A homogeneous disk of mass m and of radius r free to rotate without friction about its horizontal axis, is used used as a pulley.
An extensible string of negligible mass is wound around this pulley and a mass M is connected to its free end. When the system is released the center of inertia G of mass M moves downward with an acceleration and the pulley thus aquires an angular acceleration O".
At any date t the velocity of G is equal to that of any point C on the periphery of the pulley (sincethe string ddoesn't slide over the pully). Show that the acceleration a of G is equal to the tangential acceleration of C .
Deduce a relation between a and O"
I don't know what this is about but the torque providing the tangential acceleration is provided by the falling weight Mg times the radius R.
That torque is equal to I(alpha) where alpha is angular acceleration. I for a solid disc is 1/2mR^2 (I think, might want to check on me). So:
MgR = 1/2mR^2(alpha). therefore alpha = 2Mg/mR. The tangential acceleration =R(alpha) which is 2Mg/m. How you would derive this some other way I don't know.
To determine the relationship between the acceleration (a) of the center of inertia G and the angular acceleration (Ω") of the pulley, we can consider the forces acting on the system.
1. Inertia of the pulley:
Since the pulley is rotating about its own axis, it experiences an angular force due to its inertia. This force can be expressed as I * Ω", where I is the moment of inertia of the pulley and Ω" is the angular acceleration.
2. Tension in the string:
The tension in the string is responsible for both the linear acceleration of G and the angular acceleration of the pulley. The tension can be divided into two components:
- T₁, the tangential component, which contributes to the angular acceleration of the pulley.
- T₂, the radial component, which contributes to the linear acceleration of the center of inertia G.
3. Weight of the mass M:
The weight of the mass M provides a downward force, which also contributes to the linear acceleration of G.
Now let's analyze the conditions at point C on the periphery of the pulley:
- At point C, the tangential component of tension (T₁) is solely responsible for the angular acceleration of the pulley. T₁ is given by T₁ = I * Ω", as explained earlier.
- The linear acceleration of point C is purely tangential and is equal to the tangential component of tension divided by the mass of the pulley. Therefore, the tangential acceleration of C is given by ac = T₁ / m, where m is the mass of the pulley.
Since the velocity of G is equal to that of C (due to the non-sliding nature of the string), we have vG = vC. Differentiating this equation with respect to time t, we get aG = aC, where aG and aC represent the accelerations of G and C, respectively.
Hence, we can conclude that the acceleration (a) of the center of inertia G is equal to the tangential acceleration (ac) of point C.
Finally, substituting the expression for T₁ = I * Ω" and ac = T₁ / m, we can deduce the relationship between a and Ω":
a = I * Ω" / m
This equation relates the linear acceleration of the center of inertia G to the angular acceleration of the pulley.