JAMB - Physics (2008 - No. 39)

A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed of 7 ms-1. What is the speed, if the force is tripled?

4.0 ms^-1

6.6 ms^-1

12.1 ms^-1

21.0 ms^-1

The force which is centripetal
F = \(\frac{MV^2}{R}\)
= \(\frac{5 \times 7^2}{3.5}\)
= 70

Since force is 70 N when speed is 7m/s

Speed = ? when force 70N is tripled → 70 * 3 = 210N

210 = \(\frac{5×v^2}{3.5}\)
210 × 3.5 = 5 * v\(^2\)
735 ÷ 5 = v\(^2\)
v\(^2\) = 147
v =12.12m/s