Question

A uniform cylindrical spool of mass M and radius R unwinds an essentially
massless rope under the weight of a mass m. If R = 12 cm, M = 400 gm and m = 50 gm, find the speed of m after it has descended 50 cm starting from rest.

Solve the problem twice: once using Newton's laws for torques, and once by application of energy conservation principles.

Ok, while the first part may have confused me, I may have gotten close to the answer:

I(spool) = 0.5MR^2
= 0.00288 kg m^2

Torque = R *(Fg + Ft - Ft)sin(angle)
= R * (mg)sin(90)
= Rmg
= -5.88 x 10^-2 N-m (I converted everything to meters and kg)

So then:
Torque = I * alpha
alpha = Torque/I

But since we want to find the speed of mass m after it has fallen 50 cm, alpha is kind of a useless value to us. So I used the conversion that

a(tangential) = alpha * radius
alpha = a(t)/radius

Torque/I = a(t)/radius
a(t) = radius * Torque/I
= -20.4 m/s

I then solved for the v(final) of mass m after 50 cm of falling using the equation:
v(f) = sqrt(vi^2 + 2*a(t) * delta(x))
and got -4.52 m/s

However, I am really confused about how to even approach the problem using energy conservation principles. How would you go about doing that at all?

Answer 1

Using energy principles,
(KE=kinetic energy, PE=gravitational potential energy)

KE+PE=constant, so
KEi+PEi = KEf+PEf

KEi=0 (both are stationary)
KEf=(1/2)Iω^2+(1/2)mv^2

PEi=0
PEf=0+mgh (weight descended)

Equate energies:
0 = (1/2)Iω^2+(1/2)mv^2 + 0+mgh

(note that h=-0.50
Everything is given, except v and ω
However, v and ω are related by
v=Rω, or ω=v/R
so we can solve for v in:

(1/2)[(1/2)MR^2](v/R)^2+(1/2)mv^2 + mgh = 0
Using
m=.05 kg,
M=0.4 kg
R=0.12 m
h=-0.5 m
g=9.81 m/s²
you will get v
(between 1.4 and 1.5 m/s)