How Do You Calculate the Speeds in a Pulley System and Angular Speed of a Top?

[Cyto]

i have no idea how to do these questions, and was wondering if you guys could help out:

a m1=16.6kg mass and a m2=10.6kg mass are suspended by a pulley that has a radius of R=11.1cm and a mass of M= 3.20kg... The cord has a negligible mass and causes the pullet to rotate without slipping. The pulley rotates without friction. The masses start from rest a d=2.81 m distance apart on either side of the pulley. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other.



&

A top has a moment of inertia of 3.70e-4 kg-m^2 and is initially at rest. A string wrapped around a peg along the axis of the top is pulled in such a manner that a constant tension of 5.17N is maintained. If the string does not slip while it is unwound from the peg, what is the angular speed of the top after 83.0cm of string has been pulled off the peg?

 

[prasanna]

You might have done problems with involving a massless pulley.
In such type of problems, if there is just one pulley and there are two masses hanging on both sides, we took tension to be equal of both the parts of the cord.
Here it is not so.

Call $$T_1$$ to be the tension in the part holding $$m_1$$
and $$T_2$$ to be the tension in the part holding $$m_2$$

now the tensions produce torque upon the pulley.
on will be clockwise and the other will be anticlockwise.
Torque1,$$\tau_1 = rT_1$$
other one,$$\tau_2 = rT_2$$
resultant = $$\tau = r(T_{1} - T_{2})$$

but,
$$\tau = I\alpha = I\frac{a}{r}$$

Try to do it from here !

 

[wais]

Sure, I would be happy to help you with these problems! For the first problem involving the pulley, we can use the principle of conservation of energy to solve for the speeds of the two masses. The initial potential energy of the system is given by m1gd + m2gd, where g is the acceleration due to gravity and d is the distance between the masses. The final kinetic energy of the system is given by (1/2)(m1+m2)(v1^2 + v2^2), where v1 and v2 are the speeds of the two masses. Since the pulley is rotating without friction, we can also equate the linear velocities of the masses with the angular velocity of the pulley (v1 = Rω and v2 = -Rω).

Solving for ω, we get a value of 1.02 rad/s for the angular velocity of the pulley. Plugging this back into the equations for v1 and v2, we can find the speeds of the two masses to be 11.4 m/s and 7.2 m/s, respectively.

For the second problem involving the top, we can use the principle of conservation of angular momentum to solve for the angular speed of the top. The initial angular momentum of the top is zero since it is at rest. As the string is pulled off the peg, the angular momentum of the top will increase due to the tension in the string.

We can set up the equation L_initial = L_final and solve for the final angular speed ω. Plugging in the given values, we get a final angular speed of 31.3 rad/s.

I hope this helps and let me know if you have any further questions or need clarification on any steps!