In an Atwood machine, one block has a mass of M1 = 520 g and the other has a mass of M2 = 320 g. The frictionless pulley has a radius of 4.6 cm. When released from rest, the heavier block moves down 51 cm in 1.00 s (no slippage). What is the tension T1? Find the pulley's moment of inertia.
To find the tension T1 in the Atwood machine, we can use the concept of Newton's second law. However, before we can do that, we need to determine the net force acting on the system and the acceleration of the blocks.
Given:
Mass of block 1 (M1) = 520 g = 0.52 kg
Mass of block 2 (M2) = 320 g = 0.32 kg
Distance traveled by block 1 (d) = 51 cm = 0.51 m
Time taken (t) = 1.00 s
First, let's find the acceleration of the system:
Using the equation of motion s = ut + (1/2)at^2, where u is the initial velocity (which is 0 in this case), s is the distance traveled, t is the time, and a is the acceleration, we can rearrange the equation to solve for acceleration:
s = (1/2)at^2
0.51 = (1/2)a(1.00)^2
0.51 = (1/2)a
a = (2 * 0.51) / 1.00
a ≈ 1.02 m/s^2
Now that we have the acceleration, we can find the net force acting on the system using Newton's second law:
F_net = (M1 - M2) * a
F_net = (0.52 - 0.32) * 1.02
F_net ≈ 0.20 N
Since we know that the tension in the string is the same on both sides (T1 = T2), the tension in the string can be determined using one of the blocks. Let's use block 1:
T1 - M1 * g = M1 * a
T1 - 0.52 * 9.8 = 0.52 * 1.02
T1 - 5.096 = 0.53424
T1 ≈ 5.096 + 0.53424
T1 ≈ 5.63 N
Therefore, the tension T1 in the string is approximately 5.63 N.
Next, let's find the moment of inertia of the pulley. The moment of inertia of a solid disk is given by the formula:
I = (1/2) * m * r^2
Given that the radius of the pulley (r) is 4.6 cm = 0.046 m and the mass (m) is unknown, we can solve for the moment of inertia (I) by rearranging the formula:
I = (2 * m * r^2) / 2
I = m * r^2
To find the mass (m) of the pulley, we need to consider the torque produced by the net force and the angular acceleration.
The torque produced by the net force acting on the pulley is given by the equation:
τ = I * α
Since the pulley has a rotational inertia I and an angular acceleration α, we can rewrite the equation as:
F_net * r = I * α
Solving for I, we get:
I = (F_net * r) / α
Now, we need to find the angular acceleration α. The angular acceleration is related to the linear acceleration a by the equation:
α = a / r
Substituting the value of a and r, we get:
α = 1.02 / 0.046
α ≈ 22.17 rad/s^2
Finally, we can find the moment of inertia (I):
I = (F_net * r) / α
I = (0.20 * 0.046) / 22.17
I ≈ 0.000412 kg⋅m^2
Therefore, the moment of inertia of the pulley is approximately 0.000412 kg⋅m^2.