NEET 2013 Questions (MCQ) on Rotational Motion

God used beautiful mathematics in creating the world.

– Dirac, Paul  Adrien Maurice

The following two questions on rotational motion, which were included in the NEET 2013 question paper, are worth noting:

(1) A small object of uniform density rolls up a curved surface with an initial velocity “v”. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

(1) solid sphere

(2) hollow sphere

(3) disc

(4) ring

Initially the object has rotational and translational kinetic energy but zero gravitational potential energy. At the maximum height of 3v²/4g it has zero kinetic energy since the entire kinetic energy is converted into gravitational potential energy. Thus we have

            ½ I ω² + ½ mv² + 0 = mg(3v²/4g) where ‘I’ is the moment of inertia of the object about the axis of rotation, ‘ω’ is the angular velocity, ‘m’ is the mass and ‘v’ is the linear velocity of the object.

Simplifying, ½ I ω² = ¼  mv²

Since ω =v/R the above equation becomes

            Iv²/R² = ½ mv²

Therefore I = mR²/2

This means that the object is a disc [Option (3)].

(2) A rod PQ of M  and length L is hinged at end P. The rod is held horizontally by a massless string tied to point Q as shown in the figure. When the string is cut, the initial acceleration of the rod is

(1) g/L

(2) 2g/L

(3) 2g/3L

(4) 3g/2L

When the string is cut, the rod rotates about the end P and the torque responsible for the rotation is MgL/2.

[The weight Mg of the rod acts through the centre of gravity located at the middle of the rod. The distance of the line of action of the weight from the axis of rotation (the lever arm for the torque) is L/2]

Since torque is equal to Iα where I is the moment of inertia and α is the angular acceleration we have

            Iα = MgL/2 ……………..(i)

The moment of inertia of the rod about the axis of rotation through its end is ML²/3 as given by the parallel axes theorem.

[Moment of inertia of a uniform rod about a central axis perpendicular to its length is ML²/12. Moment of inertia about a parallel axis through the middle is ML²/12 + M(L/2)² = ML²/3].

Substituting for I in equation (i),  we have

            (ML²/3)α = MgL/2

Therefore α = 3g/2L

Two Questions (MCQ) on Rotation of Rigid Bodies

“Learn from yesterday, live for today, hope for tomorrow. The important thing is to not stop questioning.”

– Albert Einstein

Today we will discuss a couple of questions involving rigid body rotation. The first question may appear to be familiar to you as it has appeared in various entrance exam question papers. The second one is not so common but you must certainly work it out yourself before going through the solution given here.

(1) A thin straight uniform rod AB (Fig.) of length L and mass M, held vertically with the end A on horizontal floor, is released from rest and is allowed to fall. Assuming that the end A (of the rod) on the floor does not slip, what will be the linear velocity of the end B when it strikes the floor?

(a) √(3L/g)

(b) √(2L/g)

(c) √(3gL)

(d) √(2g/L)

(e) √(3g/L)

When the rod falls, its gravitational potential energy gets converted into rotational kinetic energy. Therefore we have

MgL/2 = ½ I ω² …………….. (i)

where I is the moment of inertia of the rod about a normal axis passing through its end and ω is the angular velocity of the rod when it strikes the floor.

[Note that initially the centre of gravity of the rod is at a height L/2 and that’s why the initial potential energy is MgL/2]

The moment of inertia of the rod about the normal axis through its end is given by

I = ML²/3

[The moment of inertia of the rod about an axis throgh its centre and perpenicular to its length is ML²/12. On applying parallel axes theorem, the moment of inertia about a parallel axis through the end is ML²/12 + M (L/2)² = ML²/3]

Substituting for I in Eq.(i), we have

MgL/2 = ½ (ML²/3) ω²

Therefore ω = √(3g/L)

The linear velocity v of the end B of the rod is given by

v = ωL = √(3gL)

(2) A thin straight uniform rod AB (Fig.) of length L and mass M is pivoted at point O, distant L/4 from the end A. The friction at the hinge is negligible and the rod can rotate freely (about the hinge) in a vertical plane. Initially the rod is held horizontally and is released from rest. The linear velocity of the end B of the rod when it momentarily attains vertical position A₁B₁ is

(a) √(3L/4g)

(b) √(27gL/14)

(c) √(18gL/7)

(d) √(3g/4L)

(e) √(2gL)

The gravitational potential energy of the rod gets converted into rotational kinetic energy when the rod is release from its horizontal position. The centre of gravity of the rod is lowered through a distance L/4 and hence the decrease in the gravitational potential energy is MgL/4. The gain in rotational kinetic energy is ½ I ω² where I is the moment of inertia of the rod about the axis of rotation passing through the point O and ω is the angular velocity of the rod when it attains the vertical position. Therefore, from the law of conservation of energy we have

MgL/4 = ½ I ω² …………….. (i)

Here I = ML²/12 + M (L/4)² = M [(L²/12) + (L²/16)]

Substituting in Eq.(i) we have

g = [(L/6) + (L/8)] ω² = (7L/24)ω²

Or, ω = √(24g/7L)

Since the end B of the rod is at distance 3L/4 from the axis of rotation, the linear velocity v of the end B of the rod is given by

v = ω×(3L/4) = [√(24g/7L)] (3L/4)

