A massless string is wrapped around a cylinder of mass 0.400 kg and radius 0.100 m. The string is attached to the ceiling. The cylinder is released. As it falls, the string unwinds. What is the magnitude of the acceleration of the cylinder as it falls in m/s2?


A.3.27
B.6.54
C.4.91
D.7.61
E.9.81

B

For the cylinder moment of inertia (rotational inertia) is I = m•R²/2,
v =ω•R = > ω= v/R
PE =KE
m•g•h = m•v²/2 +I•ω²/2 = m•v²/2 + (m•R²/2) • ( v/R)²/2 =
= 3•m•v²/4, =>
v² = 4•g•h/3
From kinematics:
v² = 2•a•h, therefore
4•g•h/3 = 2•a•h,
a =4•g/6 = 2•g/3 = 9.8•2/3 = 6.54 m/s² .
Note! Mass and radius are unnecessary for solution

Well, let's see here. A massless string, huh? That's like having a ghost string, right? Pretty spooky. Anyway, this ghost string is wrapped around a cylinder. I hope the cylinder doesn't mind having a ghost string wrapped around it!

Now, as the cylinder falls, the string unwinds. But don't worry, the cylinder won't get tangled up in any ghostly knots. Phew!

To find the magnitude of the acceleration of the cylinder as it falls, we can use Newton's second law. Remember that F = ma? Well, in this case, the only force acting on the cylinder is its weight, which is equal to the mass of the cylinder times the acceleration due to gravity (9.81 m/s^2).

So, we can set up the equation F = ma, where F is the weight of the cylinder (mg), m is the mass of the cylinder (0.400 kg), and a is the acceleration we're trying to find.

Now, we can solve for a: a = F/m = (mg)/m = g = 9.81 m/s^2.

So, the magnitude of the acceleration of the cylinder as it falls is 9.81 m/s^2. That's option E.

Just remember, when it comes to ghost strings and falling cylinders, things can get a little spooky, but the laws of physics are always there to keep us grounded!

To solve this problem, we can use the concept of rotational dynamics. The torque acting on the cylinder can be found by considering the tension in the string and the moment of inertia of the cylinder.

1. Calculate the moment of inertia:
The moment of inertia (I) of a solid cylinder about its central axis is given by the formula:

I = (1/2) * m * r^2

where m is the mass of the cylinder and r is the radius.

Plugging in the values, we get:
I = (1/2) * 0.400 kg * (0.100 m)^2 = 0.002 kg·m^2

2. Find the torque:
The torque (τ) acting on the cylinder is given by the product of the tension in the string (T) and the radius of the cylinder (r):

τ = T * r

3. Equate the torque and the moment of inertia, and find the tension:
Since there is no net torque acting on the cylinder, the torque due to the tension in the string must be equal to the torque required to rotate the cylinder. Equating τ and I, we get:

T * r = I * α

where α is the angular acceleration.

Since the string is unwinding, the linear acceleration (a) of the cylinder is related to the angular acceleration by the equation:

a = α * r

Substituting this into the equation, we have:

T * r = I * (a / r)

4. Solve for the acceleration:
Rearranging the equation, we get:

a = T * r^2 / I

Substituting the given values, we get:

a = T * (0.100 m)^2 / 0.002 kg·m^2

5. Calculate the tension:
The tension in the string can be calculated using Newton's second law for rotational motion:

τ = I * α = T * r

Substituting the values, we get:

T = τ / r = (m * g) / r

where g is the acceleration due to gravity.

Substituting the given values, we have:

T = (0.400 kg * 9.81 m/s^2) / 0.100 m = 3.924 N

6. Calculate the acceleration:
Substituting the tension value into the equation for acceleration, we get:

a = (3.924 N) * (0.100 m)^2 / 0.002 kg·m^2

Simplifying the equation, we get:

a ≈ 19.62 m/s^2

The magnitude of the acceleration of the cylinder as it falls is approximately 19.62 m/s^2.

The closest answer choice to this value is option C. 4.91 m/s^2.

To find the magnitude of the acceleration of the cylinder as it falls, we need to consider the forces acting on the system.

First, we have the force of gravity pulling the cylinder downward, given by the formula F = m * g, where m is the mass of the cylinder and g is the acceleration due to gravity. In this case, m = 0.400 kg.

Next, we need to consider the tension force in the string. As the cylinder falls, the string unwinds, causing a tension force on the cylinder in the upward direction.

Finally, we need to consider the rotational motion of the cylinder. As the string unwinds, the cylinder experiences a torque due to the force of gravity acting on its weight. This causes the cylinder to rotate, which affects its acceleration.

To find the acceleration, we can use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the difference between the gravitational force and the tension force.

Now, let's calculate the values for the forces:

Gravitational force: F_gravity = m * g = 0.400 kg * 9.81 m/s^2 = 3.924 N

Tension force: The tension force in the string can be calculated using the torque equation. The torque is given by the formula tau = I * alpha, where I is the moment of inertia and alpha is the angular acceleration. Since the string is massless, the moment of inertia of the cylinder is given by I = 1/2 * m * r^2, where m is the mass of the cylinder and r is its radius.

Using the relationship between linear acceleration and angular acceleration, alpha = a / r, where a is the linear acceleration of the cylinder and r is the radius, we can rewrite the torque equation as tau = (1/2 * m * r^2) * (a / r) = 1/2 * m * a * r.

The tension force in the string is equal to the torque divided by the radius: F_tension = tau / r = 1/2 * m * a.

Now, let's put it all together and find the acceleration:

Net force on the cylinder: F_net = F_gravity - F_tension = m * g - 1/2 * m * a

Since the cylinder is accelerating downward, the net force is positive. Therefore, we can set up the equation:

m * g - 1/2 * m * a = m * a

Simplifying the equation, we get:

g - 1/2 * a = a

Rearranging the equation to solve for a, we have:

1/2 * a = g

a = 2 * g = 2 * 9.81 m/s^2 = 19.62 m/s^2

Therefore, the magnitude of the acceleration of the cylinder as it falls is 19.62 m/s^2.

Since none of the provided answer options match the calculated acceleration, it seems an error might have occurred either in reading the question, copying the options, or in the calculation itself. Please double-check the information and calculations to arrive at the correct answer.

A.3.27 i'm 100% sure :D