## Saturday, October 14, 2006

### M.C.Q. on Circular Motion, Moment of Inertia & Rigid Body Rotation

The section on rotational motion may appear to be uninteresting to some of you, but your attitude is to be corrected. Once you understand the basic principles, this section becomes interesting and easy to score high marks. Let us begin with a M.C.Q. meant for checking your understanding of basic principles:
A small pendulum bob of mass ‘m’ is tied to one end of a light inextensible string and is revolved in a vertical circle of radius ‘r’. If T is the tension in the string and ‘v’ is the speed of the stone at the highest point, the net force on the stone at the highest point is
(a) mg+T (b) T-mg (c) mg (d) mg + T + mv2/r (e) mg – T +mv2/r

Even though this is a simple question, you are likely to pick out the wrong answer if you have not understood the basic things. The net force is the centripetal force acting on the bob and is equal to mg + T. You should note that at the highest point both the tension and the gravitational pull are towards the centre of the circle and they add up to produce the net force.
If you were asked to find out the net force at the lowest point of the circle, your answer would be (T-mg) because T acts towards the centre (upwards) and mg as usual acts downwards.
Suppose the speed of the bob at the highest point were just sufficient to keep the bob moving along the circle. Then T would be zero. The speed v1 of the bob will be given by mv1 2 /r = mg since the centripetal force is then supplied by the gravitational pull alone. This critical speed at the top of the circle is therefore v1 = √(rg). What will be the speed of the bob at the bottom of the vertical circle if the bob were just to move along the circular path? The kinetic energy of the bob at the bottom is higher by mg×2r because it loses gravitational potential energy when it falls from top to bottom. The total kinetic energy at he bottom is ½ mv12 + mg×2r = ½ mgr + 2mgr = (5/2) mgr. If v2 is the speed of the bob at the bottom, we have ½ mv22 = (5/2) mgr. The critical speed at the bottom of the circle is
v2 = √(5rg). You should remember this: If a body is to move in a vertical circle it should have a minimum horizontal speed of √(5rg) at the bottom of the circle. So if you arrange a simple pendulum and push the bob to have a horizontal speed of √(5rg), the bob will just move along a vertical circle.
When the bob is moving along a vertical circle, the difference between the tensions in the sting when the bob is at the top and bottom of the circle is 6mg. You may prove this by way of a simple exercise for you.
Now consider the following mcq which will be quite easy for you now:
A small steel sphere is suspended using light inextensible string to form a simple pendulum. This bob is initially at rest. What is the minimum speed with which another identical steel sphere should hit the pendulum bob so that the bob just moves along a vertical circle? Treat the collision as elastic.
(a) √(rg) (b) √(2rg) (c) √(2.5rg) (d) √[(5/4)rg] (e) √(5rg)
The correct option is the last one: √(5rg). Since the spheres are identical, the sphere which is moving will come to rest at the instant of hitting the pendulum bob, transferring entire kinetic energy to the bob and the bob will move forward with the same speed. Since the critical speed (at the bottom) for the motion along a vertical circle is √(5rg), this is the minimum speed of the moving sphere.
Consider another multiple choice question:
A stone of mass ‘m’ is projected with a velocity ‘u’ making an angle of 45˚with the horizontal. The magnitude of the angular momentum of the projectile about an axis perpendicular to the plane of projection and passing through the point of projection is
(a) mu2/√2g (b) mu3/4√2 g (c) mu3/√2 g (d) mu2/2g (e) zero
The maximum height ‘H’ reached by the projectile is (u2sin2θ)/2g with usual notations. The velocity of the projectile at the maximum height is ucosθ, which is the horizontal component of its velocity that remains unaltered through out its motion. The magnitude of the angular momentum of the projectile at the highest point is = magnitude of linear momentum×lever arm = mucosθ×H = mucosθ × u2sin2θ/2g = mu3cos45×(sin245)/2g = mu3/4√2 g [Option (b)].
Let us now disuss the following questions which appeared in Kerala Engineering Entrance 2001 test paper:
(1) If ‘I’ is the moment of inertia and ‘E’ is the kinetic energy of rotation of a body, then its angular momentum will be
(a) √(EI) (b) 2EI (c) E/I (d) √(2EI) (e) EI
Since the rotational kinetic energy E = ½ I ω2 we can write E = ½ I2 ω2/I from which the angular momentum Iω = √(2EI).
It will be convenient If you remember the expression for rotational kinetic energy as E = L2/2I where L is the angular momentum. This equation is similar to that for translational kinetic energy in the form E = p2/2m where ‘p’ is the linear momentum.
(2) A circular thin disc of mass 2 kg has a diameter 0.2 m. Calculate the moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc (in kgm
2).
(a) 0.01 (b) 0.03 (c) 0.02 (d) 3 (e) 2
The moment of inertia of the disc about its central axis, perpendicular to its plane is MR2/2. Therefore the moment of inertia about a parallel axis passing through its edge, on applying the parallel axis theorem is MR2/2 + MR2 = (3/2)MR2 = (3/2) ×2×(0.1)2 = 0.03 kgm2.
(3) A torque of 50 Nm acting on a wheel at rest rotates it through 200 radians in 5 seconds. Calculate the angular acceleration produced (in rad/sec2)
(a) 8 (b) 4 (c) 16 (d) 12 (e) 10
The torque given in the question is not required for working it out. It just serves the purpose of a distraction. Just as you remember the equation, s = ut + ½ at2 in linear motion, you should remember its angular counter part as θ = ω0t + ½ αt2 where θ is the angular displacement, ω0 is the initial angular velocity (which is zero in the problem), and α is the angular acceleration. Substituting the values given, we have, 200 = 0 + ½×α×25 from which α = 16 rad/sec2.
The following M.C.Q. appeared in the All India Pre-Medical/Dental Entrance Exam.(C.B.S.E.)-2004 question paper:
A round disc of moment of inertia I2 about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia I1 rotating with an angular velocity ‘ω’ about the same axis. The final angular velocity of the combination of discs is
(a) I2ω/(I1 + I2) (b) ω (c) I1ω/(I1+ I2) (d) (I1 + I2)ω/I1
This is a simple question, the answer of which is based on the law of conservation of angular momentum. The initial angular momentum of the system is I1 ω. The final angular momentum is (I1 + I2)ω' where ω' is the final angular velocity of the combination of the discs. On equating the initial and final angular momenta, we obtain, ω' = I1ω/(I1 + I2).
The following question has appeared in different forms in Medical and Engineering Entrance tests:
If the earth suddenly shrinks to one-eighths its present volume without changing its mass and spherical shape, the duration of the day will
(a) decrease by 6 hours (b) increase by 6 hours (c) decrease by 12 hours (d) decrease by 18 hours (e) remain unchanged

