A string is wrapped around a uniform solid cylinder of radius r. The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m. Note that the positive y direction is downward and counterclockwise torques are positive.
Find the magnitude, alpha, of the angular acceleration of the cylinder as the block descends.
Express your answer in terms of the cylinder's radius r and the magnitude of the acceleration due to gravity g.
Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M r^2
m g - T = m a
T*r = torque = I*alpha = (1/2) m r^2 * alpha
T = (1/2) m*r*alpha
m g = ma + (1/2) m r * alpha
= ma + (1/2) m a
(since alpha = a/r)
a = (2/3) g
alpha = (2/3)(g/r)
Well, let's unravel this problem, shall we?
First, we need to consider the forces acting on the system. We have the weight of the block, which is mg, pulling it downwards. We also have tension in the string, which exerts a force on the cylinder.
The tension in the string can be divided into two components. The vertical component, Tvert, cancels out the weight of the block. The horizontal component, Thorz, produces a torque on the cylinder, which causes it to rotate.
Now, let's look at the torque equation. The torque produced by the tension is equal to the product of the force (Thorz) and the radius (r) of the cylinder. This torque is responsible for the angular acceleration (alpha) of the cylinder.
So, we have alpha = Torque / Moment of Inertia.
The moment of inertia (I) for a solid cylinder rotating about its axis is given by I = (1/2) * m * r^2.
The torque produced by the tension can be calculated using the equation: Torque = F * r.
Since the force (Thorz) is the horizontal component of the tension, we can use trigonometry to find it. The angle formed between the vertical and horizontal components is the same as the angle formed between the string and the radius of the cylinder. This angle can be approximated by sin(theta) = radius (r) divided by the length of the string (L).
So, the force (Thorz) can be calculated using the equation: Thorz = T * sin(theta).
Substituting all these equations into alpha = Torque / Moment of Inertia, we get:
alpha = (F * r) / ((1/2) * m * r^2)
Since we have an expression for the force in terms of the tension and the angle (Thorz = T * sin(theta)), we can substitute it into the equation:
alpha = (T * sin(theta) * r) / ((1/2) * m * r^2)
Now, we can substitute sin(theta) = r / L:
alpha = (T * r * r / L) / ((1/2) * m * r^2)
Since L is greater than r, we can approximate it by L = 2 * pi * r:
alpha = (T * r * r / (2 * pi * r)) / ((1/2) * m * r^2)
Simplifying further:
alpha = (T * r) / (2 * pi * m * r)
Now, we just need to find an expression for the tension in terms of the given variables.
Considering the forces in the system, Tvert is equal to mg, and Thorz is equal to the horizontal tension component.
But since Tvert cancels out mg, we have:
T * cos(theta) = mg
Now, we can approximate the cosine of the angle by cos(theta) = (L - r) / L:
T * [(L - r) / L] = mg
Simplifying further:
T = (mg * L) / (L - r)
Substituting this expression for T back into the equation for alpha, we get:
alpha = ((mg * L) / (L - r) * r) / (2 * pi * m * r)
Simplifying one more time:
alpha = (mg * r) / (2 * pi * m * (L - r))
And there you have it! The magnitude of the angular acceleration (alpha) of the cylinder as the block descends is (mg * r) / (2 * pi * m * (L - r)).
Now, isn't it amazing how a simple string can lead to such complex equations? It's almost like a magician's trick, isn't it?
To find the magnitude, α, of the angular acceleration of the cylinder, we need to analyze the forces and torques acting on the system.
The forces acting on the system are:
1. The tension force, T, in the string.
2. The weight force, mg, acting on the cylinder and the block.
The torque acting on the cylinder can be calculated as follows:
1. The torque due to the tension force is given by τT = Tr, where r is the radius of the cylinder.
2. The torque due to the weight force is given by τmg = mg * 0 = 0, since the weight force acts at the center of the cylinder.
Since the block is descending, the tension force T will be greater than the weight force mg. Therefore, we can write the equation for the net torque as:
τnet = τT - τmg
τnet = Tr - 0
τnet = Tr
According to Newton's second law for rotational motion, τ = Iα, where I is the moment of inertia of the cylinder and α is the angular acceleration.
The moment of inertia, I, can be calculated as I = 0.5 * m * r^2, assuming the density of the cylinder is uniform.
So we can write: Tr = Iα
Tr = 0.5 * m * r^2 * α
Now, we can solve for the angular acceleration, α:
α = (Tr) / (0.5 * m * r^2)
α = (2Tr) / (m * r^2)
α = (2 * T * r) / (m * r^2)
α = (2T) / (m * r)
To find T, we can use the equilibrium of forces along the vertical axis:
T - mg = 0
T = mg
Plug in T = mg into the equation for α:
α = (2mg) / (m * r)
α = 2g / r
Therefore, the magnitude, α, of the angular acceleration of the cylinder as the block descends is given by α = 2g / r, where g is the magnitude of the acceleration due to gravity and r is the radius of the cylinder.
To find the magnitude of the angular acceleration (α) of the cylinder as the block descends, we can use the principles of rotational dynamics.
Step 1: Free-Body Diagram
Let's start by drawing a free-body diagram for the cylinder and block system. The forces acting on the system are:
1. The weight of the cylinder acting downward (mg)
2. The weight of the block acting downward (mg)
3. The tension force in the string acting upward.
Step 2: Torque Equation
Since the cylinder is free to rotate about its axis, we can use the torque equation:
τ = Iα
where τ is the net torque exerted on the cylinder, I is the moment of inertia of the cylinder, and α is the angular acceleration.
Step 3: Moment of Inertia
The moment of inertia of a solid cylinder about its axis of rotation is given by:
I = (1/2)mr^2
where m is the mass of the cylinder and r is its radius.
Step 4: Analyzing Torques
The net torque acting on the cylinder is the difference between the torque due to the tension force and the torque due to the weight of the block. Since the cylinder and block have the same mass, the torques due to their weights cancel each other out.
The torque due to the tension force (τ_tension) can be calculated as follows:
τ_tension = T · r
where T is the tension in the string and r is the radius of the cylinder.
Step 5: Equating Torques
Setting up the torque equation:
τ_net = τ_tension - τ_blockweight = Iα
τ_net = T · r - 0 = (1/2)mr^2α
Simplifying the equation:
T · r = (1/2)mr^2α
Step 6: Solving for α
Substituting the value of T from the free-body diagram:
mg · r = (1/2)mr^2α
Canceling out the r term and simplifying:
g = (1/2)α
Finally, solving for α:
α = 2g
Therefore, the magnitude of the angular acceleration of the cylinder as the block descends is 2g, where g is the magnitude of the acceleration due to gravity.
Let T be the string tension. That and the angular acceleration alpha are the unknowns. Separate equations for the acceleration of the block and the angular acceleration of the cylinder will allow you to solve for both variables. The moment of inertia of the cylinder in (1/2) M r^2
m g - T = m a
T*r = torque = I*alpha = (1/2) m r^2 * alpha
T = (1/2) m*r*alpha
m g = ma + (1/2) m r * alpha
= ma + (1/2) m a
(since alpha = a/r)
alpha = (2/3) g
Note that the magnitude of the linear acceleration of the block is 23g , which does not depend on the value of r .