In the figure, block 1 has mass m1 = 450 g, block 2 has mass m2 = 560 g, and the pulley is on a frictionless horizontal axle with a radius R = 5.4 cm. When released from rest, block 2 falls 73 cm in 5.2 seconds without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on block 2) and (c) tension T1 (the tension force on block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s^2. (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number Units
The Correct Answer and Explanation is:
Here is the step-by-step solution to the problem.
Initial Setup and Conversion to SI Units
First, we identify the given information and convert all units to the standard SI system (meters, kilograms, seconds) to ensure consistency in our calculations.
- Mass of block 1, m₁ = 450 g = 0.450 kg
- Mass of block 2, m₂ = 560 g = 0.560 kg
- Radius of the pulley, R = 5.4 cm = 0.054 m
- Distance block 2 falls, Δy = 73 cm = 0.73 m
- Time elapsed, t = 5.2 s
- Initial velocity, v₀ = 0 m/s (since it’s released from rest)
- Acceleration due to gravity, g = 9.81 m/s²
(a) What is the magnitude of the acceleration of the blocks?
Since the cord connecting the blocks is inextensible, both blocks move with the same constant linear acceleration, a. We can determine this acceleration using the kinematic equation for displacement under constant acceleration:
Δy = v₀t + ½at²
Given that the system starts from rest, the initial velocity v₀ is 0. The equation simplifies to:
Δy = ½at²
Now, we can solve for the acceleration a:
a = 2Δy / t²
a = 2 * (0.73 m) / (5.2 s)²
a = 1.46 m / 27.04 s²
a ≈ 0.0540 m/s²
Answer for (a):
- Number: 0.0540
- Units: m/s^2
(b) What is tension T₂ (the tension force on the block 2)?
To find the tension T₂, we apply Newton’s second law (ΣF = ma) to block 2. Since m₂ > m₁, block 2 accelerates downwards. The forces acting on it are gravity (m₂g) pulling down and tension (T₂) pulling up. We’ll define the downward direction as positive.
ΣF₂ = m₂g – T₂ = m₂a
Solving for T₂:
T₂ = m₂g – m₂a = m₂(g – a)
T₂ = 0.560 kg * (9.81 m/s² – 0.0540 m/s²)
T₂ = 0.560 kg * (9.756 m/s²)
T₂ ≈ 5.46 N
Answer for (b):
- Number: 5.46
- Units: N
(c) What is tension T₁ (the tension force on the block 1)?
Similarly, we apply Newton’s second law to block 1, which accelerates upwards. The forces on block 1 are tension (T₁) pulling up and gravity (m₁g) pulling down. We’ll define the upward direction as positive.
ΣF₁ = T₁ – m₁g = m₁a
Solving for T₁:
T₁ = m₁g + m₁a = m₁(g + a)
T₁ = 0.450 kg * (9.81 m/s² + 0.0540 m/s²)
T₁ = 0.450 kg * (9.864 m/s²)
T₁ ≈ 4.44 N
Note: Because the pulley has mass, T₁ and T₂ are not equal. The difference creates the torque that rotates the pulley.
Answer for (c):
- Number: 4.44
- Units: N
(d) What is the magnitude of the pulley’s angular acceleration?
The cord does not slip on the pulley, which means the linear acceleration of a point on the rim of the pulley is the same as the linear acceleration of the blocks. The relationship between linear acceleration (a) and angular acceleration (α) is:
a = αR
Solving for α:
α = a / R
α = (0.0540 m/s²) / (0.054 m)
α = 1.00 rad/s²
Answer for (d):
- Number: 1.00
- Units: rad/s^2
(e) What is its rotational inertia?
The rotational inertia (I) of the pulley can be found using Newton’s second law for rotation (Στ = Iα). The net torque (τ_net) on the pulley is generated by the difference in tensions T₁ and T₂. Tension T₂ pulls in the direction of rotation (clockwise), while T₁ opposes it.
τ_net = T₂R – T₁R = (T₂ – T₁)R
Now, we set this equal to Iα:
(T₂ – T₁)R = Iα
Solving for the rotational inertia I:
I = (T₂ – T₁)R / α
Using the more precise, unrounded values for the tensions to avoid errors:
I = (5.463 N – 4.439 N) * (0.054 m) / (1.00 rad/s²)
I = (1.024 N) * (0.054 m) / (1.00 rad/s²)
I ≈ 0.0553 kg·m²
Answer for (e):
- Number: 0.0553
- Units: kg*m^2
