AP Physics B & C - Multiple Choice Practice Questions Involving Rotational Motion

Example isn't another way to teach, it is the only way to teach.

– Albert Einstein

Questions (Multiple Choice and Free Response) involving rotational motion were discussed on various occasions on this site. Essential formulae to be remembered in this section were discussed in the post dated 20^th January 2008. You can access all the posts related to this section by clicking on the label ‘rotational motion’ below this post or by trying a search using the search box provided on this page. In both cases you need to use ‘older posts’ links to obtain all related posts.

Some of the questions on rotational motion may prove to be some what difficult and time consuming for many among you. Once you master the basic principles, you will become more confident. Today we will discuss a few more multiple choice practice questions involving rotational motion:

(1) A metre stick of mass 0.2 kg can be balanced on a knife edge at the 60 cm mark (see figure) when a piece of rock (specific gravity = 3) fully immersed in water is suspended from the 70 cm mark. If the piece of rock were in air, suspended from the 70 cm mark itself, what would be the position of the knife edge for the balance?

(a) 56 cm

(b) 58 cm

(c) 62 cm

(d) 64 cm

(e) 66 cm

The weight of the metre stick acts through its centre of gravity, which is at the 50 cm mark. The torque due to the weight of the metre stick tries to rotate it in the anticlockwise sense where as the torque due to the apparent weight of the piece of rock in water tries to rotate the metre stick in the clockwise sense. These torques have equal magnitudes since the metre stick is horizontal. Since the lever arms are equal (0.1 m each), the apparent weight of the piece of rock must be equal to the weight of the metre stick which is equal to 0.2 kgwt.

[Note that we got the above result by equating the magnitudes of the torques: 0.2×0.1 = w₂×0.1 where w₂ is the apparent weight (or, the weight inside water) of the piece of rock].

If the weight of the piece of rock in air is w₁, the loss of weight in water is w₁ – w₂ and the specific gravity, which is equal to 3 is given by

w₁/( w₁ – w₂) = 3

Or, w₁/( w₁ – 0.2) = 3

Therefore, w₁ = 0.3 kgwt.

When the piece of rock is in air, the knife edge should be at distance x from the centre of gravity such that

0.2×x = 0.3×(0.2 – x)

This gives x = 0.06/0.5 = 0.12 m = 12 cm

The position of the knife edge is therefore 50 cm + 12 cm = 62 cm.

(2) A sphere rolls without slipping along a horizontal surface. If the velocity of the centre of mass of the sphere is v, what is the velocity of the point of contact (with the horizontal surface) of the sphere?

(a) 2 v

(b) v

(c) v/2

(d) v/4

(e) zero

When the sphere rolls, it has an angular motion (rotation) and a linear motion (translation). If the the horizontal surface is perfectly smooth, the sphere will slip fully and will just spin about its central axis. In other words, it will have angular motion alone and the velocity of the contact point will be v, directed backwards.

When the surface is rough, the sphere rolls forward and the centre of mass moves forward with velocity v. Since the sphere is rigid, the contact point (lowest point) of the sphere too has to move forward with the velocity v. Since the contact point has an additional backward velocity v due to the spin of the sphere, the resultant velocity of the contact point on the sphere is zero [Option (e)].

[In the above question what is the velocity of the top point of the sphere? You can easily show that it is 2 v, forward].

(3) For opening a door 1 m wide, the minimum force required is 20 N. If the door is pushed at a distance of 0.4 m from the line of the hinges, what is the minimum force required for opening it?

(a) 8 N

(b) 10 N

(c) 20 N

(d) 40 N

(e) 50 N

The force required for opening the door will be minimum if it is applied perpendicular to the plane of the door, with the point of application of the force farthest away from the line of the hinges (at the edge of the door, opposite to the line of the hinges). This ensures that the torque produced by the force is maximum. Evidently the minimum torque required for opening the door is 20×1 Nm = 20 Nm.

[Remember that torque = force×lever arm. It is the torque (and not the force) that matters for opening the door].

If the force is applied at a distance of 0.4 m from the line of the hinges, the minimum force F required for opening the door is given by

F×0.4 = 20 Nm

Therefore, F = 50 N.

