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: $KE_{rot} = \frac{1}{2}I\omega^2$.
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:
$ mgh = 9mv^2 $The mass m cancels out.
$ gh = 9v^2 $Now, substitute $g = 10 \text{ m/s}^2$ and $h = 3.6 \text{ m}$:
$ (10 \text{ m/s}^2) \times (3.6 \text{ m}) = 9v^2 $ $ 36 \text{ m}^2/\text{s}^2 = 9v^2 $Solve for $v^2$:
$ v^2 = \frac{36 \text{ m}^2/\text{s}^2}{9} = 4 \text{ m}^2/\text{s}^2 $Solve for v:
$ v = \sqrt{4 \text{ m}^2/\text{s}^2} = 2 \text{ m/s} $Conclusion
The speed of the 2m mass when it has descended through a height of 3.6 m is 2 m/s.
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Important Questions from Mechanics
- The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is _________ m/s.
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- Given below are two statements :
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II : The time period of revolution of the satellite is $T = 2\pi \sqrt{\frac{R_e}{g}}$ (for satellite very close to the earth surface), where $R_e$ radius of earth and g acceleration due to gravity.
In the light of the above statements, choose the correct answer from the options given below :
