Question

The mass of a bicycle rider along with the bicycle is $100\, kg$. He wants to cross over a circular turn of radius $100 \, m $ with a speed of $10\,m$$s^{-1}$. If the coefficient of friction between the tyres and the road is $0.6$, the frictional force required by the rider to cross the turn, is

• A. $300\, N$
• B. $600\,N$
• C. $1200\, N$
• D. $150\, N$

Answer 1

Answer: (B)

Centripetal force
$=\frac{mv^{2}}{r}$ $=\frac{100\times10\times10}{100}$ $=100\, N$
Required frictional force to cross the turn,
$=\mu mg=0.6\times100\times10$ $=600 \,N$
As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.