Two masses, m1=3kg and m2=2kg, are suspended with a massless rope over a pulley of mass M = 10kg. The pulley turns without friction and may be modeled as a uniform disk of radius R=.1m. You may neglect the size of the masses. The rope does not slip on the pulley. The system begins at rest.

If m1 starts at a height h=1m above the ground, what is its speed v when it hits the ground?

I've tried a few different ways to get it but I can't seem to get the answer. I've tried using the sum of the torques as well as trying to use energy. I can't seem to figure it out, any help is much appreciated!

The acceleration should come out as 0.1g actually.

So, v = sqrt(2*a*s) = sqrt(2*0.1g*1) = 1.40m/s

In vector form

m1•¬a = m1•¬g - ¬T1
m2•¬a = m2•¬g - ¬T2
I•¬ε =¬M

Projections on the vertical axis:
m1•a = m1•g - T1,
m2•a = m2•g – T2,
(mR^2/2)•(a/R) = (T1-T2)•R.
Solving this system for acceleration a, we obtain

a= (m1-m2)•g/(m1 +m2+m/2) =(3-2)/(3+2+10/2) = 0.1 m/s^2.
Since
s =v^2/2•a.
v =sqrt(2•a•s) = sqrt(2•0.1•1)= 0.45 m/s.

To solve this problem, we can apply the principle of conservation of mechanical energy. We'll consider the potential energy and kinetic energy of the system.

Let's start by calculating the potential energy of mass m1 when it is at height h above the ground. The potential energy is given by the formula:

PE1 = m1 * g * h

where m1 is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height above the ground.

Now, as the mass m1 falls, it will lose potential energy and gain an equal amount of kinetic energy. At the same time, the pulley will start rotating, and mass m2 will also gain kinetic energy. We can relate the linear speed v of mass m1 to the angular speed ω of the pulley using the relationship v = R * ω, where R is the radius of the pulley.

To determine the angular speed ω, we need to consider mechanical equilibrium for the system. The sum of the torques acting on the pulley must be zero since there are no external torques acting on it. The torque due to the gravitational force acting on mass m1 will cause the pulley to rotate in the counterclockwise direction, while the torque due to the gravitational force acting on mass m2 will cause the pulley to rotate in the clockwise direction.

Using the equation for the torque τ = I * α, we can write the torques caused by m1 and m2 as:

τ1 = m1 * g * R
τ2 = -m2 * g * R

where I is the moment of inertia of the pulley and α is the angular acceleration. Since the pulley is a uniform disk, the moment of inertia is given by I = (1/2) * M * R².

The sum of the torques is:

τ = τ1 + τ2 = (m1 - m2) * g * R

Since the sum of the torques is zero, we have:

(m1 - m2) * g * R = I * α

Substituting the value of I and rearranging, we get:

(m1 - m2) * g * R = (1/2) * M * R² * α

Simplifying further:

(m1 - m2) * g = (1/2) * M * R * α

Now, we can relate the angular acceleration α to the linear acceleration a of mass m2 using the relationship α = a / R. Substituting this into the equation, we get:

(m1 - m2) * g = (1/2) * M * a

Lastly, we can relate the linear acceleration a to the speed v using the formula a = v² / (2 * R). Substituting this, we have:

(m1 - m2) * g = (1/2) * M * v² / (2 * R)

Simplifying further:

2 * (m1 - m2) * g * R = M * v²

Now, we can solve for the speed v:

v² = (2 * (m1 - m2) * g * R) / M

Finally, taking the square root of both sides, we get:

v = √((2 * (m1 - m2) * g * R) / M)

Substituting the given values:

m1 = 3 kg
m2 = 2 kg
M = 10 kg
R = 0.1 m
g ≈ 9.8 m/s²

v = √((2 * (3 - 2) * 9.8 * 0.1) / 10)
v = √(0.196)
v ≈ 0.442 m/s

Therefore, the speed of mass m1 when it hits the ground is approximately 0.442 m/s.

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