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?
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 for rotational motion. The torque on the yo-yo is equal to the moment of inertia times the angular acceleration. Since the yo-yo unwinds without slipping, the torque comes from the tension in the string. The torque can be calculated as the product of the tension T and the radius of the yo-yo, R.
The torque can be written as:
τ = Iα
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Since the yo-yo unwinds without slipping, the angular acceleration is related to the linear acceleration of the center of mass, a, and the radius, R:
α = a/R
Substituting this into the torque equation, we get:
T R = I (a/R)
Simplifying, we find:
T = (I/R^2) a
Since the moment of inertia for the yo-yo is given as 1/2 MR^2, we have:
T = (1/2 MR^2 /R^2) a
T = (1/2 M) a
The tension in the string is equal to half the mass of the yo-yo times the acceleration of the center of mass.
b. To find the tension in the string as the disk unwinds and falls, we can use the equation from part a:
T = (1/2 M) a
c. While descending, the center of mass of the yo-yo moves straight down. This is because the weight of the yo-yo acts vertically downward, and there are no other horizontal forces acting on the yo-yo. Therefore, there is no net horizontal force to cause any horizontal motion.
To answer these questions, let's break them down one by one:
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 of motion. The net force acting on the yo-yo is the tension in the string (T) minus the weight of the yo-yo (mg), where m is the mass of the yo-yo and g is the acceleration due to gravity. Therefore, we have the equation:
T - mg = ma
Since the yo-yo is falling, its acceleration is downward, so we can substitute -a for a. Rearranging the equation, we get:
T - mg = - ma
Now, we can solve for a:
a = (mg - T) / m
This gives us the downward acceleration of the center of mass of the yo-yo as it unrolls from the string in terms of g (acceleration due to gravity), mass (M), and tension (T).
b. To find the tension in the string as the yo-yo unwinds and falls, we can consider the torque being applied to the yo-yo. The angular acceleration of the yo-yo is related to its linear acceleration by the equation:
a = Rα
Where R is the radius of the yo-yo and α is the angular acceleration of the yo-yo. Since the yo-yo is unwinding without slipping, the linear acceleration at the center of mass of the yo-yo is equal to the radius times the angular acceleration (a = Rα). Therefore, we can substitute a = Rα into the equation derived in part a:
(Rα) = (mg - T) / m
Simplifying the equation, we get:
T = mg - m(Rα)
Now, we can substitute the moment of inertia for the yo-yo (I = 1/2MR^2) and the relationship between linear and angular acceleration (a = Rα) to get the final equation:
T = mg - (1/2)Ia / R
This equation gives us the tension in the string as the yo-yo unwinds and falls in terms of g (acceleration due to gravity), mass (M), radius (R), and moment of inertia (I).
c. While descending, the center of mass of the yo-yo moves straight down. This is because the only force acting on the yo-yo in the horizontal direction is the tension in the string, which is always directed radially inward towards the metal pole. Therefore, there is no net horizontal force on the yo-yo, and it does not move to the left or right. The vertical force due to gravity causes the yo-yo's center of mass to move straight down.