The following questions are specifically for the AP Physics C aspirants:

(4) The angular displacement θ of a fly wheel at the instant t is given by

θ =2t³ – 4 t² + 8

The angular acceleration of the fly wheel after 2s in is

(a) 20 rad/s

(b) 16 rad/s

(c) 12 rad/s

(d) 8 rad/s

(e) 6 rad/s

The angular acceleration α is given by

α = d²θ/dt²

Since dθ/dt = 6 t² – 8 t, we have

α = 12 t – 8

The angular acceleration after 2 s = (12×2) – 8 = 16 rad/s

(5) A cylinder of moment of inertia 2 kgm² has angular displacement θ given by

θ =3t² – 4 t

The torque acting on the cylinder

(a) decreases linearly with time

(b) increases linearly with time

(c) decreases non-linearly with time

(d) increases non-linearly with time

(e) remains constant

The angular acceleration of the cylinder is α = d²θ/dt² = 6 rad/s².

The torque τ is given by

τ = I α where I is the moment of inertia.

Therefore, τ =2×6 = 12 Nm

Since this is constant (independent of time), option (e) is correct.

[The expression for the angular displacement θ is similar to the expression s = ut + ½ at² for the linear displacement in uniformly accelerated motion. Therefore from the form of the expression itself you should be able to understand that the angular acceleration and hence the torque, is constant].

(6) Two identical solid hemispheres, each of mass m and radius r are welded together as shown. What is the moment of inertia of this system about the axis AB which is perpendicular to the plane surfaces of the hemispheres? [Moment of inertia of a solid sphere of mass M and radius R about any diameter is (²/₅)MR²]

(a) (¹⁴/₅)mr²

(b) (⁷/₅)mr²

(c) (⁶/₅)mr²

(d) (¹⁴/₅)mr²

(e) (¹⁴/₅)mr²

Since the moment of inertia of a solid sphere of mass M and radius R about any diameter is (²/₅)MR², its moment of inertia I about a tangent is (by the theorem of parallel axes) given by

I = (²/₅)MR² + MR² = (⁷/₅)MR²

The moment of inertia of a hemisphere about a tangent such as AB must be half of this. Since the axis AB is the common tangent to the two hemispheres, the total moment of inertia of the system containing the two hemispheres must be (⁷/₅)MR².

In the above expression M is the mass of the entire sphere and we have

M = 2m

Also R = r as given in the question.

Therefore, the answer is (⁷/₅)(2m)r² = (¹⁴/₅)mr²

You will find few more practice questions (with solution) in this section here.

Practice Questions (MCQ) on Magnetic Fields for AP Physics B & C

Questions involving magnetic fields posted on this site earlier can be accessed by clicking on the label ‘magnetic field’ below this post or by trying a search for ‘magnetic field’ using the search box provided on this page. Answering questions on magnetic field will not be generally difficult; but occasionally you will find questions consuming too much of your precious time. Today we will discuss a few more multiple choice questions involving magnetic fields:

(1) Two straight long vertical wires W₁ and W₂ carry steady currents I and 3I respectively as shown. P is a point midway between the wires. The resultant magnetic flux density at P due to the currents in the wires is B. The direction of B is

(a) vertically upwards

(b) vertically downwards

(c) horizontal and directed normally into the plane of the figure (away from the reader)

(d) horizontal and directed normal to the plane of the figure, towards the reader

(e) horizontal and directed towards the wire W₂

The magnetic field at P due to the current I in the wire W₁ is directed normally into the plane of the figure (away from the reader). The magnetic field at P due to the current 3I in the wire W₂ opposes the above field. Since the magnitude of the field due to the wire W₂ is three times that due to the wire W₁, the resultant field will be directed normal to the plane of the figure, towards the reader [Option (d)].

(2) In the above question if the current 3I in the wire W₂ is switched off, what will be the magnetic field at P?

