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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Monday, August 29, 2011

### AP Physics C – Circular Motion and Rotation – Multiple Choice Questions Based on Some Interesting Aspects of Rolling

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

Rolling motion is very much related to our daily life since the wheels used in transportation move between places by rolling. All of you know that the invention of wheels for transportation was a turning point in the history of mankind. It is therefore important that in your AP Physics course you are required to understand the special features of rolling motion.

Even though you come across rolling bodies very commonly in your daily life, many among you might be unaware of some of the interesting aspects of rolling. For instance, some of you might not have noted that a smooth sphere cannot roll on a smooth surface, whether the surface is horizontal or inclined. For a body to roll along a surface, there has to be friction between the body and the surface. If friction is absent, the body will just slide along the surface. If the friction is insufficient, the body will roll as well as slide (slip) along the surface. In pure rolling there will be no slipping.

This post is meant for making you aware of similar aspects of rolling, by working out a few multiple choice practice questions. Here are the questions: (1) The adjoining figure shows a disc of radius R rolling without slipping along a horizontal surface. Assume that the centre of mass C of the disc is displaced along the positive x-direction. If the angular velocity of the disc is ω, what is the velocity VA of the point A of the disc at the instant it is in contact with the horizontal surface?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(c) ωR directed along the negative x-direction

(d) 2ωR directed along the negative x-direction

(e) Zero

Since the disc is rolling without slipping, the contact point A of the disc (to be more precise, the line of contact through A) must be at rest.

The translational velocity Vcm of the centre of mass C of the disc is ωR which is directed parallel to the horizontal surface. Since the disc is a rigid body all points of the disc must move forward (along the positive x-direction) with this translational velocity. But all points of the disc have an additional linear velocity Vr because of the rotation of the disc about its central axis passing through C.

We have Vr = ωr where r is the distance of the point from the centre of mass C. In the case of the contact point A of the disc the distance from the centre of mass is R so that Vr = ωR and is directed along the negative x-direction.

Thus the contact point has the common translational velocity Vcm = ωR directed along the positive x-direction and the linear velocity (due to rotation about C) Vr = ωR directed along the negative x-direction, with the result that it is at rest. The correct option is (e).

(2) What is the velocity VB of the topmost point B of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(c) Zero

(d) ωR directed along the negative x-direction

(e) 2ωR directed along the negative x-direction

The topmost point B of the disc has translational velocity Vcm = ωR directed along the positive x-direction as in the case of all other points of the disc. The additional linear velocity on account of the rotation is Vr = ωR. This too is directed along the positive x-direction in the case of the topmost point A. Therefore, the resultant velocity of the topmost point is 2ωR directed along the positive x-direction [Option (b)].

(3) What is the velocity VD of the point D (at the left edge) of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) (√2) directed along the positive x-direction

(c) Zero

(d) ωR directed vertically upwards

(e) (√2)ωR directed at 45º to the horizontal The velocity of the point D has two parts (as explained in the solution of question No.1):

(i) The translational velocity Vcm = ωR directed horizontally (along the positive x-direction).

(ii) The additional linear velocity on account of the rotation given by Vr = ωR, directed vertically upwards.

The resultant velocity of the point D is (√2)ωR, directed at an angle of 45º as shown in the adjoining figure [Option (e)].

[Note that the two velocities of equal magnitude ωR at right angles give a resultant velocity of magnitude (√2)ωR inclined equally (at 45º in this case) to both]. (4) Consider any arbitrary point P on a thin disc rolling on any surface. If the centre of mass is C and the point of contact with the surface is A at any instant, the instantaneous velocity of the point P is directed

(a) along the radius through P

(b) parallel to the surface

(c) perpendicular to the line AP

(d) along the line CP

(e) perpendicular to the line CP

You can forget about the velocities Vcm and Vr we considered in solving the previous questions and imagine that the disc is rotating about a parallel axis through the point of contact A (or, in other words, rotating about the line of contact) with angular velocity ω. You can easily obtain the velocities of the points we discussed above. Try yourself!

You will easily arrive at the answer to the present problem: [Perpendicular to the line AP given in option (c)].

[The magnitude of the velocity of point P will be ωr’ where r’ is the distance AP].

(5) A solid sphere and a hollow sphere made of two different metals, but of the same mass and radius are given to you. Which one of the following methods will be suitable for identifying them?

(a) Arrange two simple pendulums of equal length with the hollow sphere and the solid sphere as bobs and compare their periods of oscillation.

(b) Supply equal charges to them and compare the electric potentials on them.

(c) Allow them to fall freely under gravity from the same height and compare their times of fall.

(d) Allow them to roll down from the top of the same inclined plane and compare the times taken to reach the bottom of the incline.

(e) Compare their apparent loss of weight when fully submerged under water.

In all cases except (d) the quantities measured will not distinguish the hollow sphere from the solid sphere. The time for rolling down the incline will be greater for the hollow sphere since its moment of inertia is greater. It has more laziness (inertia) to roll!

