Question

A massless rope passes over a frictionless pulley with a radius of 0.15 m. It has a 13 kg mass (ml) on the left and a 2.8 kg mass (m2) on the right, both hanging freely. Initially, 13 kg mass is 2.0 m higher than the 2.8 kg mass. Assuming the system starts from rest and acceleration due to gravity is 9.81 m/s², what is the tension in the rope? Bonus: What is the angular velocity of the pulley when the masses are even with each other? (hint: you will need to use something outside of this unit to answer this.)

thanks!​

Answer 1

To find the tension in the rope, we can use the principles of Newton's second law and torque. We'll break down the problem into two parts: the motion of the masses and the motion of the pulley.

1. Motion of the masses:

Let's consider the downward direction as positive. The 13 kg mass (m1) experiences a downward force of its weight (m1 * g) and an upward tension force in the rope (T). The 2.8 kg mass (m2) experiences a downward force of its weight (m2 * g). The net force on each mass is given by:

m1 * g - T = m1 * a

m2 * g = m2 * a

Since the masses are connected by a rope, they have the same acceleration:

m1 * g - T = m2 * g

Now we can solve for T:

T = (m1 - m2) * g

Substituting the given values:

T = (13 kg - 2.8 kg) * 9.81 m/s²

T = 103.07 N

The tension in the rope is approximately 103.07 N.

2. Angular velocity of the pulley:

The angular acceleration of the pulley can be calculated using torque. The torque exerted on the pulley is equal to the net torque applied to it, which is caused by the tension in the rope. The torque can be calculated using the equation:

τ = I * α

Since the pulley is a uniform disc, its moment of inertia (I) can be calculated as:

I = (1/2) * m * r²

Where:

m is the mass of the pulley (assumed to be negligible since it's massless)

r is the radius of the pulley

The torque equation becomes:

τ = (1/2) * m * r² * α

We can substitute α with the linear acceleration (a) divided by the radius of the pulley (r):

α = a / r

The torque equation becomes:

τ = (1/2) * m * r * a

The torque τ is also equal to the tension in the rope multiplied by the radius of the pulley (r):

τ = T * r

Now we can set up an equation using the torque equation and the equation for tension:

T * r = (1/2) * m * r * a

The mass m of the pulley cancels out, and we can solve for the linear acceleration a:

a = 2 * T / r

Substituting the values we already know:

a = 2 * 103.07 N / 0.15 m

a ≈ 1373.8 m/s²

To find the angular velocity ω, we can use the equation:

ω² = ω₀² + 2 * α * θ

Since the pulley starts from rest, ω₀ = 0. The angular displacement θ is given by the distance traveled by the 13 kg mass:

θ = 2.0 m / r

Substituting the values into the equation:

ω² = 0 + 2 * 1373.8 m/s² * (2.0 m / 0.15 m)

ω² ≈ 36618.67 rad²/s²

Finally, taking the square root of both sides, we can find the angular velocity ω:

ω ≈ 191.18 rad/s

Therefore, the angular velocity of the pulley when the masses are even with each other is approximately 191.18 rad/s.