“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)Rω 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[(R5 – R15)/(R3 – R13)].
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].
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