A thin, light string is wrapped around the rim of a 4.00kg solid uniform disk that is 30.0cm in diameter. A person pulls on the string with a constant force of 100N tangent to the disk. Find the linear acceleration of its center of mass. If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the angular acceleration of the disk about its center of mass and what will be the linear acceleration of its center of mass?

Since the pulled string exerts the only force on the disc (assuming that gravity is ignored here), you can use

F = m a for the acceleration of the center of mass, in both cases.

Pulling the string also starts the disk spinning at an acclerating spin rate. Use the relation
Torque = Force * R = I* alpha
for the angular rate of acceleration, alpha. I, the moment of inertia, will be different for the solid disc and the thin walled disc, despite the equal masses.

Well, it seems like we have a fun circular problem here!

To find the linear acceleration of the center of mass of the solid uniform disk, we first need to find the moment of inertia of the disk. The moment of inertia for a solid uniform disk rotating about its central axis is (1/2) * m * r^2, where m is the mass of the disk and r is the radius of the disk.

In this case, the mass is given as 4.00kg, and the radius is half of the diameter, so r = 0.30m/2 = 0.15m. Substituting these values, we get the moment of inertia as (1/2) * 4.00kg * (0.15m)^2 = 0.045kg * m^2.

Now, we can use Newton's second law for rotational motion, τ = I * α, where τ is the torque applied, I is the moment of inertia, and α is the angular acceleration.

In this case, the torque applied is the force applied tangentially multiplied by the radius (τ = F * r). So, τ = 100N * 0.15m = 15Nm.

Now we can rearrange the equation to solve for α: α = τ / I. Substituting the values, we get α = 15Nm / 0.045kg * m^2 = 333.33 rad/s^2.

So, for the solid uniform disk, the angular acceleration about its center of mass is 333.33 rad/s^2.

To find the linear acceleration of the center of mass, we can use the equation a = α * r, where a is the linear acceleration, α is the angular acceleration, and r is the radius.

Substituting the values, we get a = 333.33 rad/s^2 * 0.15m = 49.9995 m/s^2 (rounded to 5 significant figures).

So, the linear acceleration of the center of mass of the solid uniform disk is approximately 50.0 m/s^2.

Now, let's move on to the hollow thin-walled cylinder!

For a hollow thin-walled cylinder, the moment of inertia is given by I = m * r^2, where m is the mass and r is the radius.

In this case, the mass is still 4.00kg, and the radius remains the same at 0.15m. Substituting these values, we get I = 4.00kg * (0.15m)^2 = 0.09kg * m^2.

We already know the angular acceleration of the disk, which is 333.33 rad/s^2. So, we've got that covered!

To find the linear acceleration of the center of mass, we can again use the equation a = α * r, where a is the linear acceleration, α is the angular acceleration, and r is the radius.

Substituting the values, we get a = 333.33 rad/s^2 * 0.15m = 49.9995 m/s^2 (rounded to 5 significant figures).

So, the linear acceleration of the center of mass of the hollow thin-walled cylinder is also approximately 50.0 m/s^2.

In conclusion, the angular acceleration of the disk about its center of mass for both the solid uniform disk and the hollow thin-walled cylinder is the same, which is 333.33 rad/s^2. However, the linear acceleration of the center of mass is also the same for both cases, which is approximately 50.0 m/s^2.

Hope that answers your question in a fun and circular way!

To find the linear acceleration of the center of mass, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass multiplied by the acceleration.

1. Find the moment of inertia for the solid disk:
The moment of inertia for a solid disk can be calculated using the formula: I = (1/2) * m * r^2, where m is the mass and r is the radius. In this case, the radius (diameter/2) is 30 cm / 2 = 15 cm = 0.15 m.
I = (1/2) * 4.00 kg * (0.15 m)^2 = 0.09 kg * m^2

2. Calculate the torque applied by the pulling force:
The torque, τ, is equal to the force applied multiplied by the perpendicular distance from the center of rotation to the point where the force is applied. Since the force is tangent to the disk, the perpendicular distance is equal to the radius of the disk.
τ = F * r = 100 N * 0.15 m = 15 N * m

3. Calculate the angular acceleration of the solid disk:
The torque is also equal to the moment of inertia multiplied by the angular acceleration, τ = I * α. Rearranging this equation, we can solve for α.
α = τ / I = 15 N * m / 0.09 kg * m^2 ≈ 166.67 rad/s^2

4. Calculate the linear acceleration of the center of mass of the solid disk:
The linear acceleration, a, is related to the angular acceleration, α, and the radius, r, via the equation a = r * α.
a = 0.15 m * 166.67 rad/s^2 = 25 m/s^2

Now, let's consider the hollow thin-walled cylinder.

