If we did all the things we are capable of, we would literally astound ourselves.

– Thomas Alva Edison

Normally you can expect questions on circular motion and rotation in the free response category and/or multiple choice category in the AP Physics Examination. The ^{th} January 2008. A free response question and a few multiple choice questions in this section were discussed afterwards. You can access all those posts by clicking on the

(1) A cubical wooden block of side *L* and mass *M* is placed on a rough horizontal surface. The friction between the block and the surface is sufficient so that the block can be toppled without sliding by applying a horizontal force on the block as shown in the adjoining figure. What is the *minimum* force *F* required for toppling the block?

(a) *Mg*

(b) 2* Mg*

(c) *Mg*/2

(d) *Mg*/√2

The cube will topple by rotating about the edge P (Fig.). When the torque produced by the applied force *F *is* just enough *to topple the block, we can equate the toppling torque due to *F* to the restoring torque due to the weight *Mg* about the edge P:

*FL = MgL/*2

Therefore, *F = Mg/*2

(2) In uniform circular motion of a particle which one among the following does not remain constant?

(a) Kinetic energy

(b) Speed

(c) Angular momentum

(d) Momentum

(e) None

(3) A simple pendulum of length *L *suspended from the ceiling is displaced and let go so that the bob of the pendulum moves with uniform angular velocity *ω* along a horizontal circle of radius *R* (Fig.). The bob has mass *M* and its size is negligible compared to *R*. The tension in the string of the pendulum is

(a) *Mg*

(b) *MgL* cos θ* *

(c) *M *(*Rω*^{2} + *g* cos θ)* *

(d) *M *(*R*^{2}*ω*^{4} + *g*^{2} cos^{2}θ)^{1/2}

(e)* M *(*R*^{2}*ω*^{4} + *g*^{2})^{1/2}

(This question is meant for AP Physics C aspirants. But AP Physics B aspirants too will ‘enjoy’ it).

There are three forces acting on the bob:

(i) The weight *Mg* of the bob acting vertically downwards.

(ii) The centrifugal force *MRω*^{2} acting radially outwards.

(iii) The tension *T *in the string.

Evidently the tension *T* is the resultant of the other two forces so that we have

*T = *[(*MRω*^{2})^{2} + (*Mg*)^{2}]^{1/2}^{}

Thus *T = **M *(*R*^{2}*ω*^{4} + *g*^{2})^{1/2}

(4) *For AP Physics C*

The moment of inertia of a thin uniform circular ring of mass *M *and radius *R *about an axis passing through the edge and parallel to the plane of the ring is

(a) 2 *MR*^{2}* *

(b) *MR*^{2}

(c) 3*MR*^{2}/2

(d) *MR*^{2}/4

(e) 3*MR*^{2}/4

Since the moment of inertia of a thin ring about the central axis passing perpendicular to the plane of the ring is *MR*^{2}, its moment of inertia about any diameter is *MR*^{2}/2. (This follows from the perpendicular axes theorem).

The moment of inertia about an axis passing through the edge and parallel to the plane of the ring (as obtained by applying the parallel axis theorem) is *MR*^{2}/2 + *MR*^{2 }= 3*MR*^{2}/2.

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