Physics 1a, Fall 2014
Quiz 4 Solutions
Problem 1
(a) (2 Points)
We know that by conservation of energy, Ugrav = Trot + TCM where Trot = 21 Iω 2 , TCM = 12 M v 2
and U = M gy is the potential. But v = ωr since the spool spins without slipping, so:
M gh =
1
(I + M r2 )ω 2
2
Now we also know that:
Z
Z
Z
I = r02 dm = r02 ρdV = 2 ∗ 2π
R
r
03
0
So we know that M gh = 14 M (2r2 + R2 )ω 2 so ω 2 =
4gh
2r 2 +R2 .
(1)
M/2
1
dr0 = M R2
2
πR
2
(2)
Then:
TCM = M gh
2r2
2r2 + R2
(3)
Trot = M gh
R2
+ R2
(4)
and
2r2
(b) (1 Point)
As we said in part a, since the spool is not slipping on the rail, v = ωr. Then:
s
r
4gh
4gh
=
v0 = r
2
2
2r + R
2 + (R/r)2
(5)
(c) (2 Points)
We know that the magnitude of the force of friction is given by |Ff | = |ṗ| = M |v̇| and the
magnitude of the torque due to friction is given by |τf | = |L̇| = I|ω̇| and also that |τf | = R|Ff |.
Combining these, we obtain:
I|ω̇| = M R|v̇|
(6)
Since some of the angular speed is essentially being transformed into translational speed, we
know that a decrease in ω will increase v so we may write:
I ω̇ = −M Rv̇
v̇ = −
(7)
R
ω̇
2
(8)
vf − v0 = −
R
R vf
v0
(ωf − ω0 ) = −
−
2
2 R
r
(9)
Quiz 4 Solutions – Fall 2014
2
vf =
2v0
R
2
R
1+
= v0
+
3
2r
3 3r
(10)
Note that for R = r, this equation gives vf = v0 , as it should.
Problem 2
(a) (1 Point)
We are given that the magnitude of the restoring torque is given by τ = −κθ. But we know
that τ = L̇ = I ω̇ = I θ̈.
So we then have that:
I θ̈ = −κθ
(11)
p
k/m, we see that
1
M
dr0 = M R2
πR2
2
(12)
so, comparing
with the equation mẍ = −kx which gives the frequency
p
ω0 = κ/I. Furthermore:
Z
I=
r02 dm =
Z
r02 ρdV = 2π
Z
R
r03
0
So:
r
ω0 =
2κ
M R2
(13)
(b) (2 Points)
We now assume a solution of the form θ(t) = Asin(ω0 t + δ) and solve for A, δ. One immediately
sees that δ = 0 from the θ(t = 0) = 0 condition. Furthermore, clearly θ̇ = Aω0 cos(ω0 t), so
θ̇(t = 0) = Aω0 . But we are given that θ̇(t = 0) = Ω0 , so:
A=
Ω0
ω0
(c) (1 Point)
p
The equation of motion will now be ω00 = κ/I 0 where this time I 0 = I + mr2 . So
r
r
κ
2κ
0
ω0 =
=
1
2
2
2
M
R
+ 2mr2
2 M R + mr
(14)
(15)
(d) (1 Point)
By the equations of motion:
θ0 = A0 sin(ω00 t + δ 0 ).
(16)
But θ0 must equal θ at t = π/ω0 , so θ0 (t = π/ω0 ) = 0. Then we can see that:
0 = A0 sin(ω00 π/ω0 + δ 0 )
(17)
which immediately give us that δ 0 = −ω00 π/ω0 .
Furthermore, given that there is no net torque when the putty falls and sticks on the disk,
angular momentum is conserved. Then:
I θ̇(t = π/ω0 ) = I 0 θ̇0 (t = π/ω0 )
(18)
where I 0 = I + mr2 . Therefore, this equation for θ̇ gives us:
−IΩ0 = I 0 A0 ω00 cos(ω00 π/ω0 − ω00 π/ω0 )
(19)
Quiz 4 Solutions – Fall 2014
3
0 I
This gives that A0 = − Ω
ω 0 I 0 . Or:
0
|A0 | =
Ω0
M R2
ω00 M R2 + 2mr2
(20)
Note that this solution makes sense in the asymptotic limits of M >> m, whereby the amplitude
will be the same as in part b, and also in the limit of m >> M , whereby the amplitude will
approach 0 (since dropping a very heavy object on a rotating disk will certainly dampen the
motion.)