') = v02 + 2Rg "
cos(! ') =
v02 2
+
3Rg 3
Question 4.1.5: (5p) What is the minimal velocity v0 at which the skier stops touching the ground even at the top of the hill ? If the skier is to stop touching the ground almost at the top of the hill then ! =0 and therefore cos(! )=1.
Plugging this into the previous equation we get:
cos(! ') = 1 =
# 2&
v02 2
+ ! v02 = 3Rg %1" ( = Rg ! v0 = Rg
$ 3'
3Rg 3
5