Kerala Medical Entrance 2008 Questions on Rotational Motion

Here are the two questions on angular motion which appeared in the Kerala Medical Entrance 2007 question paper:

(1) When a ceiling fan is switched off, its angular velocity reduces to half its initial value after it completes 36 rotations. The number of rotations it will make further before coming to rest is (Assume angular retardation to be uniform)

(a) 10

(b) 20

(c) 18

(d) 12

(e) 16

You have to use the equation, ω² = ω₀² + 2αθ for finding the angular acceleration α and hence the number of further rotations. Note that this equation is the rotational analogue of the equation v² = v₀² + 2as (or, v² = u² + 2as) in linear motion.

Since the angular velocity has reduce to half of the initial value ω₀ after 36 rotations, we have

(ω₀ /2) ² = ω₀² + 2α×36 from which α = – ω₀²/96

[We have expressed the angular displacement θ in rotations itself for convenience]

If the additional number of rotations is x, we have

0 = (ω₀ /2) ² + 2αx = (ω₀ /2) ² + 2×(– ω₀²/96)x

This gives

x = 12

(2) Two particles starting from a point on a circle of radius 4 m in horizontal plane move along the circle with constant speeds of 4 ms^–1 and 6 ms^–1 respectively in opposite directions. The particles will collide with each other after a time of

(a) 3 s

(b) 2.5 s

(c) 2.0 s

(d) 1.5 s

(e) 3.5 s

This is a very simple question. The sum of the distances to be traveled by the two particles is 2πR = 2π×4 = 8π.

The relative speed is 10 ms^–1 so that the time required for the particles to collide is 8π/10 = 2.5 s.

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