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A paper disc of radius R rotates at a constant angular speed about an axis passing through the centre of the disc. P and Q are two particles of the disc. P is on the circumference of the disc and Q is at a distance of R/2 from the centre.​

What is the ratio centripetal acceleration of p/ centripetal acceleration of Q

A. 1/2

B. 1

C. 2

D. 4

1 Answer

6 votes

Final answer:

The ratio of the centripetal accelerations between a particle on the circumference of the disc and another at half the radius from the centre is 2/1, since centripetal acceleration is directly proportional to the radius for a given angular speed.

Step-by-step explanation:

The student is asking about the ratio of centripetal accelerations of two particles, P and Q, located on a disc that is rotating at a constant angular speed. The centripetal acceleration formula, ac = v²/r or ac = rω², where v is the tangential speed, r is the radius, and ω (omega) is the angular velocity, is relevant here.

Since the angular velocity is constant for both the particles P (on the circumference) and Q (at a distance of R/2 from the centre), their centripetal accelerations are proportional to the radii. For P, the radius is R, and for Q it is R/2. Thus, using the formula ac = rω², the centripetal acceleration for P is acP = Rω² and for Q it is acQ = (R/2)ω² = Rω²/2. The ratio of P's to Q's centripetal acceleration is therefore (Rω²) / (Rω²/2) = 2/1. So, the correct answer is C. 2.

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