The two blocks in the figure below, m1 = 1.2 kg and m2 = 4.8, are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.38 N·m. If the blocks are released from rest, how long does it take the 4.8 kg block to reach the floor from a height of h = 1.0 m?
To find the time it takes for the 4.8 kg block to reach the floor from a height of 1.0 m, we can follow these steps:
Step 1: Find the acceleration of the system
The net force acting on the system is the weight of the 4.8 kg block minus the tension in the rope. We can calculate this as:
F_net = m2 * g - T
Where m2 is the mass of the 4.8 kg block and g is the acceleration due to gravity.
Step 2: Find the tension in the rope
The tension in the rope can be found by using the torque equation for the pulley:
τ = I * α
Where τ is the torque, I is the moment of inertia of the pulley, and α is the angular acceleration of the pulley.
Given that the torque exerted by friction at the axle is 0.38 N·m, we can set this equal to the torque produced by the tension in the rope:
0.38 N·m = (1/2) * m * r^2 * α
Where r is the radius of the pulley, which is half its diameter.
Step 3: Calculate the angular acceleration
Solve the torque equation for α:
α = (2 * 0.38 N·m) / (m * r^2)
Step 4: Find the tension in the rope
Using the equation τ = I * α, we can solve for the tension in the rope:
τ = I * α
T = (1/2) * m * r^2 * α
Step 5: Calculate the acceleration of the system
By substituting the tension into the net force equation, we can find the acceleration:
F_net = m2 * g - T
m2 * a = m2 * g - (1/2) * m * r^2 * α
a = g - (1/2) * r^2 * α
Step 6: Determine the time it takes for the block to reach the floor
Using the kinematic equation for an object accelerating from rest:
h = (1/2) * a * t^2
We can solve for t:
t = sqrt((2 * h) / a)
To find the time it takes for the 4.8 kg block to reach the floor from a height of 1.0 m, we can use the principles of mechanics and energy conservation.
First, let's consider the motion of the blocks and the pulley. The system is governed by the forces acting on the two blocks and the pulley.
1. Tension force: The tension in the rope is the same on both sides of the pulley. We can denote it as T.
2. Gravitational force: Each block experiences a gravitational force due to their respective masses. The force on the 1.2 kg block is m1 * g, and the force on the 4.8 kg block is m2 * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Now, let's set up the equations of motion for the system. We will consider the motion of each block separately.
For the 1.2 kg block:
1. The tension force T points upwards.
2. The gravitational force m1 * g points downwards.
3. The net force acting on the block is the difference between the tension and gravitational force: T - m1 * g.
4. Using Newton's second law, the acceleration of the 1.2 kg block is given by a1 = (T - m1 * g) / m1.
For the 4.8 kg block:
1. The tension force T points downwards.
2. The gravitational force m2 * g points downwards.
3. The net force acting on the block is the sum of the tension and gravitational force: T + m2 * g.
4. Using Newton's second law, the acceleration of the 4.8 kg block is given by a2 = (T + m2 * g) / m2.
Now, since the two blocks are connected by a rope passing over a pulley, their accelerations are related:
a2 = -a1
This is because if one block goes down, the other block goes up, and vice versa.
Next, let's consider the rotational motion of the pulley. The torque exerted by the friction at the axle opposes the motion of the pulley and can be given by the equation:
τ = I * α
where τ is the torque, I is the moment of inertia of the pulley, and α is the angular acceleration of the pulley.
We are given the torque (0.38 N·m), and we can calculate the moment of inertia of the pulley using its mass and dimensions.
The moment of inertia of a solid disc around its central axis (perpendicular to the plane of the disc) is given by:
I = (1/2) * m * R²
where m is the mass of the pulley and R is its radius (half of the diameter).
Now, let's find the angular acceleration of the pulley. The pulley's angular acceleration is related to the linear accelerations of the blocks by the following equation:
α = (a2 - a1) / R
Substituting the value of α into the torque equation, we have:
τ = I * (a2 - a1) / R
Now, we can see that the torque is opposing the motion of the pulley, so its direction is opposite to the linear accelerations of the blocks. Therefore, we can rewrite the equation as:
|(a2 - a1) / R| = τ / I
Now, we have a relationship between the linear accelerations of the blocks and the torque exerted by the friction at the axle.
Finally, we can relate the linear accelerations to the displacement h. The distance traveled by each block when it reaches the floor can be calculated using the equations of motion:
s1 = (1/2) * a1 * t²
s2 = (1/2) * a2 * t²
where s1 and s2 are the distances traveled by the 1.2 kg and 4.8 kg blocks, respectively, and t is the time taken to reach the floor.
Since the displacement of the blocks is equal to h, we have:
s1 + s2 = h
Substituting the equations for s1 and s2, and using the relationship between the linear accelerations obtained earlier, we can solve for t:
[(1/2) * (T - m1 * g) * t² + (1/2) * (T + m2 * g) * t²] / R = h
Simplifying the equation, we get:
[T * (m1 + m2) + (m2 - m1) * g] * t² = 2 * R * h
Now, we can solve for t:
t = sqrt[2 * R * h / (T * (m1 + m2) + (m2 - m1) * g)]
Substituting the given values for R (12 cm diameter) and h (1.0 m), as well as the masses m1 (1.2 kg) and m2 (4.8 kg), we can then calculate the tension force T and substitute it back into the equation to find the time t.
Note: It is important to double-check the units used in the calculations and ensure that they are consistent (e.g., meters for distance, kilograms for mass, and seconds for time).