The following two questions were included from rotational motion in the EAMCET 2009 (Medical) question paper. (The first question has appeared in many entrance exam question papers. It was included in the EAMCET 2009 Engineering question paper also). Here are the questions with solution:

(1) A rod of length ‘l’ is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

(1) √(g/l)

(2) √(3gl)

(3) 3√(g/l)

(4) √(3g/l)

When the rod falls its gravitational potential energy mgl/2 gets converted into rotational kinetic energy of the rod. (Note that ‘m’ is the mass of the rod and initially the centre of gravity of the rod is at a height l/2 with respect to the horizontal plane).

Therefore we can write

½ Iω^{2} = mgl/2 where I is the moment of inertia of the rod about an axis passing through the end (at P) of the rod and perpendicular to the length of the rod and ‘ω’ is the angular velocity of the rod when it hits the horizontal plane.

Here I = ml^{2}/3.

[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml^{2}/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml^{2}/12 + m(l/2)^{2} = ml^{2}/3].

Substituting for I we have

½ (ml^{2}/3)ω^{2} = mgl/2

Since ω = v/l where ‘v’ is the velocity with which the rod hits the ground, we have

½ (ml^{2}/3)(v/l)^{2} = mgl/2

This gives v = √(3gl)

(2) A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and traveling with uniform speed ‘v’ collide with the two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m = M/6 and v = L ms^{–1}, then the time taken by the rod to rotate through π/2 is

(1) 1 sec

(2) 2π sec

(3) π sec

(4) π/2 sec

Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by

Iω = mvL/2 + mvL/2 where ‘I’ is the moment of inertia of the rod carrying the masses m and m at its ends.

[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum. Before the collision the two masses have angular momentum about the central axis. These are shown on the right hand side of the above equation].

Since v = L the above equation gets modified as

Iω = mL^{2}^{}

After the collision, the rod and the masses move together and the total angular momentum is given by

Iω = [(ML^{2}/12) + 2m(L/2)^{2}] ω

[The first term within the square bracket above is the moment of inertia of the rod and the second term is the moment of inertia of the two masses].

From the above two equations, we have

mL^{2} = [(ML^{2}/12) + mL^{2}/2^{ }] ω

Since m = M/6 the above equation becomes

M/6 = [(M/12) + (M/12)] ω = (M/6) ω

Therefore ω = 1 radian /sec and the time taken by the rod to rotate through π/2 radian is π/2 sec.

You will find many questions on rotational motion on this site. You can access all of them by clicking on the label ‘rotation’

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