(a) B

(b) – B

(c) – B/2

(d) B/2

(e) – B/3

The direction of the magnetic field produced by the current I in the wire W₁ is opposite to the net field B in question No.(1). So the sign has to be negative. Since the currents I and 3I in the wires W₁ and W₂ produce opposing fields at P (in question no.1), the resultant field B has magnitude proportional to 2I. When the current 3I in the wire W₂ is switched off, the field at P is proportional to I so that the magnitude is reduced to half the net value in question No.(1).

The field at P in this case is therefore – B/2.

The following questions are specifically meant for AP Physics C aspirants:

(3) The magnetic flux density at the centre of a plane circular coil of radius R carrying a current I is B₀. If the current in the coil is reversed, the magnetic flux density on the axis of the coil at distance R from the centre will be

(a) B₀/√8

(b) – B₀/√2

(c) – B₀/√8

(d) – B₀/2

(e) B₀/2

The magnitude of the magnetic flux density B at a point on the axis of the coil at distance x from the centre of the coil is given by

B = μ₀nR²I/2(R² + x²)^3/2 where n is the number of turns in the coil.

The magnetic field B₀ at the centre (x = 0) of the coil is given by

B₀ = μ₀nI/2R

The field (B₁) on the axis at distance R from the centre (x = R) is given by

B₁ = μ₀nR²I/2(R² + R²)^3/2 = μ₀nR²I/4√2 R³

Or, B₁ = μ₀nI/4√2 R

Therefore, B₁ = B₀/2√2 = B₀/√8

Since the current in the coil is reversed, the direction of the magnetic field must be reversed so that the correct option is – B₀/√8.

(4) An electron moving with velocity (30 i + 40 j) ms^–1 enters a magnetic field of – 2 k tesla where i, j and k are unit vectors along the x, y and z directions respectively. Then

(a) the speed of the electron will not change; but the path will become parabolic

(b) the speed alone will change

(c) the speed will not change; but the path will become helical

(d) the speed will change and the path will become circular

(e) the speed will not change; but the path will become circular

Outside the magnetic field the path of the electron is straight and is contained in the xy plane. The magnetic field is directed along the negative z-direction an hence the electron enters normally into the magnetic field. A magnetic force acts perpendicular to the direction of motion of the electron.

The path of the electron within the magnetic field will therefore be circular. But the speed will be unchanged since a stationary magnetic field cannot change the kinetic energy of the electron. The correct option is (e).

[If the initial velocity of the electron had an z-component also (for instance, v = 30 i + 40 j + 20 k), the electron would enter the magnetic field at an angle other than 90º and then the path would be helical].

AP Physics B & C - Multiple Choice Practice Questions on Electric Circuits

Few weeks have been elapsed after my last post on this blog. I have been extremely busy during the last few weeks and now I am a bit relieved.

Today I’ll give you a few multiple choice practice questions on electric circuits. You will find a few earlier posts in this section elsewhere on this blog, which you can access by clicking on the label ‘direct current circuit’ below this post. The questions I give below are meant for checking your grasp of the fundamental principles:

(1) A battery of constant emf V volt and negligible internal resistance is connected across a non uniform wire of nichrome having length L and resistance R. Pick out the correct statement from the following:

(a) The currents at all sections of the wire are not the same

(b) The potential drop per unit length of the wire is constant

(c) The current density along the wire is uniform

(d) The drift velocity of the charge carriers in the wire is non-uniform

(e) All the above statements are wrong

The expression for the electric current I in the wire is

I = navq where n is the number of charge carriers per unit volume, a is the area of cross section, v is the drift velocity of the charge carriers and q is the charge on each carrier.

[In conductors the charge carriers are electrons and q = e, the electronic charge].

The current everywhere is the same and hence the statements (a), (b) and (c) are wrong. The only correct statement is (d) since the drift velocity as given from the above equation will be different, depending on the area of cross section of the wire.

(2) A cell of emf V volt and internal resistance r ohm is connected across an external resistance 5r. The terminal voltage (terminal potential difference) of the cell is

(a) V/6

(b) 5V/6

(c) 6V/5

(d) V/5

(e) V/4

Since the external resistance is across the cell, the terminal potential difference is the same as the potential difference across the external resistance.

The current through the external resistance is V/(r+5r) = V/(6r).

