what are the forces acting on a string that is wounded around a uniform disk of Radius R and mass R? i know tension is in the string...what else?
and I of the disk is 1/2 MR2... so how would you show that the tension in the string is one third the weight of the disk?
and...show that the magnitude of the acceleration of the center of mass is 2g/3
and show that the linear speed v, of the centre of mass is (4gh/3) ^1/2.. how would u derive ur answer for the linear speed by using instead energy conservation principles
The tension in the string is the force exerted down by the falling disk and up by whatever is holding the disk. They are equal and opposite. If the disk is a yo-yo, that force is exerted by your finger on the string to give the tension.
Force up on disk = T
Force down on disk = m g
a = acceleration down = (mg -T)/m
Torque on disk = T r
Torque = I alpha
so
alpha = T r/I
but alpha = a/r by geometry
T r/I = (m g - T)/(m r)
m T r^2 = (m g-T) I
T (m r^2 + I) = m g I
but I = .5 m r^2
T (1.5 m r^2) = .5 m^2 g r^2
T = (.5/1.5) m g = (1/3) m g
now you should be able to do the rest
for the energy part
KE = .5 m v^2 + .5 I omega^2
but omega = v/r
To answer your first question, let's consider a string wound around a uniform disk of radius R and mass M. When the disk rotates, there are two significant forces acting on the string:
1. Tension in the string: The tension in the string is the primary force acting on it. It is responsible for providing the centripetal force required to keep the disk moving in a circular path. The tension force acts tangentially to the disk's edge and towards the center of the disk.
2. Weight of the disk: The weight of the disk acts vertically downward due to gravity. It is important to note that the gravitational force does not directly affect the string, but it affects the equilibrium of the entire system.
Now, let's move on to your second question about showing that the tension in the string is one-third of the weight of the disk. We can solve this using Newton's second law and considering the rotational dynamics of the disk.
1. Newton's second law (in the tangential direction):
The net force acting on the disk can be expressed as the difference between the tension force and the gravitational force, which is equal to the mass of the disk (M) multiplied by the acceleration at the center of mass (a).
Tension - Weight = Ma
2. Equating centripetal acceleration and angular acceleration:
The centripetal acceleration (a_c) of the disk can be related to the angular acceleration (α) and radius (R) using the equation: a_c = Rα.
3. Relating angular acceleration to the linear acceleration of the center of mass:
The linear acceleration (a) of the center of mass can be related to the angular acceleration (α) using the equation: α = a/R.
Substituting the equations in step 2 and 3, we get: a_c = Ra/R = a.
4. Substituting the value of a in the equation from step 1:
Tension - Weight = Ma
Tension = Ma + Weight
Tension = M(a + g)
Since the magnitude of the acceleration at the center of mass is 2g/3, we substitute that value:
Tension = M(2g/3 + g)
Tension = Mg(2/3 + 1)
Tension = Mg(5/3)
Tension = (5/3)Mg
From this, we can conclude that the tension in the string is indeed one-third of the weight of the disk.
Moving on to your third question about the magnitude of the acceleration of the center of mass, we have already established that a = 2g/3.
Finally, let's address your fourth question regarding the linear speed of the center of mass and how to derive it using energy conservation principles.
To derive the linear speed of the center of mass using energy conservation principles, we can consider the conversion of gravitational potential energy to kinetic energy.
1. Initially, when the disk is at rest, all of its mass is at a vertical distance h above the reference plane.
Gravitational Potential Energy (initial) = Mgh
2. When the disk starts rotating, its center of mass moves horizontally. At the highest point of the rotation (where the string is parallel to the reference plane), all of the gravitational potential energy is converted to kinetic energy.
Kinetic Energy (final) = (1/2)Mv^2
3. Equating the initial gravitational potential energy and the final kinetic energy:
Mgh = (1/2)Mv^2
Canceling out the mass (M) on both sides and solving for v:
v^2 = 2gh
Taking the square root of both sides, we get:
v = √(2gh)
So, the linear speed (v) of the center of mass is equal to √(2gh/3), as derived using energy conservation principles.
I hope this explanation helps! Let me know if you have any further questions.