Question

A particle moving in a horizontal circle of radius 0.5 m completes half rotation. The work done by the centripetal force of 5 N on the particle (in J) is:

• A. 2
• B. 5
• C. 2.5
• D. 3
• E. 0

Hint:

Centripetal force is a "no-work" force because it is always perpendicular to the direction of motion.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This question asks for the work done by the centripetal force on a particle in uniform circular motion. The definition of work is crucial here.
Step 2: Key Formula or Approach:
1. Work done (W) by a constant force (F) is given by \( W = \vec{F} \cdot \vec{d} = Fd\cos\theta \), where \( \vec{d} \) is the displacement and \( \theta \) is the angle between the force and displacement vectors. 2. In uniform circular motion, the centripetal force is always directed towards the center of the circle. 3. The instantaneous displacement of the particle is always tangential to the circle. Step 3: Detailed Explanation:
For a particle moving in a circle, at any instant: - The centripetal force vector \( \vec{F}_c \) points radially inwards, towards the center of the circle. - The instantaneous velocity vector \( \vec{v} \), and thus the infinitesimal displacement vector \( d\vec{s} \), is tangent to the circular path. A tangent to a circle is always perpendicular to the radius at the point of tangency. This means the angle between the centripetal force vector and the displacement vector is always \( 90^\circ \). The work done by the centripetal force over an infinitesimal displacement \( d\vec{s} \) is: \[ dW = \vec{F}_c \cdot d\vec{s} = F_c \, ds \, \cos(90^\circ) \] Since \( \cos(90^\circ) = 0 \), the work done over any small part of the path is zero. \[ dW = 0 \] To find the total work done over half a rotation (or any part of the rotation), we integrate dW, which will also be zero. \[ W = \int dW = \int 0 = 0 \] Step 4: Final Answer:
The work done by the centripetal force is always zero, regardless of the distance traveled. The information about the radius, force magnitude, and half-rotation is extraneous. The answer is 0 J. The question was likely cancelled because the correct answer, 0, was not provided as an option.