Challenge Problems:
Yo-Yo
A yo-yo is pulled by its string on the floor without slipping. Depending on the angle of
the string with respect to the horizontal, the yo-yo will travel towards the puller or away
from the puller. Find the angle from the horizontal the string has to be pulled which
demarcates the forward movement and backward movement.
F
r1
r2
!
r
Solve for the acceleration, either angular or translational.
Solution 1:
Consider rotation to occur about a pivot at the contact point with the horizontal. Let F be the
pulling force on the string. The torque about the that point is given by r x F.
Using some simple trig we can express the vectors for r and F in terms of rectangular
coordinates.
�r =r1 sin θî + (r2 − r1 cos θ) ĵ
F� =F cos θî + F sin θ ĵ
\vec r =& r_1 \sin \theta \hat i + (r_2-r_1\cos \theta) \hat j \\
\vec F =& F \cos \theta \hat i + F \sin \theta \hat j
Now just do the cross product to get the torque and simplify. ( k̂ would be out of the paper.)
�τ =�r × F� = (xFy − yF x )k̂
=[r1 sin θF sin θ − (r2 − r1 cos θ)(F cos θ)]k̂
=(r1 F sin2 θ − r2 F cos θ + r1 F cos2 θ)k̂
=(r1 F − r2 F cos θ)k̂
\vec \tau =& \vec r \times \vec F = (x F_y - y F_x) \hat k \\
=& [r_1 \sin \theta F \sin \theta - (r_2 -r_1 \cos \theta)(F \cos \theta)] \hat k \\
=& (r_1 F \sin^2 \theta - r_2 F \cos \theta + r_1 F \cos^2 \theta) \hat k\\
=& (r_1 F - r_2 F \cos \theta) \hat k
Let the moment of inertia of the yo-yo be expressed in terms of the outer radius:
Icm = cmr22
Where c is just a constant less than one depending on the mass distribution of the yoyo.
Using the parallel axis theorem to get the I about the pivot point at the contact between
the surface and the rim:
I = (c + 1)mr22
Thus the angular acceleration is
α=
F(r1 − r2 cos θ)
(c + 1)mr22
\alpha = \frac{ F(r_1
- r_2
\cos \theta) }{(c+1)mr_2^2}
Here positive α would correspond to rolling to the left, away from the puller. This occurs
when r1/r2 is greater than cos θ.
Solution 2:
F
r1
f
r2
θ
r
Consider the rotation about the centre of mass and let the static friction force be
represented by f.
The net force on the object on the c.m. is
Fnet = F cos θ − f = ma
The net torque about c.m. is (magnitude only)
τ = Fr1 − f r2 = Icm α
\tau = fr_2 - Fr_1 = I_{\rm cm} \alpha
Substituting for f and I we get
τ = Fr1 − (F cos θ − ma)r2 = cmr22 (−a/r2 )
\tau = Fr_1 -(F\cos \theta -ma )r_2 = cmr_2^2(- a/r_2)
\tau = (F\cos \theta -ma )r_2 - Fr_1 = cmr_2^2 (-a/r_2)
Notice that α = −a/r2 in order for the signs to be consistent.
Solving for a
a=−
a
F(r1 − r2 cos θ)
(c + 1)mr2
= -\frac{F(r_1 - r_2 \cos \theta)}{(c+1)mr_2}
This is consistent with solution 1. When cos θ is greater than r1/r2 then a is positive.
Double Atwoodʼs Machine:
The two pulleys in the diagram shown below are identical and are uniform disks
of mass m. Find an expression for the linear acceleration of the 4m mass.
2m
T1
m
m
m
2m
m
4m
T2
4m
2mg
is equivalent to
4mg
First apply Newtonʼs second law to the three objects assuming that the acceleration of
the 2m mass is equal in magnitude and opposite in direction to that of the 4m mass.
−2ma =T 1 − 2mg
4ma =T 2 − 4mg
Iα =T 1 R − T 2 R
-2ma =& T_1 - 2 mg \\
4ma =& T_2 - 4mg \\
I \alpha =& T_1R - T_2R
The last equation needs help by expressing I and α in terms of m and a, then
simplifying.
a
1
(2mR2 ) =(T 1 − T 2 )R
2
R
ma =(T 1 − T 2 )
\frac{1}{2} (2mR^2 )\frac{ a}{R} =&( T_1 - T_2 )R\\
ma =&( T_1 - T_2 )
We have 3 equations and 3 unknowns. Eliminate the Tʼs and solve for a:
−6ma =(T 1 − T 2 ) + 2mg
−6ma =ma + 2mg
-6ma =& (T_1 - T_2) + 2mg \\
-6ma =& ma + 2mg
2
a=− g
7 .
Bouncing Ball
A ball bounces on the floor repeatedly. At each bounce thereʼs a change in momentum.
What is the average force on the ball from the floor during the multiple bounces?
Assume that each bounce the kinetic energy after the bounce is η times that before the
bounce where η <1.
y
h
!h
vi
vf = –vi"!
t
The average force is equal to the impulse divided by the time interval between bounces.
Measure the time interval between the two highest points of the bounce. The starting
height is h and by conservation of energy the next highest point it reaches is ηh.
∆t = tfall + trise
�
�
= 2h/g + 2ηh/g
�
√
= 2h/g(1 + η)
\Delta t =&~ t_{\rm fall} + t_{\rm rise}\\
=& \sqrt{2h/g} + \sqrt{2\eta h/g}\\
=& \sqrt{2h/g}(1+\sqrt{\eta})
The change in momentum is equal to the impulse:
∆p = pf − pi
= mvf − mvi
= − mηvi − mvi
�
�
= 2gηh + 2gh
�
√
= 2gh(1 + η)
\D
Now\Delta p =&~ p_{\rm f} - p_{\rm i}\\
=&~ m v_{\rm f}-mv_{\rm i}\\
=& -m \eta v_{\rm i}-mv_{\rm i}\\
=& \sqrt{2g\eta h} +\sqrt{2g h}\\
=&\sqrt{2g h}(1+\sqrt{\eta})
Finally divide impulse by time interval to get the average force.
∆p
∆t
�
√
m 2gh(1 + η)
= �
√
2h/g(1 + η)
Favg =
=mg
F_{\rm avg} =& \frac{\Delta p}{\Delta t}\\
=& \frac{\sqrt{2g h}(1+\sqrt{\eta})}{\sqrt{2 h/g}(1+\sqrt{\eta})}\\
=& mg
A not completely unexpected result.