EAMCET 2009 (Medical) Questions (MCQ) on Rotational Motion

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 Iω² = 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 = ml²/3.

[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml²/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml²/12 + m(l/2)² = ml²/3].

Substituting for I we have

             ½ (ml²/3)ω² = mgl/2

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

             ½ (ml²/3)(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 = mL²

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

             Iω = [(ML²/12) + 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

             mL² = [(ML²/12) + mL²/2 ] 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’

You will find many useful questions with solution in this section at physicsplus and at AP Physics Resources.

A Question (MCQ) on Rolling

Here is a question which checks your understanding of basic things in angular motion and simple harmonic motion:

A solid cylinder of mass M and radius R is resting on a horizontal platform which is parallel to the XZ plane. The cylinder can rotate freely about its own axis, which is along the X-direction. The platform is given a linear simple harmonic motion of angular frequency ‘ω’ and amplitude ‘A’ in the Z-direction. If there is no slipping between the cylinder and the platform, the maximum torque acting on the cylinder is

(a) 2MR²Aω² (b) MR²Aω² (c) 2MRAω²

(d) MRAω² (e) MRAω²/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_max = ω²A

[ If you don’t remember the above expression, you may use the simplest form of simple harmonic motion of amplitude ‘A’ and angular frequency ‘ω’ and differentiate it twice:

z = A sin ωt (We write the displacement as ‘z’ since it is in the Z-direction).

a = d²z/dt² = –ω²A sin ωt

Therefore, the maximum acceleration is ω²A].

The maximum angular acceleration (α_max) of the cylinder is given by

α_max = a_max/R = ω²A/R.

The maximum torque (τ_max) on the cylinder is given by

τ_max = α_maxI, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR²/2).

Therefore, maximum torque τ_max = (ω²A/R) (MR²/2) = MRAω²/2

You will find an interesting MCQ on rolling at physicsplus: MCQ on Rolling Bodies

Two Multiple Choice Questions on Moment of Inertia

(1) Three identical thin rods each having mass 2 kg and length 1 m are joined to form an equilateral triangle. The moment of inertia of this system about an axis perpendicular to the plane of the triangle and passing through one corner of the triangle is (in kg m²)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

The moment of inertia of the rods AB and CB about an axis perpendicular to the plane of the triangle and passing through the corner B is ML²/3 + ML²/3 = 2 ML²/3 where M is the mass of each rod and L is the length. [Note that the moment of inertia of a thin rod about a normal axis through its centre of mass is ML²/12 and hence its moment of inertia about a normal axis through one end, according to the theorem of parallel axes, is (ML²/12) + M(L/2)² = ML²/3].
The moment of inertia of the rod AC about the axis through B is
ML²/12 + M[(√3/2)L]² = ML²/12 + 3 ML²/4 = 5ML²/6, in accordance with the theorem of parallel axes. [The distance of AC from B is L sin60° = (√3/2)L].
The moment of inertia of the entire triangle about the axis through B is the sum of the moments of inertia of the three rods and is equal to 2 ML²/3 +5ML²/6 = 9ML²/6 = 3ML²/2. Substituting for M (=2kg) and L (=1m), we obtain the answer as 3 kgm².

(2) The moment of inertia of a body does not depend upon its
(a) mass (b) axis of rotation (c) shape (d) angular velocity (e) size
This is a very simple question and you should not have any doubt in picking out option (d) as the correct answer.

Rotational Motion – Rolling bodies

If you release from rest differently shaped regular bodies such as disk, ring, hollow sphere, solid sphere, hollow cylinder(pipe) and solid cylinder from the top of an inclined plane, thereby allowing them to roll down the plane, you will find that the solid sphere always arrives at the bottom first and the ring and the pipe arrive last. The acceleration ‘a’ of a body of radius ‘R’ rolling down an inclined plane (without slipping) is given by a= g sinθ/[1+( k² / R²)] where ‘g’ is the acceleration due to gravity, ‘θ’ is the angle of the plane and ‘k’ is the radius of gyration given by I= Mk² where ‘I’ is the moment of inertia of the body (of mass M) about the central axis about which it is rolling down. Since the moment of inertia of a solid sphere about its diameter is (2/5) MR², we have (2/5)MR² = Mk². Therefore, k²/R² = 2/5.
The ring and the pipe have k²/R² = 1, since I=MR² = Mk². If you find the values of k²/R² in the case of differently shaped rolling bodies, you will realize that it is minimum (2/5) for the solid sphere and maximum (equal to1) for a thin ring and a thin pipe. Since (k²/R²) appears in the denominator in the expression for ‘a’, the acceleration down the plane is maximum in the case of a solid sphere and minimum in the case of a ring or a pipe. So the solid sphere arrives at the bottom first and the ring and the pipe arrive last.
You should remember that the moment of inertia of a disk (and that of a solid cylinder) about its rolling axis is ½ MR² and that of a hollow sphere is (2/3) MR²
It is interesting to note that all solid spheres will arrive at the bottom together, irrespective of their mass and size. Similarly, all thin rings and thin pipes will arrive together, irrespective of their mass and size. Generally speaking, bodies of a given shape will arrive together, irrespective of their mass and size.
Now, consider the following M.C.Q.:

