## Friday, September 02, 2011

### All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2011 Questions (MCQ) on Circular Motion and Rotation

Today we will discuss three questions on circular motion and rotation, which appeared in AIPMT 2011 question paper. Even though difficult questions can be easily set from this section, these questions are relatively simple. Most question setters ask simpler questions to medical degree aspirants compared to engineering degree aspirants! Here are the questions and their solution:

(1) The instantaneous angular position of a point on a rotating wheel is given by the equation θ(t) = 2t3 – 6t2. The torque on the wheel becomes zero at

(1) 2 s

(2) 1 s

(3) 0.2 s

(4) 0.25 s

The torque on the wheel will be zero when the angular acceleration is zero.

Now, angular acceleration α is given by

α = d2θ/dt2

We have dθ/dt = 6t2 – 12t and

d2θ/dt2 = 12t – 12

Therefore, torque is zero when 12t – 12 = 0.

This gives t = 1 s

(2) A particle moves in a circle of radius 5 cm with constant speed and time period 0.2π s. The acceleration of the particle is

(1) 5 m/s2

(2) 15 m/s2

(3) 25 m/s2

(4) 36 m/s2

The acceleration ‘a’ of the particle is given by

a = ω2R where ω is the angular velocity and R is the radius of the circular path.

Since ω = 2π/T where T is the time period, we have

a = (2π/T)2 R = (2π/0.2π)2 ×(5×10–2) m = 5 m/s2

(3) The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I0. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

(1) I0 + ML2

(2) I0 + ML2/2

(3) I0 + ML2/4

(4) I0 + 2ML2

By the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia about a parallel axis through the centre of mass and the product Ma2 where M is the mass of the body and a is the separation between the two axes.

Therefore, the moment of inertia of the rod about an axis passing through one of its ends and perpendicular to its length is I0 + M(L/2)2 = I0 + ML2/4

You will find similar questions (with solution) in this section here.