A yo-yo with mass M and radius R is wound with a light string. Someone drops it from rest with the string attached to a metal pole. The yo-yo falls and unwinds (spins) without slipping. The moment of inertia for the yo-yo is ½ MR^2 (a disk).
a. In terms of g, find the downward acceleration of the center of mass of the yo-yo as it unrolls from the string.
b. What is the tension in the string as the disk unwinds and falls?
c.While descending, does the center of mass of the yo-yo move to the left, the right, or straight down? Explain your answer in complete sentences.
I think straight down, but why?
T = string force up
m g = weight force down
m a = mg - T
(I guess the string is wound around the outside of this yo yo)
Torque = I alpha
T R = .5 m R^2 alpha
but alpha = a/R
so T = .5 m a
so
ma = T - .5 ma
(3/2) m a = (1/2) m g
a = g/3
T = .5 m a = (1/2)m g/3
T = mg/6
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To understand why the center of mass of the yo-yo moves straight down as it unrolls and falls, let's break down the problem and analyze each part.
a. To find the downward acceleration of the center of mass of the yo-yo as it unrolls from the string, 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 acting on the yo-yo is the force of gravity pulling it downward.
The force of gravity (Fg) is given by Fg = Mg, where M represents the mass of the yo-yo and g represents the acceleration due to gravity. Since the yo-yo falls without slipping, the frictional force between the yo-yo and the metal pole does not contribute to the acceleration of the center of mass.
Therefore, the net force acting on the yo-yo is solely the force of gravity, and it can be written as Fnet = Mg. According to Newton's second law, this net force is equal to the product of the mass and acceleration of the yo-yo's center of mass, giving us the equation:
Mg = Ma
Simplifying, we find that the acceleration of the center of mass is a = g. So, the yo-yo's center of mass has a downward acceleration equal to the acceleration due to gravity, g.
b. To determine the tension in the string as the yo-yo unwinds and falls, we can analyze the torques acting on the yo-yo. Since the yo-yo is unrolling without slipping, the tension in the string provides the torque necessary to accelerate the yo-yo's rotation.
The torque exerted by the tension in the string is given by τ = Iα, where τ represents the torque, I represents the moment of inertia of the yo-yo, and α represents the angular acceleration.
Since the yo-yo is falling without slipping, the angular acceleration is related to the linear acceleration of the center of mass. The linear acceleration, a, is equal to the angular acceleration, α, multiplied by the radius, R, of the yo-yo.
Therefore, the torque exerted by the tension in the string can be written as τ = I(Ra).
We know that the moment of inertia of the yo-yo is 1/2 MR^2, so substituting this value into the torque equation, we have τ = (1/2 MR^2)(Ra) = 1/2 M(R^2a).
Now, we can equate this torque to the net torque provided by the force of gravity. The net torque is equal to the weight of the yo-yo multiplied by the distance from the point of rotation to the center of mass, which is equal to R (since the yo-yo is a disk). Mathematically, this can be written as τ = Mga.
Equating the two torques, we get 1/2 M(R^2a) = Mga. Cancelling out the M's on both sides and simplifying, we find that a = 2g.
Now that we know the linear acceleration of the center of mass is a = 2g, we can determine the tension in the string using the equation Fnet = Ma. The net force acting on the yo-yo is the tension in the string (T) minus the force of gravity (Mg), so we have T - Mg = Ma.
Substituting the value of a, we get T - Mg = 2Mg. Rearranging the equation, we find T = 3Mg.
Therefore, the tension in the string as the disk unwinds and falls is T = 3Mg.
c. The center of mass of the yo-yo moves straight down while descending. This is because when the yo-yo is dropped, the only force acting on it is the force of gravity pulling it downward. Since the yo-yo is not initially in motion, there is no horizontal force to cause it to move left or right. As a result, the only direction the center of mass can move is straight down.
check my arithmetic
yes, no side force so no side motion