The system shown in the figure below consists of a m1 = 5.52-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m2 = 2.76-kg block.The pulley is a uniform disk of radius 7.86 cm and mass 0.592 kg. Calculate the speed of the m2 = 2.76-kg block after it is released from rest and falls a distance of 2.08 m. Calculate the angular speed of the pulley at this instant.
To calculate the speed of the m2 block after it falls a distance of 2.08m, we need to consider the conservation of energy.
First, we calculate the gravitational potential energy that is converted into kinetic energy. The gravitational potential energy is given by the formula:
PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height.
For the m2 block, the gravitational potential energy converted is:
PE = m2 * g * h
Given that m2 = 2.76kg and h = 2.08m, and using the standard acceleration due to gravity g = 9.8m/s^2, we can calculate the gravitational potential energy.
PE = 2.76kg * 9.8m/s^2 * 2.08m = 55.259 Joules
Since the system is frictionless, this potential energy is converted entirely into kinetic energy. The kinetic energy of an object is given by:
KE = 0.5 * m * v^2
where m is the mass of the object and v is its velocity.
For the m2 block, the kinetic energy is
KE = 0.5 * m2 * v^2
Since the pulley is a uniform disk, it has an associated rotational kinetic energy. The rotational kinetic energy of a rotating object is given by:
KE_rot = 0.5 * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
For a uniform disk, the moment of inertia is given by:
I = 0.5 * m * r^2
where m is the mass of the disk and r is its radius.
In this case, the moment of inertia of the pulley is:
I = 0.5 * 0.592kg * (0.0786m)^2 = 0.01742 kg*m^2
Therefore, the rotational kinetic energy of the pulley is:
KE_rot = 0.5 * 0.01742kg*m^2 * ω^2
Since energy is conserved, the total energy is the sum of the gravitational potential energy, the kinetic energy of the m2 block, and the rotational kinetic energy of the pulley:
Total Energy = PE + KE + KE_rot
55.259J = 0.5 * 2.76kg * v^2 + 0.5 * 0.01742kg*m^2 * ω^2
To find the speed of the m2 block after it is released, we can rearrange the equation to solve for v:
v = sqrt((2 * (55.259J - 0.5 * 0.01742kg*m^2 * ω^2)) / 2.76kg)
Substituting the given values, we can calculate the speed of the m2 block.
To find the angular speed of the pulley at this instant, we use the relationship between the linear speed of the edge of the pulley (v) and the angular speed of the pulley (ω):
v = ω * r
Where r is the radius of the pulley (0.0786m).
Rearranging the equation, we can solve for ω:
ω = v / r
Using the previously calculated value of v and the given value of r, we can calculate the angular speed of the pulley.