Or, v = √[(24g×9L²) /(7L×16)] = √(27gL/14)

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2009 Questions on Rotational Motion and Centre of Mass

A few questions on circular motion and rotation, which appeared in AIPMT 2011 question paper were discussed in the last post. A few more questions on rotational motion and centre of mass with their solution are given below. These questions were included in the All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 question paper:

(1) A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis perpendicular to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

(1) ωM /(M + 2m)

(2) ω(M + 2m) /M

(3) ωM /(M + m)

(4) ω(M – 2m) /(M + 2m)

This question is a popular one and has appeared in various entrance question papers. Its answer is based on the law of conservation of angular momentum:

I₁ω₁ = I₂ω₂ where I₁ and I₂ are the initial and final moments of inertia and ω₁ and ω₂ are the initial and final angular velocities respectively.

Therefore we have

MR²ω = (MR² + 2 mR²) ω₂

Therefore, ω₂ = ωM /(M + 2m)

(2) Four identical thin rods each of mass M and length L, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

(1) (²/₃) ML²

(2) (¹³/₃) ML²

(3) (¹/₈) ML²

(4) (⁴/₃) ML²

We have to use the theorem of parallel axes: I = I_CM + Ma² where I_CM is the the moment of inertia about an axis through the centre of mass of the rod and I is the moment of inertia about a parallel axis at distance a.

The moment of inertia of a thin rod of mass M and length L about an axis through the mid point of the rod and perpendicular to its length is ML²/12. The moment of inertia of the rod about a parallel axis at a distance L/2 (which is the distance of the centre of the frame from each ro.is ML²/12 + M (L/2)² = ML²/3

You must remember that moment of inertia is a scalar quantity. Since there are four rods in the square frame, the moment of inertia of the entire frame is four times the moment of inertia of one rod. Therefore the answer is (⁴/₃) ML²

(3) If F is the force acting on a particle having position vector r and τ be the torque of this force about the origin, then

(1) r.τ > 0 and F.τ < 0

(2) r.τ = 0 and F.τ = 0

(3) r.τ = 0 and F.τ ≠ 0

(4) r.τ ≠ 0 and F.τ = 0

This question may appear to be a difficult one at the first glance. But this is a very simple question if you remember that the torque τ is the ‘cross product’ (vector product) of r and F:

τ = r×F

The ‘cross product’ of two vectors is a vector perpendicular to both individual vectors. The torque vector is thus perpendicular to both r and F. Therefore the ‘dot product’ (scalar product) of r and τ as well as that of F and τ is zero [Option (2)].

(4) Two bodies of mass 1 kg and 3 kg have position vectors i + 2 j + k and –3 i – 2 j + k espectively. The centre of mass of this system has a position vector

(1) –2 i – j + k

(2) 2 i – j – 2 k

(3) – i + j + k

(4) –2 i + 2 k

If two point masses m₁ and m₂ have position vectors r₁ and r₂ their centre of mass has position vector R given by

R = (m₁ r₁ + m₂ r₂) /( m₁ + m₂)

Here we have m₁ = 1, m₂ = 3, r₁ = i + 2 j + k and r₂ = –3 i – 2 j + k

Substituting, R = [1(i + 2 j + k) + 3(–3 i – 2 j + k)] /(1+3)

Or, R = (– 8 i – 4 j + 4 k) /4 = –2 i – j + k

You can access all questions in this section by clicking on the label ‘rotation’ below this post. Make use of the buttons ‘older posts’ and ‘newer posts’ at the end of the page to navigate through all the posts.

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.

Angular Momentum- Two Multiple Choice Questions

How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though some times sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our happiness depends.

– Albert Einstein

(1) A thin uniform wooden rod AB of length L and mass M is hinged without friction at the end B. A small block of clay of mass m moving horizontally with velocity v srikes the end B of the rod and gets stuck to it. The angular velocity of the system about A just after the collision is

(a) mv/(mL + ML)

(b) 3mv/(mL + ML)

(c) (mv+ ML)/ 3mL

(d) mv/(3mL + ML)

(e) 3mv/(3mL + ML)

The angular momentum of the system about the point A just before collision is the angular momentum of the block of clay which is equal to mvL. (Note that the perpendicular distance of the line of action of the linear momentum (mv) of the block of clay from the point A is L).

The angular momentum of the system about the point A just after the collision is Iω where I is the total moment of inertia of the rod and clay and ω is the angular velocity of the system immediately after the collision.

We have I = ML²/3 + mL²

Equating the angular momenta before and after collision, we have

mvL = (ML²/3 + mL²) ω

Therefore ω = 3mv/(3mL + ML).

(2) A particle is projected at an angle of 60º with the horizontal with linear momentum of magnitude p. The horizontal range of this projectile is R. Just before the projectile strikes the ground at A, what is the magnitude of its angular momentum about an axis perpendicular the plane of motion and passing through the point of projection? Neglect air resistance.

(a) pR

(b) pR/2

(c) (√3) pR /2

(d) (√3) pR

(e) Zero

This is a very simple question. The magnitude of the linear momentum of the projectile just before it strikes the ground will be equal to p. The magnitude of the angular momentum at the moment will be p×Lever arm = p×ON = p×R sin 60º = (√3) pR /2.