When the volume of a sphere becomes one-eighths, its radius becomes half. The angular momentum of the earth is conserved in spite of the shrinkage so that we have I1ω1= I2ω2 where I1and I2 are the moments of inertia and ω1 and ω2 are the angular velocities of the earth before and after the shrinkage respectively. Substituting for I1 [= (2/5)MR2] and I2 [= (2/5)M (R2/4)] we obtain ω2 = 4ω1. Since the angular velocity changes to 4 times the initial value, the spin period of the earth (T= 2π/ω) changes to one-fourth of the initial value. So, the duration of the day will become 24/4 = 6 hours. The duration therefore decreases by 18 hours.
Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n2 hours.
Let us now consider the following interesting question which appeared in the IIT Screening 2000 question paper:
A long horizontal rod has a bead which can slide along its length, and initially placed at a distance ‘L’ from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration ‘α’. If the coefficient of friction between the rod and the bead is ‘μ’, and gravity is neglected, then the time after which the bead starts slipping is
(a) √(μ/α) (b) (μ/α) (c) 1/√(μα) (d) infinitesimal
The bead gets pressed against the rod because of the tangential acceleration of the rod at the location of the bead. The force with which the bead presses against the rod is F = ma = mLα. where ‘m’ is the mass of the bead and ‘a’ is the tangential acceleration which is equal to Lα. The frictional force therefore is μmLα. It is the centrifugal force which tries to push the bead outwards. Just when the slipping begins, we have μmLα = mv2/L. Initially ‘v’ is zero and as time elapses, its value increases in accordance with the equation, v = 0 + at = Lαt where ‘t’ is the time at which slipping begins. Substituting this value of ‘v’ in the above condition for slipping, we obtain μ = αt2 from which t = √(μ/α).