[The acceleration a of a body rolling down an incline of angle θ (with respect to the horizontal) is given by

a = (g sin θ)/[1 + (k2/R2)] where g is the acceleration due to gravity, R is the radius of the rolling body and k is the radius of gyration defined by I = Mk2 where I is the moment of inertia of the body about its central axis (axis of rolling) and M is the mass of the body.

For a solid sphere k2/R2 = 2/5 since I = (2/5)MR2.

For a thin hollow sphere k2/R2 = 2/3 since I = (2/3)MR2

For a thick hollow sphere of outer radius R and inner radius R1 the moment of inertia about the central axis is (2/5)M[(R5R15)/(R3R13)].

The quantity k2/R2 is certainly more than 2/5 in this case also. Since k2/R2 appears in the denominator of the expression for acceleration, the solid sphere has greater acceleration and it reaches the bottom of the incline earlier.

In fact in the case of regularly shaped bodies such as ring, disc, cylinder, hollow sphere, solid sphere etc., made of material of uniform density, the solid sphere has the maximum acceleration while rolling down an incline and a thin ring has the least acceleration (since it has k2/R2 = 1).

If you have a solid sphere with non uniform density such that there is more concentration of material near the centre, it will accelerate faster than a solid sphere made of a material of uniform density].

## Saturday, August 6, 2011

### AP Physics B - Wave Motion (including sound) - Multiple Choice Practice Questions on Sound

Questions on sound were discussed earlier on this site. You can access all posts related sound by clicking on the label ‘sound’ below this post.
Today we will discuss a few more multiple choice practice questions on sound, relevant to AP Physics B aspirants.
(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed V/6 where V is the speed of sound in air. The wave length of sound heard by the listener is
(a) 0.5 m
(b) 0.54 m
(c) 0.66 m
(d) 0.7m
(e) 0.8 m
You might have come across questions of the above type under Doppler effect.
If the real frequency of the source of sound is n we have
V/n = λ where λ is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have
V/n = 0.6 ……………..(i)
When the source moves away from the listener at speed V/6 the n sound waves produced per second will occupy a distance V + V/6 so that the wave length as measured by the listener becomes (V + V/6)/n. Therefore we have
(V + V/6)/n = λ1 ……….(ii)
where λ1 is the new wave length.
Dividing Eq (ii) by Eq (i) we obtain
1 + 1/6 = λ1/0.6
Or, 7/6 = λ1/0.6 from which λ1 = 0.7 m.
(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is
(a) twice that of B
(b) three times that of B
(c) four times that of B
(d) half that of B
(e) one third that of B
The frequency of vibration (f) of a string of length and linear density m stretched under tension T is given by
f = (s/2) √(T/m) where s = 1,2,3,4 etc. For the fundamental mode s = 1. For the 2nd mode (or 1st overtone) s = 2. For the 3rd mode (or 2nd overtone) s = 3 and so on.
The linear density is mass per unit length and is equal to aρ where ‘a’ is the area of cross section and ρ is the density of the material of the string.
Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have
(2/2) √(T/m1) = (1/2) √(T/m2) where m1 is the linear density of A and m2 is the linear density of B.
From mthe above equation it follows that m1 = 4 m2.
Since the wires are of the same material, the area of cross section of A must be 4 times that of B.
[Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/m1 = 1/m2 to arrive at m1 = 4 m2 and the final answer a1 = 4 a2]
(3) Two sound notes of wave lengths λ1 and λ2 in air produce n beats per second. If λ1 < λ2 the speed of sound in air is
(a) λ1λ2n /(λ2 + λ1)
(b) (λ1 + λ2)n
(c) (λ1λ2 /n )(λ2 + λ1)
(d) λ1λ2n / λ1
(e) λ1λ2n /(λ2 λ1)
The frequency f1 of the sound of wave length λ1 is given by
f1 = v/λ1 where v is the speed of sound in air.
The frequency f2 of the sound of wave length λ2 is given by
f2 = v/λ2
Since λ1 < λ2 we have f1 > f2
Since the number of beats produced per second is n, we have
f1 f2 = n
Therefore, v/λ1 v/λ2 = n
Or, v (11 12) = n
This gives v = λ1λ2n /(λ2 λ1)
(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is f. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube? (a) f/3
(b) f/2
(c) f
(d) 2f
(e) 3f
The air column will vibrate with the minimum frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length L of the tube is equal to λ/2 where λ is the wave length of the fundamental note. Thus we have
L = λ/2
Or λ = 2L
[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to λ/2].
When half the length of the glass tube is immersed in water, it becomes a closed pipe of length L/2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have
L/2 = λ1 /4 where λ1 is the wave length of the resonating sound.
Or λ1 = 2L
In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as f [Option (c)].
(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?
(a) 2f/3
(b) 3f/4
(c) f
(d) 2f
(e) 3f
Initially, when both ends are open, the resonating wavelength as shown above is given by
λ = 2L
When a quarter of the glass tube is submerged in water, we have
3L/4 = λ1/4
Therefore λ1 = 3L
From the above equations λ/λ1 = 2/3
Since the frequency is inversely proportional to the wave length we have
λ/λ1 = f1/f
Therefore f1/f = 2/3 from which f1 = 2f/3