The potential difference across the external resistance is [V/(6r)] × 5r = 5V/6

(3) A current I flows through a resistive network as shown in the adjoining figure. If the power dissipated in the 8 Ω resistor is 32 watt, what is the quantity of heat generated per second in the 1 Ω resistor?

(a) 1 J

(b) 4 J

(c) 8 J

(d) 32 J

(e) 64 J

Since the power dissipated in the 8 Ω resistor is 32 watt, we have

I₁²×8 = 32 where I₁ is the current through the 8 Ω resistor.

Therefore, I₁ = 2 A.

Resistors R₁ and R₂ which make a total resistance of 12 Ω in the upper branch of the network carry the same current of 2 A. Since the total resistance of the lower branch (consisting of R₃ and R₄) of the network is 3 Ω only (which is a quarter of the resistance of the upper branch), the current through the lower branch is 4 times the current through the upper branch.

The current through the 1 Ω resistor is therefore equal to 8 A.

The heat generated per second in the 1 Ω resistor is 8²×1 = 64 joule [Option (e)].

(4) Four equal resistors connected in series across a battery of negligible internal resistance dissipates a total power of 200 milliwatt. If the parallel combination of these resistors is connected across the same battery, the total power dissipated in them will be

(a) 1.2 W

(b) 1.6 W

(c) 3.2 W

(d) 4.8 W

(e) 6.4 W

If the emf of the battery is V volt and the value of each resistor is R ohm, we have, in the first case (when the series combination is across the battery)

V²/4R = 0.2 watt ………(i)

In the second case (when the parallel combination is across the battery), if the total power dissipated is P, we have

V²/(R/4) = P …………..(ii)

Dividing Eq (i) by Eq (ii), 1/16 = 0.2/P so that P = 3.2 W.

The questions given above are meant for AP Physics B as well as AP Physics C aspirants. The following question is meant specifically for AP Physics C aspirants:

(5) The adjoining figure shows an infinite ladder network in which the resistors are of the same value, each equal to 1 Ω. The effective resistance between the points A and B is nearly

(a) infinite

(b) 2.333 Ω

(c) 2.505 Ω

(d) 2.732 Ω

(e) 2.999 Ω

Let us suppose that the effective resistance between the points A and B is R_e. Since the ladder is infinitely long, we can add to it one more section containing three 1 Ω resistors as shown in the adjoining figure.

The new terminals are A₁ and B₁ and the effective resistance of this modified infinite ladder between these terminals will still be R_e. Therefore we have

R_e = 1 + [1×R_e/(1+R_e)] + 1

[The second term on the right hand side of the above equation is the parallel combined value of 1 Ω and R_e]

Or, R_e + R_e² = 2 + 2R_e + R_e

Thus R_e² – 2R_e – 2 = 0

This quadratic equation gives R_e = [2 ±√(4+8)]/2

Or, R_e = 1 ±√3

The effective resistance has to be positive. So the answer is 1+√3 = 2.732 Ω, very nearly.

You will find similar multiple choice questions with solution here.

AP Physics C - The Concept of Potential Energy - How it Simplifies Problem Solving

The concept of potential energy such as electrostatic potential energy, gravitational potential energy, elastic potential energy and the like usually makes seemingly difficult problems in physics simple to solve. Let us consider an example in electrostatics:

Three point positive charges Q₁, Q₂ and Q₃ are arranged equidistant R from the origin O as shown in the adjoining figure. Another point positive charge q of mass m, initially at rest, is released from the origin O. Derive an expression for the velocity of the point charge when it is far away from the origin.

This question can be made simpler if there is only a single charge Q₁ instead of three charges Q₁, Q₂ and Q₃. The complexity of the question can be increased further if the three charges Q₁, Q₂ and Q₃ are at unequal distances from the origin.

Well, let us come to the question as it is. Since you are asked to determine the velocity of the charge q, you may be tempted to think of the electric field and the force (in fact, the resultant force) acting on the charge and the acceleration it produces. Equations you have often used in kinematics also may come to your mind. But the force and the accelearation are variable in this case and you realize that the method you plan to use does not seem to be workable. (A bright student will not have confusions of this sort and he will proceed in the right direction).

