Finding the mass of a pulley?

Question

By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.2 kg and m2 = 43.4 kg. The pulley can be treated as a uniform, solid, cylindrical disk. The downward acceleration of the 43.4 kg block is observed to be exactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley.

Image: http://www.webassign.net/CJ/p9-42alt.gif

Anybody have any idea how to do this?

I would appreciate any help.

Thank You.

Answer 1

Set up the force diagram and you'll see that
I*α = (T2 - T1)*R
where I = MR^2 / 2

T1 - m1*g = m1*g/2
T1 = 3*m1*g/2

m2*g - T2 = m2*g/2
T2 = m2*g/2

Also α = a/R = g/(2R)

MR^2 / 2 * g/(2R) = (m2 - 3*m1)*R*g/2

M = 2*(m2 - 3*m1)
= 13.6 kg

Answer 2

Here's the approach I would use:

1. Define some variables. Let T1 be the rope's tension on the m1 side; and let T2 be the tension on the m2 side.

2. Write "Fnet = ma" equations for each of the two masses (m1 and m2) separately. Note that, on each side, the net force consists of weight pulling down and tension pulling up.

(Using "up" as the positive direction):

Fnet1 = T1 − (m1)g = (m1)(a1)
Fnet2 = T2 − (m2)g = (m2)(a2)

3. The problem tells you that the acceleration of m2 is -g/2 (negative because it's "down"); so substitute:

Fnet2 = T2 − (m2)g = (m2)(-g/2)

And note that m1 must be accelerating UP at the same rate that m2 is accelerating DOWN; so substitue "g/2" for "a1":

Fnet1 = T1 − (m1)g = (m1)(g/2)

4. From the equations in Step 3, you should be able to solve for T1 and T2.

T1 = (m1)(g/2) + (m1)g = (m1)(3g/2)

T2 = (m2)(-g/2) + (m2)g = (m2)(g/2)

5. Note that T1 and T2 provide two different torques on the pulley.

Torque1 = (T1)(R) (counterclockwise)
Torque2 = (T2)(R) (clockwise)
(where "R" = pulley's radius (which we don't know!))

So the total torque in the clockwise direction is:

Τ = (T2−T1)R

6. Write an equation that relates the torque (Τ) to the pulley's angular acceleration (α):

Torque = (moment of inertia) × (angular acceleration)
Τ = Iα

From your book, you can finde that the moment of inertia ("I") for a solid cylinder is: I = MR²/2 (where M is the cylinder's mass). So:

Τ = (MR²/2)α

We already figured out that Τ = (T2−T1)R, so substitute:

(T2−T1)R = (MR²/2)α

6. We know that the rope is accelerating at rate g/2; this tells us something about how fast the pulley must be rotating. Write an equation for the pulley's angular acceleration (α), in terms of R (pulley's radius) and g/2.

α = a/R
α = (g/2)/R
α = g/(2R)

and subsitute for α in the previous equation:

(T2−T1)R = (MR²/2)(g/(2R))

Notice that the "R" now cancels out! So the above equation becomes:

(T2−T1) = (M/2)(g/(2)) = Mg/4

7. Substitute the values for T1 and T2 that we found before:

T1 = (m1)(3g/2)
T2 = (m2)(g/2)

(m2)(g/2) − (m1)(3g/2) = Mg/4

The "g" cancels out; so:

m2/2 − 3(m1)/2 = M/4

Now just solve for "M".