Question

A block of mass m2 hangs from a rope. The rope wraps around a pulley of rotational inertia I, radius R, and then attaches to a second mass m1, which sits on a frictionless table as shown in figure. Derive an equation of the acceleration of the blocks when they are released?

Answer 1

You do this by looking at the forces on the block and the torque on the pulley, separately.

The torque on the pulley is do to the rope tension"T" pulling down tangent to the pulley. So the pulley motion is given by;

TR = I(aa)

R=radius of pulley; I=moment of inertia, aa = ang. accel.

The forces on the block are the rope tension "T" pulling up and the block's weight "mg" down. So the block's motion is given by;

mg - T = ma ; (down is positive)

If the rope doesn't slip the linear accel. "a" is related to the ang. accel. by;
a = Raa

Then;
mg - T = mRaa

Now T is the same on the block as the pulley, for a massless rope. So it can be elliminated from the two eqs.;

mgR - Iaa = mR^2aa

aa = mgR/(I + mR^2)



Now this aa can be put back into the torque eq. to find "T"


Alternatively, you could've elliminated aa from the two eqs and solved for "T";

T = mgI/[I + mR^2]

Answer 2

T1 = m1*a ...........................(1)

m2*g - T2 = m2*a .......................(2)

from (1) and (2)

m2*g + T1 - T2 = (m1+m2)*a................(3)

now we have

T2*r - T1*r = I*alpha

alpha = a/r

so T1 - T2 = - I*a/r^2 ................(4)

from (3) and (4)

m2*g - I*a/r^2 = (m1+m2)*a

a = m2*g / (m1 +m2 + I/r^2)