Use of a non-inertial reference frame in a problem involving rotation?

[serllus reuel]

Homework Statement

A disk rotates with angular velocity ω. It has a groove cut along the diameter in which two blocks of mass m and M slide without friction. They are connected by a light string of length l, fixed by a catch with block m a distance r from the center (r + radius of M = l). The catch is then removed.
Find the initial radial acceleration of block m.


My ideal was to imagine a "centrifugal force" acting on each of the blocks. The magnitudes would then be mrω^2 and M(l-r)ω^2, from the formula for centripetal force. Then, I would treat the (rotating) diameter as an x-axis of sorts, thinking of the centrifugal forces as forces on the blocks, and using F=ma and the tension in the string to find the acceleration along the axis, which would turn out to be the radial acceleration.

Is this approach valid? Even if it is, is there an equivalent approach using only real forces? I have tried a few methods, but they generally simplify to the above method. Is there anything I have not thought of?

Finally, I want to note that this came up in the "Newton's laws" chapter of a book, so I really should be using real forces.

Thanks in advance

 

[tiny-tim]

hi serllus reuel!

(try using the X² button just above the Reply box )

My ideal was to imagine a "centrifugal force" acting on each of the blocks. The magnitudes would then be mrω^2 and M(l-r)ω^2, from the formula for centripetal force. Then, I would treat the (rotating) diameter as an x-axis of sorts, thinking of the centrifugal forces as forces on the blocks, and using F=ma and the tension in the string to find the acceleration along the axis, which would turn out to be the radial acceleration.

Is this approach valid?

yes

in that frame, the diameter is stationary, so just use F = ma in the usual way

Even if it is, is there an equivalent approach using only real forces?

yes, using a = ω²r

(in your first method, mω²r is on the LHS of F = ma as part of F; in the second method it's on the RHS, as part of ma)

 

[serllus reuel]

yes, using a = ω²r

(in your first method, mω²r is on the LHS of F = ma as part of F; in the second method it's on the RHS, as part of ma)

aha, I see.

thanks.