Uniform circular motion – compute string tension for a horizontal whirl A 1 kg stone is tied to a light string of length 1 m and whirled in a horizontal circle at constant angular speed 5 rad/s (idealized, neglecting gravity effects on the motion plane). What is the tension in the string?

Introduction / Context:
In ideal uniform circular motion, the string provides the inward (centripetal) force needed to keep a mass moving in a circle. For horizontal whirl models in introductory mechanics, the vertical weight is treated as balanced by other means or neglected so that tension equals the required centripetal force.

Given Data / Assumptions:

• Mass m = 1 kg, radius r = 1 m, angular speed ω = 5 rad/s.
• String is light and inextensible; motion is in a horizontal plane.
• Air resistance neglected; tension provides centripetal force.

Concept / Approach:

Centripetal force required is m * r * ω^2. In the horizontal whirl idealization, this equals string tension T. This is a common modeling step for teaching rotational dynamics and determining allowable speeds or required strengths of tethers and fasteners.

Step-by-Step Solution:

Verification / Alternative check:

Using linear speed v = r * ω = 1 * 5 = 5 m/s, centripetal force m * v^2 / r = 1 * 25 / 1 = 25 N, confirming the result.

Why Other Options Are Wrong:

5 N, 10 N, 15 N, and 20 N underpredict the required centripetal force for the given speed and radius; only 25 N satisfies m * r * ω^2.

Common Pitfalls:

Multiplying by radius twice (using r^2), or forgetting the square on ω; confusing tension in a conical pendulum (which must also balance weight) with the horizontal whirl idealization used here.

Final Answer:

25 N