Physics 1a, Fall 2014
Quiz 2 Solutions
Problem 1
(a) Let ML and lL be the mass and length of the left rod respectively.
The net torque should be zero. Evaluating the torques about point A:
Since the rod’s mass is distributed uniformly, the net torque due to gravity is from ML g acting
downwards at the half point of the rod. This is ML g Sinθ lL /2 directed into the paper.
The Normal force N is perpendicular to the rod, so the torque due to this is N lL but in the
opposite direction, i.e. out of the paper.
The torque due to friction Ff is zero as it is directed along the rod and when extended passes
through A. Similarly N1 passes through A and results in zero torque about A.
So, as the net torque is zero we have
ML g Sinθ lL /2 − N lL = 0
ML g Sinθ
N=
2
(1)
(2)
(b) Let MR and lR be the mass and length of the right rod respectively.
The net torque should be zero. Evaluating the torques about point B:
Since the rod’s mass is distributed uniformly, the net torque due to gravity is from MR g acting
downwards at the half point of the rod. This is MR g Cosθ lR /2 directed out of the paper.
The Normal force Ff is perpendicular to the rod, so the torque due to this is Ff lR but in the
opposite direction, i.e. into the paper.
Quiz 2 Solutions – Fall 2014
2
The torque due to the normal force N is zero as it is directed along the rod and when extended
passes through B. Similarly N2 passes through B and results in zero torque about B.
So, as the net torque is zero we have
MR g Cosθ lR /2 − Ff lR = 0
MR g Cosθ
Ff =
2
(3)
(4)
(c) For the sticks to stay in equilibrium (not slip) the friction Ff needs to be as given in Eq. 4. But
the maximum value Ff can attain is µ N where N is given by Eq. 2. So the rods won’t slip as
long as
Ff ≤ µ N
MR g Cosθ
µ ML g Sinθ
≤
2
2
MR Cosθ ≤ µ ML Sinθ
(5)
(6)
(7)
Now, as the two rods have the same linear mass density (say λ)
MR
λ lR
lR
=
=
= Cotθ
ML
λ lL
lL
(8)
(9)
Therefore we finally have
MR
Cotθ ≤ µ
ML
µ ≥ Cot2 θ
(10)
(11)
Quiz 2 Solutions – Fall 2014
3
Problem 2
(a) This solution considers the lab frame. In the rotating frame, the net force is zero as the mass
is always at rest but there is a pseudo force acting radially outwards. These two give the same
force equations.
In the lab frame then: As the mass moves in a uniform circle of radius R, the net acceleration is
m R ω 2 , directed towards the center. To get the minimum ωd for the mass to not slip down, we
need to consider the situation when the friction f is at its maximum value (µN ) and directed
“upwards” along the surface of the funnel.
The force equation along horizontal axis:
N Sinθ − f Cosθ = m R ω 2
(13)
= N (Sinθ − µ Cosθ)
(14)
N Sinθ − µ N Cosθ = m R
mR
ωd2
(12)
ωd2
The force equation along vertical axis:
N Cosθ + µ N Sinθ = m g
(15)
m g = N (Cosθ + µ Sinθ)
(16)
Dividing Eq. 14 by Eq. 16 we get:
s
ωd =
g(Sinθ − µ Cosθ)
R(Cosθ + µ Sinθ)
(17)
Quiz 2 Solutions – Fall 2014
4
(b) Once again in the lab frame: To get the maximum ωu for the mass to not slip up, we need to consider the situation when the friction f is at its maximum value (µN ) and directed “downwards”
along the surface of the funnel.
The force equation along horizontal axis:
N Sinθ + f Cosθ = m R ω 2
(19)
= N (Sinθ + µ Cosθ)
(20)
N Sinθ + µ N Cosθ = m R
mR
ωu2
(18)
ωu2
The force equation along vertical axis:
N Cosθ − µ N Sinθ = m g
(21)
m g = N (Cosθ − µ Sinθ)
(22)
Dividing Eq. 20 by Eq. 22 we get:
s
ωu =
g(Sinθ + µ Cosθ)
R(Cosθ − µ Sinθ)
(23)
(c) From the form of Eq. 17 we see that for a minimum ωd to exist (i.e. for it to be real) we need
Sinθ − µ Cosθ ≥ 0
(24)
(25)