## Thursday, June 05, 2008

### All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 Questions on Rotational Motion

The following questions appeared in All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 question paper:

(1) A thin rod of length L and mass M is bent at its mid point into two halves so that the angle between them is 90º. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

(a) ML2/12

(b) ML2/6

(c) √2 ML2/24

(d) ML2/24

You should note that moment of inertia is a scalar quantity so that the total moment of inertia of the bent rod is the sum of the moments of inertia of the two halves about the common axis. [The axis must be the same when you add the moments of inertia, for obvious reasons].

The moment of inertia of each half is (M/2)(L/2)2/3 so that the moment of inertia of the entire bent rod is 2×[(M/2)(L/2)2]/3 = ML2/12

[You will certainly remember that the moment of inertia of a thin uniform rod of mass m and length about a perpendicular axis through its mid point is m2/12. The moment of inertia about a perpendicular axis through the end, as give by the parallel axis theorem is (m2/12) + m(/2)2 = m2/3]

(2) The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, about their respective axes is

(a) 1 : √2

(b) √2 : 1

(c) √2 : √3

(d) √3 : √2

Since the moments of inertia of circular disc and circular ring about their axes are respectively MR2/2 and MR2, their radii of gyration about their axes are R/√2 and R respectively.

[Remember that I = Mk2 where I is the moment of inertia and k is the radius of gyration].

Therefore, the ratio of the radii of gyration = (R/√2)/ R = 1/√2.

(3) A roller coaster is designed such that the riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is 20 m. The speed of the car at the top of the hill is between

(a) 15 m/s and 16 m/s

(b) 16 m/s and 17 m/s

(c) 13 m/s and 14 m/s

(d) 14 m/s and 15 m/s

When the rider feels weightless, the weight of the rider is equal in magnitude to the centrifugal force.

Therefore, we have mg = mv2/r where m is the mass of the rider, v is the velocity and r is the radius of the path. From this

v = √(rg) = √(20×10) = √200 = 14.14 m/s.

The correct option therefore is (4).

You can find all posts related to rotational motion on this site by clicking on the label ‘rotational motion’ below this post.

You will find a useful post on the equations to be remembered in rotational motion and circular motion at aphysicsresources