6. (25 pts) A spring of spring constant kis attached to a point on the outer rim of a solid disc (moment
of inertia I= (1/2)M R2) with mass mand radius R. The disc can spin freely about an axis through its
center. As the disc rotates, the spring stretches, as shown below.
R
k
R
k
θ
Unstretched Stretched
a) (5 pts) By what distance does the point at which the spring is attached to the disc move when the
disc rotates by an angle θ?
s=Rθ
b) (5 pts) After rotating by angle θ, how much force does the spring apply to the disc (written in terms
of the angle)?
The spring is now stretched by the amount listed in part (a), so
F=−kx =−kRθ
c) (5 pts) What is the torque applied to the disc by the spring at this point?
τ=F r sin φ= (−kRθ)(R) = −kR2θ
Note that the force exerted by the spring at the point of attachment is tangent to the edge of the disc,
thus φ= 90◦.
d) (5 pts) Using “Newton’s 2nd Law for Rotation,” write down the (differential) equation of motion
for this system (but do not sol ve it). Write it so that all variables relating to the rotation of the disc
(angular position, angular velocity, and angular acceleration) are in terms of θ. (Hint: First write down
Newton’s 2nd Law for Rotation in general, then substitute for each of the pieces from your answers
above.)
τ=Iα
−kR2θ=1
2MR2d2θ
dt2
d2θ
dt2+2k
Mθ= 0
e) (5 pts) From your equation of motion, what should be the angular frequency of the motion if the disc
is now released?
The value of ωwas taken from the coefficient of the xor θin previous examples. The coefficient in this
case is 2k/M. Thus
ω=r2k
M