A body of mass M at the top of a smooth inclined plane starts from rest and slides down the plane. It reaches the bottom with a velocity ‘v’. If the same body were in the form of a ring and the plane were rough, the velocity with which the ring will reach the bottom on rolling down the plane would be
(a) v (b) v√2 (c) 2v (d) v/√2 (e) v/2
Note that if a body is to roll along any surface, friction is necessary and that is why the plane is said to be made rough in the problem. On a smooth plane, the body will slide down with acceleration gsinθ and will reach the bottom with a velocity ‘v’ given by v² = 0 + 2gsinθ×s, as given by the usual equation of linear motion, v² = u² + 2as.
Therefore, v = √(2sgsinθ).
On rolling, the acceleration down the plane is gsinθ/[1 + (k²/R²)] = ½ gsinθ since k²/R² = 1 for a ring. Therefore the velocity (v1) on reaching the bottom is given by v₁² = 0 + 2×½ gsinθ×s
Therefore, v₁ = √(sgsinθ) = v/√2 [Option (d)].

Let us discuss another M.C.Q.:

A sphere of radius ‘R’ is kept at the top end of a curved track. The upper end of the track is at a height ‘H’ and the lower end which is horizontal, is at a height ‘h’ above above the ground level.

When the sphere is released, it rolls down (without slipping) along the track and after leaving the lower end of the track, it moves like a projectile and lands at the point C. The horizontal distance BC is
(a) R√[(20/7)gh(H-h)] (b) √[(20/7)gh(H-h)] (c) √[(10/7)gh(H-h)]
(d) √[(20/7)h(H-h)] (e) √[(5/7)h(H-h)]
The loss of potential energy by the sphere on rolling down along the track is equal to the gain of kinetic energy (both translational and rotational) so that
Mg(H-h) = ½ Mv² + ½ I ω² where M is the mass, ‘v’ is the linear velocity, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the sphere, which is (2/5)MR². Substituting for ω = v/R, we obtain v = √[(10/7)g(H-h)]. This is the horizontal velocity of the sphere at the lower end of the track. Since its height at the lower end of the track is ‘h’, the time taken to reach the ground from there is t = √(2h/g). Note that we obtain this time by considering the vertical motion of the sphere and using the equation, h = 0 + ½ gt². The horizontal distance BC = vt = √[(20/7)h(H-h)]. So, (d) is the correct option. Note that this contains neither R nor g.
The following question is a conventional type which you will get in your class examinations as well as in entrance tests:

A solid cylinder of mass M and radius R rolls down an inclined plane of height ‘h’ without slipping after starting from rest at the top. The speed of its centre of mass when it reaches the bottom is
(a) √(2gh) (b) √(4gh/3) (c) √(3gh/4) (d) √(2gh/3) (e) √(3gh/2)
Equating the gravitational potential energy of the cylinder at the top of the plane to the sum of the translational and rotational kinetic energies at the bottom, we have,Mgh = ½ Mv² + ½ I ω² = ½ Mv² + ½ ×½ MR² × (v/R)² = (3/4)Mv², from which v = √(4gh/3).

Here is another typical M.C.Q. involving rotational motion:
A spherical ball rolls on a table without slipping. The fraction of its total energy which is associated with rotational motion is
(a) 3/5 9b) 2/3 (c) 2/5 (d) 3/7 (e) 2/7
The rolling ball has translational and rotational kinetic energies giving it total kinetic energy equal to ½ Mv² + ½ I ω² = ½ Mv² + ½ ×(2/5)MR² × (v/R))² =½ Mv² + (1/5)Mv² = (7/10)Mv².
The rotational kinetic energy is (1/5)Mv² and the total kinetic energy is (7/10)Mv². The required ratio is therefore (1/5) / (7/10) which is 2/7 [Option (e)].

You will see interesting questions (with solution) on rotational motion here as well as here.

If you would like to have still more multiple choice questions (with solution) on rotational motion, you can have them here.

