Solving Connected Masses Pulley Problem

This physics problem involves two masses connected by a string over a rotating pulley. We need to find the speed of the masses after a specific descent. We can solve this using the principle of conservation of mechanical energy.

Energy Conservation Approach

The total mechanical energy (potential + kinetic) of the system is conserved, minus the work done by non-conservative forces (which are assumed negligible here). The kinetic energy includes both the translational kinetic energy of the masses and the rotational kinetic energy of the pulley.

Initial State (Rest)

Let's set the reference level for potential energy at the initial position of the 2m mass. Initially, the system is at rest, so all kinetic energy is zero.

• Initial Potential Energy ($PE_i$): The m mass is at a height h above the 2m mass. Let the initial height of 2m be 0. So, $PE_i = mgh$.
• Initial Kinetic Energy ($KE_i$): $KE_i = 0$.

Final State (After 3.6 m Descent)

The 2m mass descends by $h = 3.6$ m. The m mass ascends by the same height. Let the final speed of the masses be v.

• Final Potential Energy ($PE_f$): The 2m mass is now at height 0. The m mass is at height 2h relative to the initial 2m position. $PE_f = m g (2h)$. However, it's simpler to consider the *change* in potential energy. The potential energy decreases by $(2m)gh$ and increases by $mgh$. Net change = $-mg h$.
• Final Kinetic Energy ($KE_f$): This includes translational kinetic energy of both masses and rotational kinetic energy of the pulley.
  • Translational KE: $KE_{trans} = \frac{1}{2}mv^2 + \frac{1}{2}(2m)v^2 = \frac{3}{2}mv^2$.
  • Rotational KE: $KE_{rot} = \frac{1}{2}I\omega^2$.
  The pulley is a disc, so its moment of inertia is $I = \frac{1}{2} M_{pulley} r^2$. Given $M_{pulley} = 30m$, $I = \frac{1}{2} (30m) r^2 = 15mr^2$. Since the string does not slip, the angular velocity $\omega$ is related to the linear velocity $v$ by $\omega = v/r$. Substituting this into rotational KE: $KE_{rot} = \frac{1}{2} (15mr^2) (\frac{v}{r})^2 = \frac{1}{2} (15mr^2) \frac{v^2}{r^2} = \frac{15}{2}mv^2$. Total Final KE: $KE_f = KE_{trans} + KE_{rot} = \frac{3}{2}mv^2 + \frac{15}{2}mv^2 = \frac{18}{2}mv^2 = 9mv^2$.

Applying Conservation of Energy

The loss in potential energy equals the gain in kinetic energy.

Change in PE = Gain in KE

The net potential energy lost is when the heavier mass ($2m$) descends by $h=3.6$ m while the lighter mass ($m$) ascends by $h=3.6$ m. The effective potential energy change driving the motion is $(2m)gh - mgh = mgh$.

So, $mgh = KE_f$.

Substituting the values:

The mass m cancels out.

Now, substitute $g = 10 \text{ m/s}^2$ and $h = 3.6 \text{ m}$:

Solve for $v^2$:

Solve for v:

Conclusion

The speed of the 2m mass when it has descended through a height of 3.6 m is 2 m/s.