pysics rotational motion

Question

concrete mixture is made by mixing cement,stone and sand in a rotating cylindrical drum.if the drum rotates too fast,the ingredients remain stuck to the wall of the drum and proper mixing of the ingredients does not take place.the maximum rotational speed of the drum(in rpm)to ensure proper mixing is close to:(take the radius of the drum to be 1.25m and its axle to be horizontal)

OPTIONS

1)1.3

2)0.4

3)27.0

4)8.0

Answer 1

Any object that rotates in a circle must have a centripetal acceleration at all times that points toward the center of the circular path (otherwise, it would not stay going in a circle and start moving off in a straight line).  For an object moving in a vertical circle inside a horizontal tube, at the top of its circular motion, there are two forces that can provide the centripetal force (directly downward, since that is where the center of the circle is at that point):  the normal force of the inside wall of the tube (which would push down at that point) and gravity (which always acts down)

 

According to the problem, we do not want the concrete pressing against the inside wall of the tube at the top (so it does not stick to the wall of the drum).  Therefore, the normal force cannot contribute to holding the concrete in the circle.  Thus, only gravity can provide the centripetal force.  Since the magnitude of the gravitational force is for a given parcel of wet concrete is fixed, the weight of that parcel is the maximum downward force it can provide, and thus the maximum centripetal it can provide at the top of the concrete's circular path.  This means the maximum centripetal acceleration the object can have at the top of its motion, without running into the wall, is the acceleration of gravity, g.

 

Centripetal acceleration (a_c) is related to rotational speed (ω) by:

 

a_c = rω²

 

where

 

r = radius of the circular path

 

However, this is only true if ω is in rad/s.  So, if we set up this expression with g as the (maximum possible) centripetal acceleration at the top of the circular path, then:

 

g = rω²

 

This allows solving for ω, but the value will be in rad/s.  This can be converted to rpm (rev/min) by dimensional analysis:

 

rad/s * rev/rad * s/min --> rev/min

 

So ω in rad/s, calculated above, times (1 rev)/((2π)rad) * (60s)/(1min) = ω in rpm

 

Answer 2

This problem is a variation of the standard vertical uniform circular motion problem.   The standard problem asks for the rotational speed such that the normal force (from the track, string etc.) is zero.  The result turns out to be

 

    ω² R = g = 9.81 m/s² .      Here ω is the rotational speed in radi/s.    Plugging in R =  1.25  and solving for ω gives

 

   ω =  2.8 rad/s.     To get to rpm,  this must be divided by 2pi  (to convert to revolutions) and multiplied by 60 (to convert to minutes).   After these operations ,  the rotational speed comes out to be 26.73 rpm.  This is close to 27 which is the third answer.