## Pages

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

## Saturday, May 9, 2009

### AP Physics B&C- Multiple Choice Practice Questions on Circular Motion and Rotation

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 essential points to be remembered in circular motion and rotation were discussed in the post dated 20th 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 label ‘rotation’ or ‘circular motion’ below this post.

Today we will discuss a few more multiple choice practice questions on circular motion and rotation: (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

The momentum of the particle changes because of the change in its direction. [option (d)]. (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 (2 + g cos θ)

(d) M (R2ω4 + g2 cos2θ)1/2

(e) M (R2ω4 + g2)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 MRω2 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 = [(MRω2)2 + (Mg)2]1/2

Thus T = M (R2ω4 + g2)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 MR2

(b) MR2

(c) 3MR2/2

(d) MR2/4

(e) 3MR2/4

Since the moment of inertia of a thin ring about the central axis passing perpendicular to the plane of the ring is MR2, its moment of inertia about any diameter is MR2/2. (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 MR2/2 + MR2 = 3MR2/2.

You will find a few more multiple choice questions (with solution) in this section here.