Questions on Rotational motion

In most of the Medical and Engineering Entrance test papers you will encounter at least a couple of questions on rotational motion. Let us consider the following questions:

(1) A solid cylinder initially at rest rolls down an inclined plane of angle θ and height ‘h’ without slipping. The linear velocity with which it will reach the bottom of the plane is

(a) √(3gh/4) (b) √(4gh/3) (c) √(4ghsinθ/3) (d) √(5gh/4) (e) √(2ghsinθ/3)

If ‘v’ and ‘ω’ are the linear and angular velocities respectively, ½Mv² + ½Iω² = Mgh where I is the moment of inertia and M is the mass of the cylinder. Substituting I = ½ MR² and ω = v/R , we get v = √(4gh/3). Note that the angle θ mentioned in the problem is just a distraction.

(2) A simple pendulum bob of mass ‘m’ is drawn to one side so that the string is horizontal. It is then let free. When it crosses the mean position, the tension in the string is

(a) mg (b) 0.5mg (c) 1.5mg (d ) 3mg (e) 5mg

In problems of this type, you should note that the centripetal force (which is equal to mv²/r) acting on the body executing circular motion is the net force on the body. Therefore, centripetal force, mv²/r = T – mg, where r is the length of the pendulum and T is the tension in the string.Therefore, T= mv²/r + mg. But, ½ mv² = mgr, on equating the initial gravitational potential energy of the bob to its kinetic energy in the mean position. From this, mv²/r = 2mg and hence T = 2mg + mg = 3mg.

Let us consider the following two questions involving the rotation of the earth:
(1) If the radius of the earth is changed to 1/√3 times the present value, the duration of the day (in hours) will be
(a) 72 (b) 41.6 (c) 24 (d) 12 (e) 8
This question is set to check your understanding of the law of conservation of angular momentum: 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 contraction respectively. Substituting for I₁ (= 2MR²/5) and I₂ [= 2M (R²/3)/5] we obtain ω₂ = 3ω₁. Since the angular velocity changes to 3 times the initial value, the spin period of the earth (T= 2π/ω) changes two one-third of the initial value. So, the duration of the day will become 24/3 = 8 hours.
Questions of this type are often found in Medical and Engineering Entrance Test papers. Generally, if the radius of the earth becomes ‘n’ times the present value, the duration of the day becomes 24n² hours. Remember this equation and write the answer in no time!
(2) If your weight while standing on the earth’s surface at the equator is to become zero, the earth should spin at nearly ------ times the present speed.
(a) 12 (b) 14 (c) 17 (d) 24 (e) 37
This is a simple question. If you are to become weightless due to the spin of the earth, the gravitational pull on you is to be balanced by the centrifugal force so that, mg = mRω². From this ω=√(g/R) and the spin period T=2π/ω=2π√(R/g). On substituting for R = 6400 km and g = 9.8ms^-2, the period works out to be 84.6 minutes. This is one-seventeenth the present period of 24 hours. So, the earth should spin at 17 times the present speed. [Option (c)].

Rotation of Rigid Bodies:
Let us consider a couple of questions involving rigid body rotation.
(1) A meter scale is held vertically with one end on the floor and is allowed to fall. Assuming that the end on the floor does not slip, what will be the linear velocity of the other end when it strikes the floor?
(a) 2.7m/s (b) 3.1m/s (c) 5.4m/s (d) 9.8m/s (e) 11.2m/s
When the meter scale is allowed to fall, its gravitational potential energy gets converted into rotational kinetic energy so that we have
mgl/2 = ½ I ω ² where ‘l’ is the length(1metre for a metre scale), ‘m’ is the mass, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the scale. l/2 appears in the potential energy expression since the centre of gravity of the scale is initially at a height l/2. You should note that the moment of inertia of the scale is about the end in contact with the floor and is equal to ml²/3.
From the above equation, ω =√(3g/l).
The linear velocity of the feree end of the scale = ωl = √(3gl). On substituting for l(=1) and g(=9.8) the linear velocity is 5.4m/s [Option (c)].
(2) An impulsive force F acting for a short time interval ∆t is applied at one end of a thin uniform bar of mass M and length L, in a direction perpendicular to the lengthof the bar. The angular velocity with which the bar will rotate is
(a) F∆t/4ML (b) F∆t/2ML (c)2F∆t/ML (d) 4F∆t/ML (e) 6F∆t/ML
The impulse received by the bar is F∆t which is equal to the linear momentum supplied. The bar will rotate about its centre of mass. The ‘lever arm’ for the angular momentum is L/2 so that we have, (L/2)F∆t = Iω = ML²ω /12. So, ω =6F∆t/ML