1. Find the moment of inertia for the hollow thin-walled cylinder:
The moment of inertia for a hollow thin-walled cylinder can be calculated using the formula: I = m * r^2, where m is the mass and r is the radius. In this case, the radius is 0.15 m.
I = 4.00 kg * (0.15 m)^2 = 0.09 kg * m^2 (same as the solid disk)

2. Calculate the angular acceleration of the hollow thin-walled cylinder:
Since the moment of inertia remains the same, the angular acceleration also remains the same: α ≈ 166.67 rad/s^2

3. Calculate the linear acceleration of the center of mass of the hollow thin-walled cylinder:
The linear acceleration can be calculated using the same equation as before: a = r * α.
a = 0.15 m * 166.67 rad/s^2 ≈ 25 m/s^2 (same as the solid disk)

So, the angular acceleration of the solid disk about its center of mass remains the same when replaced by a hollow thin-walled cylinder, while the linear acceleration of the center of mass also remains the same.

To find the linear acceleration of the center of mass of the solid uniform disk, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is provided by the tension in the string.

1. Calculate the tension in the string:
The tension in the string is equal to the force applied tangent to the disk by the person, which is 100N in this case.

2. Calculate the moment of inertia of the solid uniform disk:
The moment of inertia of a solid uniform disk rotating about its center of mass can be calculated using the formula:
I = (1/2) * m * r^2
where I is the moment of inertia, m is the mass of the disk, and r is the radius of the disk.

In this case, the mass of the disk is 4.00kg and the radius is half of the diameter, which is 30/2 = 15.0cm or 0.15m.

I = (1/2) * 4.00kg * (0.15m)^2 = 0.09kg*m^2

3. Calculate the linear acceleration of the center of mass:
The linear acceleration of the center of mass can be calculated using the formula:
F = m * a
where F is the net force, m is the mass of the disk, and a is the linear acceleration.

Since the tension in the string is the net force, we can substitute that into the formula:
100N = 4.00kg * a

Solving for a, we find:
a = 100N / 4.00kg = 25m/s^2

So, the linear acceleration of the center of mass of the solid uniform disk is 25m/s^2.

Now, let's consider the hollow thin-walled cylinder. The mass and diameter of the cylinder are the same as the solid disk.

To find the angular acceleration of the disk about its center of mass, we need to consider the moment of inertia of the hollow thin-walled cylinder.

The moment of inertia of a hollow thin-walled cylinder can be calculated using the formula:
I = m * (rOuter^2 + rInner^2) / 2
where m is the mass of the cylinder, rOuter is the outer radius, and rInner is the inner radius of the cylinder.

Since the mass and diameter of the cylinder are the same as the solid disk, the mass is still 4.00kg and the radius is still 0.15m.

The outer radius of the cylinder is the same as the radius of the solid disk, which is 0.15m.

The inner radius of the cylinder can be calculated by subtracting the thickness of the wall from the outer radius. However, the thickness of the wall is not provided in the question. Let's assume it to be negligible, which means the inner radius is also 0.15m.

Substituting the values into the formula, we get:
I = 4.00kg * (0.15m^2 + 0.15m^2) / 2 = 0.09kg*m^2

The angular acceleration of the disk about its center of mass can be calculated using the formula:
τ = I * α
where τ is the torque applied to the disk, I is the moment of inertia, and α is the angular acceleration.

In this case, the torque is provided by the tension in the string, which is applied tangentially, and it can be calculated using the formula:
τ = F * r
where F is the net force and r is the radius of the disk.

The radius of the disk is still 0.15m, and the net force is still 100N.

Substituting the values into the formula, we get:
τ = 100N * 0.15m = 15Nm

So, τ = I * α
15Nm = 0.09kg*m^2 * α

Solving for α, we find:
α = 15Nm / 0.09kg*m^2 = 166.67 rad/s^2

Thus, the angular acceleration of the disk about its center of mass is 166.67 rad/s^2.

Since the mass and diameter are the same as the solid disk, the linear acceleration of the center of mass of the hollow thin-walled cylinder will be the same as that of the solid disk, which is 25m/s^2.