Rotational motion Help

[choi626]

1. Homework Statement

Two masses, m1 and m2, are connected by light cables to the perimeters of two cylinders of radii, r1 and r2, respectively. The cylinders are rigidly connected to each other but are free to rotate without friction on a common axle. The moment of inertia of the pair of cylinders is I = 45 kgm^2.
r1= 0.5m
r2= 1.5m
m1= 20kg
a) Find the mass of m2 so that the system is in equilibrium
b) the mass is removed and the system released from rest. Determine the angular acceleration, the cable supporting m1, and the linear speed of m1 at the time it has descended 1m.

3. The Attempt at a Solution
a) T=0
r1F1 - r2F2 = 0
m2= 6.7kg.

b)T=I(alpha)
then how do i go on?
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[Jhero]

Hi there,

To find the angular acceleration, you can use the equation T=I(alpha) as you mentioned. However, you will first need to find the net torque on the system. This can be done by considering the forces acting on m1 and m2. Since the system is in equilibrium, the net force on each mass must be zero. This means that the tension in the cables must be equal to the weight of each mass.

Once you have found the net torque, you can plug it into the equation T=I(alpha) to solve for alpha. This will give you the angular acceleration of the system.

To find the cable supporting m1, you can use the equation for centripetal force, F=mv^2/r, where v is the linear speed of m1. You can solve for v using the equation for kinetic energy, KE=1/2mv^2, and the fact that the initial kinetic energy is zero since the system is released from rest.

Finally, to find the linear speed of m1 at the time it has descended 1m, you can use the equation s=ut+1/2at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time. You can solve for t using the equation for linear speed, v=u+at, and the fact that the final speed is zero since the system comes to rest at 1m.

I hope this helps! Let me know if you have any further questions.