Perhaps you might have started thinking in terms of the electric potential and the potential energy if you were asked to determine the kinetic energy of the charge q when it is far away from the origin O. Electrostatic potential is a scalar quantity (unlike electric field, which is a vector quantity) and you can easily manipulate it. In the present problem, you can easily find the net electrostatic potential (due to the charges Q₁, Q₂ and Q₃) at the origin and hence the electrostatic potential energy of the charge q. The electrostatic potential energy at infinity (far away from the origin) is zero and so the change in the potential energy of the charge q is equal to its potential energy at the origin O. By equating this to ½ mv², you get the required velocity v of the point charge when it is far away from the origin O.

Here is how you will proceed:

The electric potential V at the origin (due to the charges Q₁, Q₂ and Q₃) is given by

V = (1/4πε₀)(Q₁/R + Q₂/R + Q₃/R)

Or, V = (1/4πε₀)[(Q₁+ Q₂+ Q₃)/R]

[The electrostatic potential at any point is the work done by an external agency to bring a unit positive charge from infinity (infinite distance) to the point. Therefore, the electrostatic potential V at a point distant r from a point positive charge Q is given by

V = – _∞∫^r [(1/4πε₀)(Q/r²)]dr

The quantity inside the square bracket is the electrostatic force acting on the unit positive charge. When the charge Q is positive, the electric field produced by it and the displacement dr of the unit positive charge are opposite in direction and this is why the sign of the expression for V is negative. The above integral gives

V = (1/4πε₀)(Q/r)

This shows that the potential energy is zero when r = ∞].

The electrostatic potential energy of the charge q when it is at the origin is (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R].

The loss of potential energy when the charge q moves from the origin O to a point far away from the origin is (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R] since the potential energy at infinity is zero.

The charge q gains an equivalent kinetic energy ½ mv² and hence we have

½ mv² = (1/4πε₀)[(Q₁+ Q₂+ Q₃)q/R]

The velocity v of the point charge when it is far away from the origin O is therefore given by

v = [(1/2πε₀)(Q₁+ Q₂+ Q₃)q/(mR)]^1/2

* * * * * * * * * * * * * * * * * *

Now consider a simple question from gravitation:

An object of mass m is located at a point P very far away from the moon. The gravitational field of the moon is negligible at the point P and the object is initially at rest. The object is given a gentle push and it moves towards the moon. Determine the speed with which the object will strike the moon’s surface. Assume that the mass and radius of the moon are M and R respectively and the moon’s gravity alone influences the motion of the object.

You can apply the concept of gravitational potential energy to obtain the answer easily as given below:

Initial gravitational potential energy U₁ of the object (at infinite distance from the centre of the moon, which we take as the origin) = 0.

[We have U = – GMm/r where G is gravitational constant. With distance r = ∞, U = 0]

Final gravitational potential energy U₂ of the object at the moon’s surface (at distance R) is given by

U₂ = – GMm/R

The loss of gravitational potential energy = U₁ – U₂ = 0 – (– GMm/R) = GMm/R

This must be equal to the gain in kinetic energy ½ mv² where v is the speed with which the object hits the moon’s surface.

Therefore, ½ mv² = GMm/R, from which v = √(2GM/R)

Note that this is the expression for escape velocity from the surface of the moon. An object at infinity, gently pushed from rest must hit the surface (of the moon or any heavenly body) with the surface value of escape velocity since a body projected from the surface with escape velocity will reach infinite distance before coming to rest.

* * * * * * * * * * * * * * * * *

The above question can almost equally well can be worked out beginning with the concept of force as follows

The gravitational force on the object of mass m at distance r = GMm/r²

Work done dW by the gravitational field in moving the object through a small distance dr along the direction of force is given by

dW = (GMm/r²)dr

We should have put the gravitational force as – GMm/R² since it is directed opposite to the direction of increase of r. Our displacement is from infinity to R and hence dr too is negative. The work dW is indeed positive.

The total work W done by the gravitational field is given by

W = _∞∫^R(GMm/r²)dr = GMm/R

This is equal to the gain in kinetic energy ½ mv². Therefore, ½ mv² = GMm/R, from which v = √(2GM/R)