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Physics 218 Honors
Exam 2
Please work 5 of the 6 problems, and cross out the one not to be graded.
1. A sawhorse is constructed of 5 identical boards as shown. Each board has a
length of 1 m. The angle between each pair of legs is 60o. How high above the
base is the center of mass? If the sawhorse were set up on an inclined plane as
shown, what is the maximum tilt angle θ for which the sawhorse would not
topple?
Solution: By symmetry, the center of mass lies along the vertical midplane of the
assembly. Each leg makes an angle of 60o with respect to the base. The cm of each leg
L sin 60°
therefore has a height hleg =
= 0.433m
2
The cm of the top board is htop = L sin 60° = 0.866m
The cm of the total assembly is then
4hleg + htop
hcm =
= 0.52m
5
The sawhorse can be tilted until its center of mass is directly above a pair of legs.
Beyond that point, the horse would topple.
hcm
tan α =
= 1.04
Let α be the angle between the cm and the base.
L cos 60°
α = 46°
The angle through which the sawhorse can be tilted is then 90° − a = 44° .
2. A bicycle has 1 m diameter wheels. It is traveling along a straight road, with a
linear velocity of 5 m/s and a linear acceleration of 1 m/s2. Calculate the vector
velocity and vector linear acceleration of dots on the wheel at points A, B, C, D, as
seen by a stationary observer on the side of the road. Give magnitude and direction in
each case in a clear manner.
Solution: The linear velocity is a vector sum of the translational velocity and the
rotational velocity. The acceleration is a vector sum of the translational acceleration,
the tangential acceleration, and the centripetal acceleration. The centripetal
acceleration is a c = v 2 / R = 50 m/s 2
r
)
v = (+5 − 5) x = 0
A: r
)
)
)
a = (+1 − 1) x + 50 y = 50 y m/s 2
B:
r
)
)
v = +5 x − 5 y m/s
r
) ) )
) )
a = +1x -1 y-50 x = - 49 x -1 y m/s 2
C:
r
)
)
v = ( +5 + 5) x = 10 x m/s
r
)
)
)
)
a = ( + 1 + 1 )x -50 y = +2 x -50 y m/s 2
D:
r
)
)
v = +5 x + 5 y m/s
r
) )
)
) )
a = +1x + 1 y + 50 x = 51x + 1 y m/s 2
3.
A uniform disk of mass 10 kg rolls without slipping up an inclined plane of angle
o
20 . Its initial velocity along the incline is 10 m/s. How far along the incline does the
disk roll before stopping? Please show a figure and neat work!
Solution:
ω
Consider forces parallel to the slope. As gravity slows the linear motion of the disk,
friction will act to slow the rotational motion as well; hence the friction force will point
up the slope. Newton’s law for forces gives
− mg sin 20° + F f = ma
Newton’s law for torques gives
− F f R = Iα
1
MR 2
2
Finally we require that there be no slipping: a = Rα.
I=
F f = Ma / 2
So
2
g sin 20° = 2.22 m/s 2
3
Now use the equation of motion along the slope:
1
x = v0 t − at 2
2
v = v 0 − at
a=
The disk comes to rest when v=0 →
t = v0 / a = 4.54 s
x = 22.7 m
4. A tetherball is hit so that it travels along a circular path around the pole as shown.
The ball has mass 1 kg, the rope is massless and has length 2 m. The linear
velocity of the ball is 2 m/s. What is its angular frequency, what is its moment of
inertia about the pole, and what is the tension in the rope? (you can treat the
tetherball as a point object.)
Solution: We must first find out what is the angle θ between the rope and the pole.
The circular orbit of the ball on a radius R = L sin θ requires a centripetal acceleration
v2/R in the horizontal plane. The direction in which the ball is free to move is the
circle of radius L about the pivot. We must use Newton’s law on the forces in this
direction of equilibrium:
v2
mg sin θ = m cosθ
R
Putting in the dependence of R on θ enables us to solve for R:
v2
g ( R / L) =
R
R4 =
(
 R2 
1 − 2 
L 

v4 2
L − R2
g2
1/ 2
)
2 2
4
v 4 − 1 + 1 + 4g L / v
= 0.74 m 2
2
g2
R = 0.86 m, θ = 64o.
R2 =
v
= 2.71 rad/s I = MR 2 = 0.93 kg m 2
R
Applying Newton’s law to the vertical forces on the ball yields
− mg + T cosθ = 0
T = 13.4 N
ω=
5. Calculate how high above the table one must hit a cue ball of radius R in order
that it roll without sliding as it leaves the cue. Assume the cue stick strikes the
ball horizontally. Show your work and make a figure.
Solution: Rolling without sliding requires that a = Rα.
The cue exerts both a force F and a torque τ = F h.
Newton’s law for force yields
F = ma
Newton's law for torque yields
τ = Fh = Iα
2
The moment of inertia for a uniform sphere is I = mR 2 .
5
Putting it all together,
2
h= R
5
The ball must be stuck at a height of 7/5 R above the table surface.
6. A spherical professor (M = 100 kg) has climbed to the top of a 5 m long ladder (m
= 20 kg, uniformly distributed along its length) that leans against a wall. The angle
between ladder and floor is 60o. There is no friction with the wall, and a static
friction coefficient of 0.55 with the floor. An angry student pulls horizontally on the
ladder at its base. How hard must she pull in order that the ladder begins to slide?
Solution:
Horizontal forces:
Vertical forces:
Fw = force of wall on top of ladder,
-Ff = friction of ladder on floor,
F = force that student uses to pull ladder to the left.
-Mg on professor,
-mg on ladder,
+F⊥ of floor on ladder.
Newton’s law of forces:
Fw − F f + F = 0
− Mg − mg + F⊥ = 0
We obtain F⊥ = 1176 N. The maximum friction force available is F f = µF⊥ = 647 N
Newton’s law of torques (take pivot at the bottom of ladder):
L
Fw L cos 60° − MgL sin 60° − mg sin 60° = 0
2
We obtain Fw = 622 N.
The additional force needed if the ladder is to slide is F = 25 N.