Two block are connected by a rope that runs over a pulley. The block on the tables has mass 4kg, the hanging block has mass 2kg, and the pulley has mass 0.5kg and radius 0.25m. Assume that the table is friction-less. If the block are released from the rest, determine their speeds after the hanging block has dropped 0.75m.
m1 =4 kg, m2 = 2 kg, m = 0.5 kg, R = 0.25 m, h= 0.75 m.
Projections on the horizontal and vertical axis:
m1•a = T1
m2•a =m2•g-T2,
I•ε =M.
If the pulley is the disk (cylinder) I =m•R²/2 ,
M = torque = (T1-T2) •R,
ε = a/R,
I•ε =M => m•R²•a/2•R =(T1-T2) •R =>
m•a/2 = (T1-T2).
m1•a + m2•a = T1 + m2•g -T2 = m2•g + (T1-T2) = m2•g +m•a/2,
a = m2•a/[m1+m2-m(m/2)] = 2•9.8/(4+2+0.125)=3.336 m/s^2,
a = v^2/2•h ,
v=sqrt(2•a•h) = sqrt(2•3.336•0.75) = 2.2 m/s^2
To determine the speeds of the blocks after the hanging block has dropped 0.75m, we need to consider the conservation of energy and the law of motion.
First, let's find the gravitational potential energy that has been converted into kinetic energy for the falling block. We can use the equation:
Potential energy (PE) = mass (m) * gravity (g) * height (h)
The mass of the hanging block is 2kg and the height it has dropped is 0.75m, so the potential energy of the hanging block is:
PE_hanging = 2kg * 9.8m/s^2 * 0.75m = 14.7 Joules
Since the hanging block is connected to the block on the table by a rope which runs over a pulley, the block on the table gains the same amount of potential energy. Therefore, the potential energy of the block on the table is also 14.7 Joules.
The total potential energy is then:
PE_total = PE_hanging + PE_table = 14.7 Joules + 14.7 Joules = 29.4 Joules
According to the law of conservation of energy, this potential energy is converted into kinetic energy. Thus, the total kinetic energy after the blocks have dropped can be expressed as:
Kinetic energy (KE) = 1/2 * mass (m) * velocity^2
Now, let's consider the rotational motion of the pulley. The kinetic energy of the pulley can be expressed as:
KE_pulley = 1/2 * moment of inertia (I) * angular velocity^2
The moment of inertia of a solid disk is given by the equation:
I = 1/2 * mass (m) * radius^2
For the pulley, the mass is 0.5kg and the radius is 0.25m. Therefore, the moment of inertia of the pulley is:
I_pulley = 1/2 * 0.5kg * (0.25m)^2 = 0.03125 kg·m^2
Since the pulley and the block on the table have the same angular velocity, we can equate the kinetic energy of the pulley to the kinetic energy of the block on the table:
KE_pulley = KE_table
Substituting the equations for kinetic energy and the moment of inertia, we have:
1/2 * 0.03125 kg·m^2 * angular velocity^2 = 1/2 * 4kg * velocity_table^2
Simplifying the equation, we get:
0.03125 kg·m^2 * angular velocity^2 = 2kg * velocity_table^2
Now, let's determine the angular velocity of the pulley by relating it to the linear velocity of the block on the table. Since the pulley's radius is 0.25m, the linear velocity of the block on the table can be expressed as:
velocity_block_on_table = radius * angular velocity
Substituting the given values, we have:
velocity_table = 0.25m * angular velocity
Now, we can substitute this expression for velocity_table into the previous equation and solve for angular velocity:
0.03125 kg·m^2 * angular velocity^2 = 2kg * (0.25m * angular velocity)^2
0.03125 kg·m^2 * angular velocity^2 = 0.03125 kg·m^2 * angular velocity^2
Therefore, the angular velocity cancels out, which means that the angular velocity of the pulley does not affect the speeds of the blocks.
Next, we can solve for the velocity of the block on the table using the equation:
KE_table = 1/2 * 4kg * velocity_table^2
Substituting the given values for the total kinetic energy (29.4 Joules), mass (4kg), and solving for velocity_table, we get:
29.4 Joules = 1/2 * 4kg * velocity_table^2
29.4 Joules = 2kg * velocity_table^2
Dividing both sides of the equation by 2kg, we have:
14.7m^2/s^2 = velocity_table^2
Taking the square root of both sides, we get:
velocity_table = sqrt(14.7m^2/s^2) ≈ 3.83m/s
Therefore, the velocity of the block on the table after the hanging block has dropped 0.75m is approximately 3.83 m/s. Since the angular velocity of the pulley does not affect the speeds of the blocks, the hanging block will also have a velocity of 3.83 m/s.