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A system consisting of two small spheres of 900 and 350 grams, joined by a thin rod can rotate freely about a vertical axis passing through the center of mass of the system. Initially the system is at rest when a moment of 5 N*m is applied. Neglecting the mass of the rod, find the angular velocity and angular acceleration of the system after 1.2 seconds. Find also the power applied.

**Solution with FisicaLab**

Select the Dynamics group and, inside this, the Circular module. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, four elements Mobile with circular movement, one element Mobile with perpendicular circular movement, two elements Force, one element Initial system, one element Final system, one element Power, one element Angular momentum, two elements Angular velocity, one element Angular acceleration, one element Moment of a force and one element Inertia. As show the image below:

The element Stationary reference system write to default the gravity value. And the time is 1.2 seconds:

**g**9.81

**t**1.2

The two elements Mobile with circular movement at top represents the initial state of the system. Taking the left as the sphere of 900 grams, which we call *Ainitial*, and the right as the sphere of 350 grams, which we call *Binitial*, we have respectively:

**Name**Ainitial

**m**900 @ g

**vt**0

**r**17.5 @ cm

**y**0

**Name**Binitial

**m**350 @ g

**vt**0

**r**45 @ cm

**y**0

Now we add these to the Initial system element, which we call *initial*:

**Name**initial

**Object 1**Ainitial

**Object 2**Binitial

**Object 3**0

**Object 4**0

The two elements Mobile with circular movement on the bottom represents the final state. Calling *Afinal* the sphere of 900 grams and *Bfinal* the sphere of 350 grams, we have:

**Name**Afinal

**m**900 @ g

**vt**vtAfinal

**r**17.5 @ cm

**y**0

**Name**Bfinal

**m**350 @ g

**vt**vtBfinal

**r**45 @ cm

**y**0

Now we add these to the Final system element, which we call *final*:

**Name**final

**Object 1**Afinal

**Object 2**Bfinal

**Object 3**0

**Object 4**0

Now add the two systems to element Angular momentum, where we put also the applied momentum:

**System i**initial

**System f**final

**M**5

Also add these to the Power element, where we put the power as an unknown:

**System i**initial

**System f**final

**P**power

Now in Angular velocity elements, put the final angular velocity as an unknown (the same for both spheres):

**Object**Afinal

**vang**vang

**Object**Bfinal

**vang**vang

We use the element Mobile with perpendicular circular movement to know the angular acceleration, applying the force produced by the applied moment. But since the system have two spheres with different mass and different radius, first we get the total mass and the effective turning radius of the final system, using the Inertia element:

**System**final

**m**mTotal

**r**radius

Now we can obtain the corresponding force for this system using the element Moment of a force:

**M**5

**f**Ft

**d**radius

Now enter these information in the element Mobile with perpendicular circular movement, which we call *system*. As tangential velocity for this system we will place 0, the initial velocity (Since the angular acceleration is constant, has the same value at initial or final state. Anyway, we don’t know the final velocity of this mobile.):

**Name**system

**m**mTotal

**vt**0

**r**radius

**at**at

**Ft**Ft

**C**>

And to the applied forces:

**f**normal

**f**centripetal

Finally in the Angular acceleration element, we assign the Mobile with perpendicular circular movement:

**Object**system

**aang**aang

Now click in the icon Solve to get the answer:

normal = 12.263 N ; centripetal = -0.000 N ; mTotal = 1.250 kg ; radius = 0.281 m ; Ft = 17.817 N ; aang = 50.794 rad/s2 ; vtAfinal = 10.667 m/s ; vtBfinal = 27.429 m/s ; vang = 60.952 rad/s ; power = 152.381 N*m/s ; at = 14.254 m/s2 ; Status = success.

Next: Example 7, Previous: Example 5, Up: Examples circular dynamics of particles [Contents][Index]