A 0.5 kg mass is swung in a horizontal circle of radius 0.50 m at speed 2.50 m/s. What is the required centripetal force?

A 0.5 kg mass is swung in a horizontal circle of radius 0.50 m at speed 2.50 m/s. What is the required centripetal force?

• A. 2.50 N
• B. 6.25 N
• C. 1.25 N
• D. 12.5 N

Answer: (B)

The inward force that keeps an object moving in a circle equals the mass times its centripetal acceleration, a_c = v^2 / r. With v = 2.50 m/s and r = 0.50 m, a_c = (2.50)^2 / 0.50 = 6.25 / 0.50 = 12.5 m/s^2. The required centripetal force is F = m a_c = 0.50 kg × 12.5 m/s^2 = 6.25 N. (Equivalently, F = m v^2 / r = 0.50 × 6.25 / 0.50 = 6.25 N.) This force points toward the center of the circle, provided by the inward pull (for example, by the string’s tension).

The inward force that keeps an object moving in a circle equals the mass times its centripetal acceleration, a_c = v^2 / r. With v = 2.50 m/s and r = 0.50 m, a_c = (2.50)^2 / 0.50 = 6.25 / 0.50 = 12.5 m/s^2. The required centripetal force is F = m a_c = 0.50 kg × 12.5 m/s^2 = 6.25 N. (Equivalently, F = m v^2 / r = 0.50 × 6.25 / 0.50 = 6.25 N.) This force points toward the center of the circle, provided by the inward pull (for example, by the string’s tension).